OP Malhotra Class 10 Maths Solutions Chapter 16 Trigonometrical Identities and Tables Exercise 16 (B)

Question 1. Find the sine, cosine, and tangent of the following angles:
(a) 15°27′
(b) 37°48′
(c) 55°17′
Answer: Using the natural sine, cosine, and tangent tables, we find the values:
(a) For 15°27′:
\( \sin 15^\circ 27' = \sin 15^\circ 24' + 3' = 0.26556 + 84 = 0.26640 \approx 0.2664 \)
\( \cos 15^\circ 27' = \cos 15^\circ 24' - 3' = 0.96410 - 23 = 0.96387 \approx 0.9639 \)
\( \tan 15^\circ 27' = \tan 15^\circ 24' + 3' = 0.27545 + 94 = 0.27639 \approx 0.2764 \)
(b) For 37°48′:
\( \sin 37^\circ 48' = 0.61291 \approx 0.6129 \)
\( \cos 37^\circ 48' = 0.79015 \approx 0.7902 \)
\( \tan 37^\circ 48' = 0.77568 \approx 0.7757 \)
(c) For 55°17′:
\( \sin 55^\circ 17' = \sin 55^\circ 12' + 5' = 0.82115 + 82 = 0.82197 \approx 0.8219 \)
\( \cos 55^\circ 17' = \cos 55^\circ 12' - 5' = 0.57071 - 120 = 0.56951 \approx 0.5695 \)
\( \tan 55^\circ 17' = \tan 55^\circ 12' + 5' = 1.43881 + 453 = 1.44334 \approx 1.4433 \) Using trigonometric tables helps us find specific angle values. Each angle has unique sine, cosine, and tangent ratios.
In simple words: We look up the given angles in special math tables to find their sine, cosine, and tangent values. Sometimes we need to add or subtract a small adjustment called 'mean difference' to get the exact value for the angle.

🎯 Exam Tip: When using trigonometric tables, always remember to add the mean difference for sine and tangent, but subtract it for cosine values, as cosine decreases with angle.

Question 2. Find the acute angle A, given
(a) sin A = 0.4919
(b) tan A = 2.7775
(c) tan A = 3.412
(d) cos A = 0.4651
(e) sin A = 0.9519
(f) cos A = 0.5757
Answer: We use the tables for sines, cosines, and tangents to find the acute angle A for each given value.
(a) Given \( \sin A = 0.4919 \):
From tables, \( \sin A \approx 0.49090 \) (for \( 29^\circ 24' \)).
Difference \( = 0.4919 - 0.49090 = 0.00100 \).
The mean difference for \( 100 \) corresponds to \( 4' \).
So, \( A = 29^\circ 24' + 4' = 29^\circ 28' \).
(b) Given \( \tan A = 2.7775 \):
From tables, \( \tan A \approx 2.77761 \) (for \( 70^\circ 12' \)). This is the closest value to \( 2.7775 \).
So, \( A = 70^\circ 12' \).
(c) Given \( \tan A = 3.412 \):
From tables, \( \tan A \approx 3.41973 \) (for \( 73^\circ 42' \)). This is the closest value to \( 3.412 \).
So, \( A = 73^\circ 42' \).
(d) Given \( \cos A = 0.4651 \):
From tables, \( \cos A \approx 0.46484 \) (for \( 62^\circ 18' \)).
Difference \( = 0.4651 - 0.46484 = 0.00026 \).
The mean difference for \( 26 \) corresponds to \( 1' \). Since cosine values decrease with angle, we subtract the mean difference from the angle.
So, \( A = 62^\circ 18' - 1' = 62^\circ 17' \).
(e) Given \( \sin A = 0.9519 \):
From tables, \( \sin A \approx 0.95159 \) (for \( 72^\circ 6' \)).
Difference \( = 0.9519 - 0.95159 = 0.00031 \).
The mean difference for \( 31 \) corresponds to \( 3' \).
So, \( A = 72^\circ 6' + 3' = 72^\circ 9' \).
(f) Given \( \cos A = 0.5757 \):
From tables, \( \cos A \approx 0.57501 \) (for \( 54^\circ 54' \)).
Difference \( = 0.5757 - 0.57501 = 0.00069 \).
The mean difference for \( 69 \) corresponds to \( 3' \). Since cosine values decrease with angle, we subtract the mean difference from the angle.
So, \( A = 54^\circ 54' - 3' = 54^\circ 51' \). Understanding how to read and interpolate values from these tables is a fundamental skill in trigonometry.
In simple words: To find the angle from its sine, cosine, or tangent value, we look it up in the math tables. Sometimes we need to adjust the angle a little using 'mean difference' to match the exact value given.

🎯 Exam Tip: Remember to add the mean difference when finding angles from sine or tangent values, but subtract it for cosine values, as cosine decreases with angle.

Question 3. Using tables, find the value of \( (2 \sin \theta - \cos \theta) \) as a decimal
(i) when \( \theta = 35^\circ \)
(ii) when \( \tan \theta = 0.2679 \)
Answer: We will use trigonometric tables to find the values of sine and cosine for the given conditions, then calculate the expression.
(i) When \( \theta = 35^\circ \):
From tables: \( \sin 35^\circ = 0.57358 \) and \( \cos 35^\circ = 0.81915 \).
Now substitute these values into the expression:
\( 2 \sin 35^\circ - \cos 35^\circ = 2(0.57358) - 0.81915 \)
\( = 1.14716 - 0.81915 \)
\( = 0.32801 \approx 0.3280 \).
(ii) When \( \tan \theta = 0.2679 \):
First, find \( \theta \) from the tangent table:
From tables, \( \tan 14^\circ 56' \approx 0.2679 \). So, \( \theta = 14^\circ 56' \).
Next, find \( \sin 14^\circ 56' \) and \( \cos 14^\circ 56' \) from tables:
\( \sin 14^\circ 56' = 0.25769 \)
\( \cos 14^\circ 56' = 0.96622 \)
Now substitute these values into the expression:
\( 2 \sin 14^\circ 56' - \cos 14^\circ 56' = 2(0.25769) - 0.96622 \)
\( = 0.51538 - 0.96622 \)
\( = -0.45084 \). Knowing how to find an angle from its tangent and then using that angle for other trigonometric functions is key here.
In simple words: For the first part, we directly find the sine and cosine of 35 degrees and put them into the sum. For the second part, we first find the angle whose tangent is 0.2679, and then use that angle to find its sine and cosine to complete the sum.

🎯 Exam Tip: Always be careful with the order of operations (multiplication before subtraction) and ensure you use the correct angle (or derived angle) for each trigonometric function.

Question 4. State for any acute angle \( \theta \)
(i) whether \( \sin \theta \) increases or decreases as \( \theta \) increases;
(ii) whether \( \cos \theta \) increases or decreases as \( \theta \) decreases.
Answer: We examine the behavior of sine and cosine functions for acute angles (angles between \( 0^\circ \) and \( 90^\circ \)).
(i) For \( \sin \theta \) as \( \theta \) increases:
We know that \( \sin 0^\circ = 0 \) and \( \sin 90^\circ = 1 \).
As the angle \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \), the value of \( \sin \theta \) increases from \( 0 \) to \( 1 \).
Therefore, \( \sin \theta \) increases as \( \theta \) increases.
(ii) For \( \cos \theta \) as \( \theta \) decreases:
We know that \( \cos 0^\circ = 1 \) and \( \cos 90^\circ = 0 \).
If we consider \( \theta \) decreasing from \( 90^\circ \) to \( 0^\circ \), the value of \( \cos \theta \) increases from \( 0 \) to \( 1 \).
Therefore, \( \cos \theta \) increases as \( \theta \) decreases. Visualizing the unit circle helps understand these trends more intuitively.
In simple words: For angles between 0 and 90 degrees, as the angle gets bigger, its sine value also gets bigger. But for the cosine value, as the angle gets smaller, its cosine value gets bigger.

🎯 Exam Tip: Remember the basic values: \( \sin 0^\circ = 0, \sin 90^\circ = 1 \) and \( \cos 0^\circ = 1, \cos 90^\circ = 0 \) to easily recall these trends.

Question 5. If \( \sin x^\circ = 0.67 \), find the value of
(a) \( \cos x^\circ \)
Answer: Given \( \sin x^\circ = 0.67 \). We need to find \( \cos x^\circ \).
First, find the angle \( x^\circ \) using sine tables:
The nearest value to \( 0.67 \) in sine tables is \( 0.67043 \), which corresponds to \( 42^\circ 6' \).
The mean difference for \( 0.67043 - 0.67 = 0.00043 \) is \( 2' \). So \( x^\circ = 42^\circ 6' - 2' = 42^\circ 4' \).
Now, find \( \cos x^\circ \) for \( x^\circ = 42^\circ 4' \):
From cosine tables, for \( 42^\circ 4' \):
\( \cos 42^\circ 4' = 0.74314 - 77 \) (mean difference for \( 4' \))
\( = 0.74237 \approx 0.7423 \).
Thus, \( \cos x^\circ = 0.7423 \). These calculations highlight the inverse relationship between sine and cosine for complementary angles.
In simple words: We first find the angle \( x \) using the given sine value from the tables. Then, we use this angle \( x \) to find its cosine value from the cosine tables.

🎯 Exam Tip: When using tables to find an angle, find the closest value, then use the mean difference to refine the angle. Ensure you subtract the mean difference when using cosine, or when refining an angle from sine/tangent values if the exact value is smaller than the table entry.

Question 6. Using trigonometric table, find the measure of the angle A when \( \sin A = 0.1822 \).
Answer: Given \( \sin A = 0.1822 \). We need to find the measure of angle A.
Using trigonometric tables for sine values:
We look for the value closest to \( 0.1822 \).
From the tables, \( \sin 10^\circ 30' = 0.18224 \). This is the nearest value.
Therefore, \( A = 10^\circ 30' \). Reading trigonometric tables accurately is essential for these types of problems.
In simple words: We just look for the number 0.1822 in the sine table. The angle next to it is our answer for A.

🎯 Exam Tip: When directly finding an angle from a given trigonometric ratio, ensure you pick the closest value in the table and apply any necessary mean difference adjustments if extremely precise interpolation is required (though often the nearest value is sufficient for direct lookups).

Question 7. In rectangle ABCD, AB = 23 cm, and \( \angle CAB = 35^\circ \). Calculate the measure of BC.

Answer: In rectangle ABCD, we are given \( AB = 23 \) cm and \( \angle CAB = 35^\circ \). We need to find the length of BC.
Consider the right-angled triangle \( \triangle ABC \), where \( \angle B = 90^\circ \).
We know that \( \tan (\text{angle}) = \frac{\text{Opposite side}}{\text{Adjacent side}} \).
For \( \angle CAB = 35^\circ \):
Opposite side \( = BC \)
Adjacent side \( = AB \)
So, \( \tan 35^\circ = \frac{BC}{AB} \).
Let \( BC = x \).
\( \tan 35^\circ = \frac{x}{23} \).
From trigonometric tables, \( \tan 35^\circ \approx 0.70021 \).
\( 0.70021 = \frac{x}{23} \)
\( x = 23 \times 0.70021 \)
\( x = 16.10483 \approx 16.11 \) cm.
Therefore, the measure of \( BC \) is approximately \( 16.11 \) cm. Using tangent is effective when you have one side and an angle, and need the opposite side in a right triangle.
In simple words: In the rectangle, the triangle ABC is a right-angled triangle. We use the tangent function, which connects the angle with the opposite and adjacent sides. By knowing the angle and one side, we can find the other side (BC).

🎯 Exam Tip: For problems involving rectangles or squares, remember that their angles are \( 90^\circ \), creating right-angled triangles that allow the use of basic trigonometric ratios (sine, cosine, tangent).

Question 8. In the figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB. Given that \( \angle AED = 60^\circ \) and \( \angle ACD = 45^\circ \); without using tables, calculate
(i) AB,
(ii) AC and
(iii) AE.

Answer: In the given figure, \( AB \) is parallel to \( DC \), and \( AD \) and \( BC \) are perpendicular to \( AB \). We are given \( AD = BC = 2 \) cm. We also have \( \angle AED = 60^\circ \) and \( \angle ACD = 45^\circ \). Since \( AB \parallel DC \) and \( AD \perp AB \), the figure \( ABCD \) is a rectangle. Also, \( AD \) and \( BC \) are perpendicular to \( CD \).
(i) To calculate AB:
Consider the right-angled triangle \( \triangle ACD \), where \( \angle D = 90^\circ \).
We are given \( \angle ACD = 45^\circ \).
We know \( \tan (\text{angle}) = \frac{\text{Opposite side}}{\text{Adjacent side}} \).
For \( \angle ACD = 45^\circ \):
Opposite side \( = AD = 2 \) cm
Adjacent side \( = CD \)
So, \( \tan 45^\circ = \frac{AD}{CD} \).
\( 1 = \frac{2}{CD} \)
\( \implies CD = 2 \) cm.
Since \( ABCD \) is a rectangle, \( AB = CD \).
Therefore, \( AB = 2 \) cm.
(ii) To calculate AC:
In the right-angled triangle \( \triangle ACD \):
We know \( \sin (\text{angle}) = \frac{\text{Opposite side}}{\text{Hypotenuse}} \).
For \( \angle ACD = 45^\circ \):
Opposite side \( = AD = 2 \) cm
Hypotenuse \( = AC \)
So, \( \sin 45^\circ = \frac{AD}{AC} \).
\( \frac{1}{\sqrt{2}} = \frac{2}{AC} \)
\( \implies AC = 2\sqrt{2} \) cm.
(iii) To calculate AE:
Consider the right-angled triangle \( \triangle ADE \), where \( \angle D = 90^\circ \).
We are given \( \angle AED = 60^\circ \) and \( AD = 2 \) cm.
The solution uses \( \tan 60^\circ = \frac{AD}{AE} \).
\( \sqrt{3} = \frac{2}{AE} \)
\( \implies AE = \frac{2}{\sqrt{3}} \) cm.
To rationalize the denominator:
\( AE = \frac{2 \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{2\sqrt{3}}{3} \) cm.
Therefore, \( AE = \frac{2\sqrt{3}}{3} \) cm. This calculation uses the specific property of 30-60-90 triangles where sides are in a specific ratio.
In simple words: First, we use trigonometry in triangle ACD to find AB and AC. Then, we use trigonometry in triangle ADE to find AE. For calculations without tables, remember the exact values for sine, cosine, and tangent of 30, 45, and 60 degrees.

🎯 Exam Tip: Always draw a clear diagram and label all known values and angles. When calculating without tables, memorize the exact values for common angles like 30°, 45°, and 60°.

Question 9. In the figure, BC = 12 cm, AB = 4 cm, \( \angle AEB = 90^\circ \), \( \angle B = 50^\circ \) and \( \angle C = 30^\circ \). Calculate the length of
(i) BE and
(ii) AC.

Answer: In the given figure, we have \( BC = 12 \) cm, \( AB = 4 \) cm, \( \angle AEB = 90^\circ \), \( \angle B = 50^\circ \), and \( \angle C = 30^\circ \). We need to calculate the lengths of BE and AC.
(i) To calculate BE:
Consider the right-angled triangle \( \triangle ABE \), where \( \angle E = 90^\circ \).
We are given \( AB = 4 \) cm and \( \angle B = 50^\circ \).
We know \( \cos (\text{angle}) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \).
For \( \angle B = 50^\circ \):
Adjacent side \( = BE \)
Hypotenuse \( = AB = 4 \) cm
So, \( \cos 50^\circ = \frac{BE}{4} \).
From trigonometric tables, \( \cos 50^\circ \approx 0.64279 \).
\( 0.64279 = \frac{BE}{4} \)
\( BE = 4 \times 0.64279 \)
\( BE = 2.57116 \approx 2.57 \) cm.
(ii) To calculate AC:
First, find AE in \( \triangle ABE \):
We know \( \sin (\text{angle}) = \frac{\text{Opposite side}}{\text{Hypotenuse}} \).
For \( \angle B = 50^\circ \):
Opposite side \( = AE \)
Hypotenuse \( = AB = 4 \) cm
So, \( \sin 50^\circ = \frac{AE}{4} \).
From trigonometric tables, \( \sin 50^\circ \approx 0.76604 \).
\( 0.76604 = \frac{AE}{4} \)
\( AE = 4 \times 0.76604 \)
\( AE = 3.06416 \approx 3.06 \) cm.
Now, consider the right-angled triangle \( \triangle AEC \), where \( \angle AEC = 90^\circ \).
We have \( AE = 3.06 \) cm and \( \angle C = 30^\circ \).
We know \( \sin (\text{angle}) = \frac{\text{Opposite side}}{\text{Hypotenuse}} \).
For \( \angle C = 30^\circ \):
Opposite side \( = AE \)
Hypotenuse \( = AC \)
So, \( \sin 30^\circ = \frac{AE}{AC} \).
\( \frac{1}{2} = \frac{3.06}{AC} \)
\( AC = 2 \times 3.06 \)
\( AC = 6.12 \) cm.
Therefore, the length of \( AC \) is approximately \( 6.12 \) cm. Breaking down complex shapes into right-angled triangles simplifies trigonometric calculations.
In simple words: First, we use the cosine function in the smaller right-angled triangle to find the length of BE. Then, to find AC, we first find AE using the sine function, and then use AE and the angle C in another right-angled triangle to find AC.

🎯 Exam Tip: When a figure contains multiple triangles, identify right-angled triangles first. Use the given angles and sides to find unknown lengths in one triangle, then use those new lengths to solve for unknowns in connected triangles.

Question 10. In the figure, in \( \triangle ABC \), \( \angle B = 90^\circ \), \( \angle C = 30^\circ \) and \( AB = 12 \) cm. BD is perpendicular to AC; find
(i) BC
(ii) AD
(iii) AC

Answer: In the given figure, for \( \triangle ABC \), we have \( \angle B = 90^\circ \), \( \angle C = 30^\circ \), and \( AB = 12 \) cm. Also, \( BD \perp AC \). We need to find BC, AD, and AC.
First, find \( \angle A \):
In \( \triangle ABC \), the sum of angles is \( 180^\circ \).
\( \angle A = 180^\circ - \angle B - \angle C = 180^\circ - 90^\circ - 30^\circ = 60^\circ \).
(i) To calculate BC:
Consider the right-angled triangle \( \triangle ABC \).
We know \( \tan (\text{angle}) = \frac{\text{Opposite side}}{\text{Adjacent side}} \).
For \( \angle C = 30^\circ \):
Opposite side \( = AB = 12 \) cm
Adjacent side \( = BC \)
So, \( \tan 30^\circ = \frac{AB}{BC} \).
\( \frac{1}{\sqrt{3}} = \frac{12}{BC} \)
\( \implies BC = 12\sqrt{3} \) cm.
(ii) To calculate AD:
Consider the right-angled triangle \( \triangle ABD \), where \( \angle D = 90^\circ \).
We know \( \angle A = 60^\circ \) (calculated above) and \( AB = 12 \) cm.
We know \( \cos (\text{angle}) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \).
For \( \angle A = 60^\circ \):
Adjacent side \( = AD \)
Hypotenuse \( = AB = 12 \) cm
So, \( \cos 60^\circ = \frac{AD}{AB} \).
\( \frac{1}{2} = \frac{AD}{12} \)
\( \implies AD = 12 \times \frac{1}{2} = 6 \) cm.
(iii) To calculate AC:
Consider the right-angled triangle \( \triangle ABC \).
We know \( \sin (\text{angle}) = \frac{\text{Opposite side}}{\text{Hypotenuse}} \).
For \( \angle C = 30^\circ \):
Opposite side \( = AB = 12 \) cm
Hypotenuse \( = AC \)
So, \( \sin 30^\circ = \frac{AB}{AC} \).
\( \frac{1}{2} = \frac{12}{AC} \)
\( \implies AC = 12 \times 2 = 24 \) cm. Always determine all angles in a triangle if possible, as it provides more options for calculations.
In simple words: We first find all the angles in the main triangle. Then, using sine, cosine, or tangent in the right-angled triangles (ABC and ABD), we find the lengths of BC, AD, and AC step by step.

🎯 Exam Tip: When a perpendicular is drawn from a vertex to the opposite side in a right-angled triangle, it creates two smaller similar triangles, allowing you to use trigonometric ratios in multiple ways.

Question 11. In the figure, the radius of a circle is given as 15 cm and chord AB subtends an angle of \( 131^\circ \) at the centre C of the circle. Using trigonometry, calculate
(i) the length of AB;
(ii) the distance of AB from the centre C.

Answer: In the given figure, the circle has its centre at C, and the radius is 15 cm. So, \( AC = BC = 15 \) cm. The chord \( AB \) subtends \( \angle ACB = 131^\circ \) at the centre. We need to calculate the length of AB and its distance from C.
Draw \( CL \perp AB \), where L is the midpoint of AB. This forms two congruent right-angled triangles, \( \triangle ALC \) and \( \triangle BLC \).
In \( \triangle ABC \), since \( AC = BC \), it is an isosceles triangle.
\( \angle CAB = \angle CBA = \frac{180^\circ - \angle ACB}{2} = \frac{180^\circ - 131^\circ}{2} = \frac{49^\circ}{2} = 24.5^\circ \).
Convert \( 24.5^\circ \) to degrees and minutes: \( 24^\circ 30' \). So \( \angle CAL = 24^\circ 30' \).
(i) To calculate the length of AB:
Consider the right-angled triangle \( \triangle ALC \), where \( \angle ALC = 90^\circ \).
We know \( \cos (\text{angle}) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \).
For \( \angle CAL = 24^\circ 30' \):
Adjacent side \( = AL \)
Hypotenuse \( = AC = 15 \) cm
So, \( \cos 24^\circ 30' = \frac{AL}{15} \).
From trigonometric tables, \( \cos 24^\circ 30' \approx 0.90996 \).
\( 0.90996 = \frac{AL}{15} \)
\( AL = 15 \times 0.90996 \)
\( AL = 13.6494 \) cm.
Since L is the midpoint of AB, \( AB = 2 \times AL \).
\( AB = 2 \times 13.6494 = 27.2988 \approx 27.3 \) cm.
(ii) To calculate the distance of AB from the centre C (which is CL):
In the right-angled triangle \( \triangle ALC \):
We know \( \sin (\text{angle}) = \frac{\text{Opposite side}}{\text{Hypotenuse}} \).
For \( \angle CAL = 24^\circ 30' \):
Opposite side \( = CL \)
Hypotenuse \( = AC = 15 \) cm
So, \( \sin 24^\circ 30' = \frac{CL}{15} \).
From trigonometric tables, \( \sin 24^\circ 30' \approx 0.41469 \).
\( 0.41469 = \frac{CL}{15} \)
\( CL = 15 \times 0.41469 \)
\( CL = 6.22035 \approx 6.22 \) cm.
Therefore, the distance of AB from the centre C is approximately \( 6.22 \) cm. Understanding how to use chord properties and trigonometry together is important here.
In simple words: We first divide the main triangle into two smaller right-angled ones. Then, using cosine, we find half the length of the chord (AL) and double it to get AB. Next, using sine, we find the distance from the center to the chord (CL).

🎯 Exam Tip: Remember that a line drawn from the center of a circle perpendicular to a chord bisects the chord. This creates two congruent right-angled triangles which are crucial for trigonometric calculations.

Question 12. In the figure, a rocket is fired vertically upwards from its launching pad P. It first rises 20 km vertically up and then travels 80 km at \( 30^\circ \) to the vertical. PA represents the first stage of its journey and AB the second; C is a point vertically below B on the same horizontal level as P. Calculate :
(i) the height of the rocket when it is at point B;
(ii) the horizontal distance of point C from point P.

Answer: A rocket is launched from P. It first goes vertically up to A, so \( PA = 20 \) km. Then it travels \( AB = 80 \) km at an angle of \( 30^\circ \) to the vertical. Point C is vertically below B and on the same horizontal level as P. We need to calculate the height of the rocket at B and the horizontal distance PC.
Draw a line from A parallel to PC, meeting BC at D. So \( AD \parallel PC \). This makes \( AD \perp BC \).
Since \( PA \) is vertical, \( \angle APC = 90^\circ \). Given that AB travels at \( 30^\circ \) to the vertical, \( \angle BAD \) would be \( 90^\circ - 30^\circ = 60^\circ \).
Also, \( DC = PA = 20 \) km.
(i) To calculate the height of the rocket when it is at point B (which is BC):
Consider the right-angled triangle \( \triangle ABD \), where \( \angle D = 90^\circ \).
We have \( AB = 80 \) km and \( \angle BAD = 60^\circ \).
We know \( \sin (\text{angle}) = \frac{\text{Opposite side}}{\text{Hypotenuse}} \).
For \( \angle BAD = 60^\circ \):
Opposite side \( = BD \)
Hypotenuse \( = AB = 80 \) km
So, \( \sin 60^\circ = \frac{BD}{80} \).
\( \frac{\sqrt{3}}{2} = \frac{BD}{80} \)
\( \implies BD = \frac{\sqrt{3} \times 80}{2} = 40\sqrt{3} \) km.
The total height \( BC = BD + DC \).
\( BC = 40\sqrt{3} + 20 \).
Using \( \sqrt{3} \approx 1.732 \):
\( BC = 40(1.732) + 20 = 69.280 + 20 = 89.28 \) km.
The height of the rocket at point B is approximately \( 89.28 \) km.
(ii) To calculate the horizontal distance of point C from point P (which is PC):
Consider the right-angled triangle \( \triangle ABD \).
We know \( \cos (\text{angle}) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \).
For \( \angle BAD = 60^\circ \):
Adjacent side \( = AD \)
Hypotenuse \( = AB = 80 \) km
So, \( \cos 60^\circ = \frac{AD}{80} \).
\( \frac{1}{2} = \frac{AD}{80} \)
\( \implies AD = 80 \times \frac{1}{2} = 40 \) km.
Since \( AD \parallel PC \) and \( AP \parallel DC \) (from construction of rectangle ADCP), \( PC = AD \).
Therefore, the horizontal distance \( PC = 40 \) km. Breaking down the rocket's path into vertical and horizontal components simplifies distance and height calculations.
In simple words: We split the rocket's angled path into vertical and horizontal parts. First, we find the vertical distance of the angled part using sine, add it to the initial vertical climb to get the total height. Then, we find the horizontal distance of the angled part using cosine, which gives us the total horizontal distance from the launch pad.

🎯 Exam Tip: When a problem involves motion at an angle, always resolve the motion into its horizontal and vertical components. This often creates right-angled triangles that can be solved using trigonometry.

Question 13. In the figure, BCDE is a rectangle, ED = 3.88 cm, AD = 10 cm and \( \angle DAC = 20^\circ 35' \). Calculate, without using Pythagoras' theorem,
(i) the length of CD;
(ii) the length of AC;
(iii) the size of angle AEB.

Answer: In the figure, BCDE is a rectangle, so \( CD = BE \) and \( ED = BC = 3.88 \) cm. Point A lies on the line passing through C and B, extended to the left of B. We are given \( AD = 10 \) cm and \( \angle DAC = 23^\circ 35' \). We need to calculate CD, AC, and \( \angle AEB \).
(i) To calculate the length of CD:
Consider the right-angled triangle \( \triangle ACD \), where \( \angle C = 90^\circ \).
We know \( \sin (\text{angle}) = \frac{\text{Opposite side}}{\text{Hypotenuse}} \).
For \( \angle DAC = 23^\circ 35' \):
Opposite side \( = CD \)
Hypotenuse \( = AD = 10 \) cm
So, \( \sin 23^\circ 35' = \frac{CD}{10} \).
From trigonometric tables, \( \sin 23^\circ 35' \approx 0.39875 + 0.00133 = 0.40008 \).
\( 0.40008 = \frac{CD}{10} \)
\( CD = 10 \times 0.40008 = 4.0008 \approx 4.001 \) cm.
(ii) To calculate the length of AC:
Consider the right-angled triangle \( \triangle ACD \).
We know \( \cos (\text{angle}) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \).
For \( \angle DAC = 23^\circ 35' \):
Adjacent side \( = AC \)
Hypotenuse \( = AD = 10 \) cm
So, \( \cos 23^\circ 35' = \frac{AC}{10} \).
From trigonometric tables, \( \cos 23^\circ 35' \approx 0.91706 - 0.00058 = 0.91648 \).
\( 0.91648 = \frac{AC}{10} \)
\( AC = 10 \times 0.91648 = 9.1648 \approx 9.165 \) cm.
(iii) To calculate the size of angle AEB:
From the rectangle properties, \( BE = CD = 4.001 \) cm. Also, \( BC = ED = 3.88 \) cm.
Since A, B, C are collinear and B is between A and C, we have \( AB = AC - BC \).
\( AB = 9.165 - 3.88 = 5.285 \) cm.
Now, consider the right-angled triangle \( \triangle AEB \), where \( \angle B = 90^\circ \).
We have \( AB = 5.285 \) cm and \( EB = 4.001 \) cm.
We know \( \tan (\text{angle}) = \frac{\text{Opposite side}}{\text{Adjacent side}} \).
For \( \angle AEB \):
Opposite side \( = AB = 5.285 \) cm
Adjacent side \( = EB = 4.001 \) cm
So, \( \tan (\angle AEB) = \frac{5.285}{4.001} \approx 1.32092 \).
From trigonometric tables, the value \( 1.32092 \) is closest to \( \tan 52^\circ 53' \) (approximately \( 1.31745 \) for \( 52^\circ 48' \) + mean difference \( 347 \) for \( 5' \)). So, \( 52^\circ 48' + 5' = 52^\circ 53' \).
Therefore, \( \angle AEB \approx 52^\circ 53' \). Trigonometric tables are a quick way to find angles from ratios.
In simple words: First, we use sine and cosine in the right triangle ADC to find the lengths of CD and AC. Then, we find the length of AB by subtracting BC from AC. Finally, in triangle AEB, we use the tangent function to find the angle AEB using the lengths of AB and EB.

🎯 Exam Tip: When using trigonometric tables, pay close attention to mean differences for angles. Remember to add the difference for tan, and subtract for cos, when interpolating to get the exact angle.

Question 14. In the figure, triangle ABC is right angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm. Without using tables find
(i) \( \tan \angle DBC \)
(ii) \( \sin \angle DBA \)

Answer: In \( \triangle ABC \), \( \angle B = 90^\circ \), \( BC = 3 \) cm, and \( AB = 4 \) cm. \( BD \perp AC \). We need to find \( \tan \angle DBC \) and \( \sin \angle DBA \) without using tables.
First, find the length of AC using Pythagoras' theorem in \( \triangle ABC \):
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (4)^2 + (3)^2 = 16 + 9 = 25 \)
\( AC = \sqrt{25} = 5 \) cm.
Now, we consider similar triangles. In \( \triangle ABC \) and \( \triangle BDC \):
\( \angle ABC = \angle BDC = 90^\circ \)
\( \angle C = \angle C \) (common angle)
Therefore, \( \triangle ABC \sim \triangle BDC \) (by AA axiom).
From similarity, the ratio of corresponding sides are equal:
\( \frac{BC}{CD} = \frac{AC}{BC} \)
\( \frac{3}{CD} = \frac{5}{3} \)
\( \implies CD = \frac{3 \times 3}{5} = \frac{9}{5} = 1.8 \) cm.
Also, \( \frac{BD}{AB} = \frac{BC}{AC} \)
\( \frac{BD}{4} = \frac{3}{5} \)
\( \implies BD = \frac{4 \times 3}{5} = \frac{12}{5} = 2.4 \) cm.
(i) To find \( \tan \angle DBC \):
In the right-angled triangle \( \triangle BDC \) (right-angled at D):
\( \tan \angle DBC = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{CD}{BD} \)
\( \tan \angle DBC = \frac{1.8}{2.4} = \frac{18}{24} = \frac{3}{4} \).
(ii) To find \( \sin \angle DBA \):
First, find AD. \( AD = AC - CD = 5 - 1.8 = 3.2 \) cm.
In \( \triangle ABD \) (right-angled at D):
\( \sin \angle DBA = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{AD}{AB} \)
\( \sin \angle DBA = \frac{3.2}{4} = \frac{32}{40} = \frac{4}{5} \). Geometric similarity helps establish relationships between sides and angles of derived triangles.
In simple words: First, we find the length of AC using Pythagoras' theorem. Then, we use similar triangles to find the lengths of BD and CD. After that, we use these lengths in the smaller right-angled triangles to find the tangent of angle DBC and the sine of angle DBA.

🎯 Exam Tip: When a perpendicular is dropped from the right angle vertex to the hypotenuse, it creates three similar triangles. This property is very useful for solving without trigonometric tables.

Question 15. In the figure, they want to find the height x of a building and the height y of the flag pole on the building. They made the measures as shown in the diagram. Find x and y. Give your answer to the nearest metre.

Answer: In the figure, BC represents the building with height \( x \), and AB represents the flag pole with height \( y \). The horizontal distance \( DC = 50 \) cm. The angle of elevation to the top of the building (B) is \( \angle BDC = 63^\circ \). The angle of elevation to the top of the flag pole (A) is \( \angle ADC = 63^\circ + 3^\circ = 66^\circ \). We need to find \( x \) and \( y \) to the nearest metre.
First, find the height of the building (x):
Consider the right-angled triangle \( \triangle BCD \), where \( \angle C = 90^\circ \).
We know \( \tan (\text{angle}) = \frac{\text{Opposite side}}{\text{Adjacent side}} \).
For \( \angle BDC = 63^\circ \):
Opposite side \( = BC = x \)
Adjacent side \( = DC = 50 \) cm
So, \( \tan 63^\circ = \frac{x}{50} \).
From trigonometric tables, \( \tan 63^\circ \approx 1.96261 \).
\( 1.96261 = \frac{x}{50} \)
\( x = 50 \times 1.96261 \)
\( x = 98.1305 \approx 98 \) m (to the nearest metre).
Next, find the total height \( AC = x + y \):
Consider the right-angled triangle \( \triangle ADC \), where \( \angle C = 90^\circ \).
For \( \angle ADC = 66^\circ \):
Opposite side \( = AC = x + y \)
Adjacent side \( = DC = 50 \) cm
So, \( \tan 66^\circ = \frac{x+y}{50} \).
From trigonometric tables, \( \tan 66^\circ \approx 2.24604 \).
\( 2.24604 = \frac{x+y}{50} \)
\( x + y = 50 \times 2.24604 \)
\( x + y = 112.302 \approx 112 \) m.
Finally, find the height of the flag pole (y):
\( y = (x + y) - x \)
\( y = 112 - 98 = 14 \) m.
Therefore, the height of the building \( x = 98 \) m and the height of the flag pole \( y = 14 \) m. Using two angles of elevation allows for solving compound height problems.
In simple words: We use the tangent function for two different angles of elevation from the ground. First, we find the height of the building (x) using the angle to its top. Then, we find the total height (building + flagpole) using the angle to the top of the flagpole. Subtracting the building's height from the total height gives us the flagpole's height (y).

🎯 Exam Tip: When dealing with combined heights (like a flag pole on a building), always consider two right-angled triangles from the observation point to the base and the top of each object, using angles of elevation.

Question 16. The upper part of tree, broken by the wind, makes an angle of \( 30^\circ \) with ground, and the horizontal distance from the root of the tree to the point where the top of the tree meets the ground is 25 metres. Find the height of the tree before it was broken, to the nearest metre.

Answer: Let TR be the original height of the tree. The tree breaks at Q, and its top part T touches the ground at S. So, \( QS \) is the broken part, and \( QR \) is the remaining part of the tree. We are given the horizontal distance from the root to where the top touches the ground, \( SR = 25 \) m, and \( \angle QSR = 30^\circ \). The total height before breaking is \( TQ + QR = QS + QR \).
First, find the length of QR:
Consider the right-angled triangle \( \triangle QSR \), where \( \angle R = 90^\circ \).
We know \( \tan (\text{angle}) = \frac{\text{Opposite side}}{\text{Adjacent side}} \).
For \( \angle QSR = 30^\circ \):
Opposite side \( = QR \)
Adjacent side \( = SR = 25 \) m
So, \( \tan 30^\circ = \frac{QR}{25} \).
\( \frac{1}{\sqrt{3}} = \frac{QR}{25} \)
\( QR = \frac{25}{\sqrt{3}} \).
Using \( \sqrt{3} \approx 1.732 \):
\( QR = \frac{25}{1.732} \approx 14.43 \) m.
Next, find the length of QS (the broken part):
In the right-angled triangle \( \triangle QSR \):
We know \( \cos (\text{angle}) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \).
For \( \angle QSR = 30^\circ \):
Adjacent side \( = SR = 25 \) m
Hypotenuse \( = QS \)
So, \( \cos 30^\circ = \frac{SR}{QS} \).
\( \frac{\sqrt{3}}{2} = \frac{25}{QS} \)
\( QS = \frac{25 \times 2}{\sqrt{3}} = \frac{50}{\sqrt{3}} \).
Using \( \sqrt{3} \approx 1.732 \):
\( QS = \frac{50}{1.732} \approx 28.868 \) m.
The total height of the tree before it was broken is \( QR + QS \).
\( \text{Height} = \frac{25}{\sqrt{3}} + \frac{50}{\sqrt{3}} = \frac{75}{\sqrt{3}} \).
To rationalize the denominator, multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \):
\( \text{Height} = \frac{75\sqrt{3}}{3} = 25\sqrt{3} \).
Using \( \sqrt{3} \approx 1.732 \):
\( \text{Height} = 25 \times 1.732 = 43.3 \) m.
To the nearest metre, the height of the tree is \( 43 \) m. This problem involves breaking a physical object and applying trigonometry to its parts.
In simple words: We consider the broken tree as a right-angled triangle. We find the height of the standing part of the tree using tangent, and the length of the broken part (which is the hypotenuse) using cosine. Adding these two lengths together gives us the original height of the tree.

🎯 Exam Tip: When a problem involves a broken object, visualize the parts forming a right-angled triangle. The broken part acts as the hypotenuse, the standing part as one leg, and the horizontal distance as the other leg.

Question 17. In the figure, ABC is an equilateral triangle of side 6 cm. D is a point in BC such that BD = 1 cm and E is the midpoint of BC. Calculate :
(i) AE,
(ii) tan ∠ADC,
(iii) ∠ADC to the nearest degree,
(iv) ∠BAD to the nearest degree.
Answer:
In an equilateral triangle ABC, all sides are equal. Here, each side is 6 cm.
E is the midpoint of BC, so BE = EC = \( \frac{BC}{2} = \frac{6}{2} = 3 \) cm.
AE is the altitude and also the median in an equilateral triangle. So, AE is perpendicular to BC.
In right-angled triangle ABE:
By Pythagoras theorem, \( AB^2 = AE^2 + BE^2 \)
\( 6^2 = AE^2 + 3^2 \)
\( 36 = AE^2 + 9 \)
\( AE^2 = 36 - 9 = 27 \)
\( AE = \sqrt{27} = 3\sqrt{3} \) cm.
Also, D is a point on BC such that BD = 1 cm.
Since E is the midpoint of BC, BE = 3 cm. Then DE = BE - BD = 3 - 1 = 2 cm.

(i) From above, \( AE = 3\sqrt{3} \) cm. This is the length of the altitude from A to BC.
\( AE = 3 \times 1.732 = 5.196 \) cm.

(ii) In right-angled triangle ADE:
We have AE = \( 3\sqrt{3} \) cm and DE = 2 cm.
\( \tan \angle ADE = \frac{AE}{DE} = \frac{3\sqrt{3}}{2} \)
\( = \frac{3 \times 1.732}{2} = \frac{5.196}{2} = 2.598 \)
This value is the tangent of ∠ADC because D, E, C are collinear and E is on BC.

(iii) To find ∠ADC to the nearest degree:
From part (ii), \( \tan \angle ADC = 2.598 \).
Using trigonometric tables, the angle whose tangent is closest to 2.598 is approximately 68°54'.
So, \( \angle ADC \approx 69^\circ \) (to the nearest degree).

(iv) To find ∠BAD to the nearest degree:
In an equilateral triangle, all angles are 60°, so \( \angle BAC = 60^\circ \).
Since AE is the median, it bisects \( \angle BAC \), so \( \angle BAE = \angle CAE = \frac{60^\circ}{2} = 30^\circ \).
In right-angled triangle ADE, we need to find \( \angle DAE \).
\( \tan \angle DAE = \frac{DE}{AE} = \frac{2}{3\sqrt{3}} \)
\( = \frac{2}{3 \times 1.732} = \frac{2}{5.196} \approx 0.3849 \)
Using trigonometric tables, the angle whose tangent is 0.3849 is approximately 21°.
So, \( \angle DAE \approx 21^\circ \).
Now, \( \angle BAD = \angle BAE - \angle DAE = 30^\circ - 21^\circ = 9^\circ \).
In simple words: First, we use the properties of an equilateral triangle and Pythagoras theorem to find the length of the altitude AE. Then, we use the tangent ratio in the right-angled triangles formed to find the required angles. An equilateral triangle is a very symmetrical shape.

🎯 Exam Tip: Remember that in an equilateral triangle, the altitude is also the median and angle bisector. This simplifies calculations significantly. Always use the properties of the triangle to establish right angles for trigonometry.

Self Evaluation And Revision (Latest Icse Questions)

Question 1. Show that \( \sqrt{\frac{1-\cos A}{1+\cos A}} = \frac{\sin A}{1+\cos A} \)
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity and try to make it look like the Right Hand Side (R.H.S.).
\( \text{L.H.S.} = \sqrt{\frac{1-\cos A}{1+\cos A}} \)
To simplify this, we multiply the numerator and denominator inside the square root by \( (1+\cos A) \). This technique is often used to eliminate a square root in the denominator.
\( = \sqrt{\frac{(1-\cos A)(1+\cos A)}{(1+\cos A)(1+\cos A)}} \)
Using the identity \( (a-b)(a+b) = a^2 - b^2 \) in the numerator and simplifying the denominator:
\( = \sqrt{\frac{1^2 - \cos^2 A}{(1+\cos A)^2}} \)
We know the fundamental trigonometric identity \( \sin^2 A + \cos^2 A = 1 \), which means \( 1 - \cos^2 A = \sin^2 A \).
\( = \sqrt{\frac{\sin^2 A}{(1+\cos A)^2}} \)
Now, we can take the square root of both the numerator and the denominator.
\( = \frac{\sqrt{\sin^2 A}}{\sqrt{(1+\cos A)^2}} \)
\( = \frac{\sin A}{1+\cos A} \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven.
In simple words: To prove this, we start with the left side, multiply the top and bottom inside the square root by \( (1+\cos A) \). This helps us use the identity \( 1-\cos^2 A = \sin^2 A \) and then take the square root to get the right side.

🎯 Exam Tip: When dealing with square roots of trigonometric expressions, multiplying by the conjugate of the denominator is a common and effective strategy. Also, remember the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \).

Question 2. Prove that : \( 1 - \frac{\cos ^2 \theta}{1+\sin \theta} = \sin \theta \)
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = 1 - \frac{\cos^2 \theta}{1+\sin \theta} \)
To combine these terms, we find a common denominator, which is \( (1+\sin \theta) \).
\( = \frac{1(1+\sin \theta) - \cos^2 \theta}{1+\sin \theta} \)
\( = \frac{1+\sin \theta - \cos^2 \theta}{1+\sin \theta} \)
We use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), which means \( \cos^2 \theta = 1 - \sin^2 \theta \). Substitute this into the expression.
\( = \frac{1+\sin \theta - (1-\sin^2 \theta)}{1+\sin \theta} \)
Now, simplify the numerator by distributing the minus sign.
\( = \frac{1+\sin \theta - 1 + \sin^2 \theta}{1+\sin \theta} \)
The \( +1 \) and \( -1 \) in the numerator cancel each other out.
\( = \frac{\sin \theta + \sin^2 \theta}{1+\sin \theta} \)
Factor out \( \sin \theta \) from the numerator.
\( = \frac{\sin \theta(1+\sin \theta)}{1+\sin \theta} \)
Since \( (1+\sin \theta) \) is a common factor in both the numerator and denominator, we can cancel it out (assuming \( 1+\sin \theta \neq 0 \)).
\( = \sin \theta \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This shows how algebraic manipulation with trigonometric identities leads to simpler forms.
In simple words: We start with the left side, put everything over a common denominator, and then use the rule \( \cos^2 \theta = 1 - \sin^2 \theta \). After simplifying, we can factor and cancel terms to get \( \sin \theta \).

🎯 Exam Tip: When simplifying expressions involving \( 1 \pm \sin \theta \) or \( 1 \pm \cos \theta \), look for opportunities to use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to substitute terms and simplify.

Question 3. Prove the following identity: \( \frac{1}{\sin \theta +\cos \theta} + \frac{1}{\sin \theta-\cos \theta} = \frac{2 \sin \theta}{1-2 \cos ^2 \theta} \)
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \frac{1}{\sin \theta +\cos \theta} + \frac{1}{\sin \theta-\cos \theta} \)
To add these two fractions, we find a common denominator, which is \( (\sin \theta +\cos \theta)(\sin \theta-\cos \theta) \).
\( = \frac{(\sin \theta-\cos \theta) + (\sin \theta+\cos \theta)}{(\sin \theta +\cos \theta)(\sin \theta-\cos \theta)} \)
Simplify the numerator by combining like terms.
\( = \frac{\sin \theta-\cos \theta + \sin \theta+\cos \theta}{\sin^2 \theta - \cos^2 \theta} \)
The \( -\cos \theta \) and \( +\cos \theta \) terms in the numerator cancel out.
\( = \frac{2 \sin \theta}{\sin^2 \theta - \cos^2 \theta} \)
Now, we use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), which implies \( \sin^2 \theta = 1 - \cos^2 \theta \). Substitute this into the denominator.
\( = \frac{2 \sin \theta}{(1 - \cos^2 \theta) - \cos^2 \theta} \)
Simplify the denominator.
\( = \frac{2 \sin \theta}{1 - 2 \cos^2 \theta} \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This process demonstrates how adding fractions and applying trigonometric identities leads to the desired form.
In simple words: We add the two fractions on the left by finding a common bottom part. Then, we use the rule that \( \sin^2 \theta \) can be changed to \( 1-\cos^2 \theta \) to simplify the bottom part and match the right side.

🎯 Exam Tip: When adding or subtracting trigonometric fractions, always start by finding a common denominator. Remember to use the difference of squares formula \( (a+b)(a-b) = a^2-b^2 \) and the Pythagorean identity to simplify.

Question 4. Prove that: \( \frac{\cos A}{1-\tan A} + \frac{\sin A}{1-\cot A} = \cos A + \sin A. \)
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \frac{\cos A}{1-\tan A} + \frac{\sin A}{1-\cot A} \)
First, we express \( \tan A \) and \( \cot A \) in terms of \( \sin A \) and \( \cos A \).
Remember that \( \tan A = \frac{\sin A}{\cos A} \) and \( \cot A = \frac{\cos A}{\sin A} \).
Substitute these into the expression:
\( = \frac{\cos A}{1-\frac{\sin A}{\cos A}} + \frac{\sin A}{1-\frac{\cos A}{\sin A}} \)
Simplify the denominators by finding common fractions.
\( = \frac{\cos A}{\frac{\cos A - \sin A}{\cos A}} + \frac{\sin A}{\frac{\sin A - \cos A}{\sin A}} \)
Now, invert the denominators and multiply.
\( = \cos A \times \frac{\cos A}{\cos A - \sin A} + \sin A \times \frac{\sin A}{\sin A - \cos A} \)
\( = \frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A} \)
Notice that \( (\sin A - \cos A) = -(\cos A - \sin A) \). We can make the denominators the same.
\( = \frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{-(\cos A - \sin A)} \)
\( = \frac{\cos^2 A}{\cos A - \sin A} - \frac{\sin^2 A}{\cos A - \sin A} \)
Now that the denominators are the same, combine the numerators.
\( = \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} \)
Use the difference of squares formula in the numerator: \( a^2 - b^2 = (a-b)(a+b) \).
\( = \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A} \)
Cancel out the common factor \( (\cos A - \sin A) \) from the numerator and denominator (assuming \( \cos A - \sin A \neq 0 \)).
\( = \cos A + \sin A \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This process highlights the importance of converting to sine and cosine and simplifying fractions.
In simple words: We change tan and cot into sin and cos. Then we combine the fractions. We use the \( a^2-b^2 \) rule to simplify the top part, and finally, we cancel out common terms to get the answer.

🎯 Exam Tip: For identities involving \( \tan \) and \( \cot \), convert them to \( \sin \) and \( \cos \) at the start. Watch out for terms like \( (a-b) \) and \( (b-a) \) that are negatives of each other, allowing for common denominators.

Question 5. Prove \( \frac{\sin A}{1+\cos A} + \frac{1+\cos A}{\sin A} = 2 \operatorname{cosec} A \).
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \frac{\sin A}{1+\cos A} + \frac{1+\cos A}{\sin A} \)
To add these two fractions, we find a common denominator, which is \( \sin A (1+\cos A) \).
\( = \frac{\sin A \times \sin A + (1+\cos A) \times (1+\cos A)}{\sin A (1+\cos A)} \)
\( = \frac{\sin^2 A + (1+\cos A)^2}{\sin A (1+\cos A)} \)
Expand \( (1+\cos A)^2 \) using \( (a+b)^2 = a^2+2ab+b^2 \).
\( = \frac{\sin^2 A + (1 + 2\cos A + \cos^2 A)}{\sin A (1+\cos A)} \)
Rearrange the terms in the numerator to group \( \sin^2 A \) and \( \cos^2 A \).
\( = \frac{(\sin^2 A + \cos^2 A) + 1 + 2\cos A}{\sin A (1+\cos A)} \)
Using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \).
\( = \frac{1 + 1 + 2\cos A}{\sin A (1+\cos A)} \)
Simplify the numerator.
\( = \frac{2 + 2\cos A}{\sin A (1+\cos A)} \)
Factor out 2 from the numerator.
\( = \frac{2(1+\cos A)}{\sin A (1+\cos A)} \)
Cancel out the common factor \( (1+\cos A) \) from the numerator and denominator (assuming \( 1+\cos A \neq 0 \)).
\( = \frac{2}{\sin A} \)
Since \( \operatorname{cosec} A = \frac{1}{\sin A} \).
\( = 2 \operatorname{cosec} A \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This shows a direct way to combine fractions and apply basic identities.
In simple words: We add the two fractions by making their bottom parts the same. After expanding and using the rule \( \sin^2 A + \cos^2 A = 1 \), we can simplify the expression to \( \frac{2}{\sin A} \), which is the same as \( 2 \operatorname{cosec} A \).

🎯 Exam Tip: When proving identities involving sums of fractions, always aim to combine them into a single fraction first. Then, look for ways to apply Pythagorean identities and algebraic factorisation to simplify the expression.

Question 6. Prove \( (1 + \tan A)^2 + (1 - \tan A)^2 = 2 \sec^2 A \).
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = (1 + \tan A)^2 + (1 - \tan A)^2 \)
Expand both terms using the algebraic identities \( (a+b)^2 = a^2+2ab+b^2 \) and \( (a-b)^2 = a^2-2ab+b^2 \).
\( = (1^2 + 2(1)(\tan A) + \tan^2 A) + (1^2 - 2(1)(\tan A) + \tan^2 A) \)
\( = (1 + 2\tan A + \tan^2 A) + (1 - 2\tan A + \tan^2 A) \)
Now, combine the like terms.
\( = 1 + 2\tan A + \tan^2 A + 1 - 2\tan A + \tan^2 A \)
The \( +2\tan A \) and \( -2\tan A \) terms cancel each other out.
\( = 1 + 1 + \tan^2 A + \tan^2 A \)
\( = 2 + 2\tan^2 A \)
Factor out 2 from the expression.
\( = 2(1 + \tan^2 A) \)
We use the trigonometric identity \( 1 + \tan^2 A = \sec^2 A \). This identity is derived from the Pythagorean identity.
\( = 2\sec^2 A \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This demonstrates how basic algebraic expansion and trigonometric identities can simplify complex expressions.
In simple words: We expand both squared terms using basic algebra rules. The middle terms \( (+2\tan A) \) and \( (-2\tan A) \) cancel out. Then we combine what's left and use the rule that \( 1+\tan^2 A \) is the same as \( \sec^2 A \) to get the final answer.

🎯 Exam Tip: Always remember the algebraic expansion formulas \( (a+b)^2 \) and \( (a-b)^2 \). Also, know the three main Pythagorean identities: \( \sin^2 \theta + \cos^2 \theta = 1 \), \( 1 + \tan^2 \theta = \sec^2 \theta \), and \( 1 + \cot^2 \theta = \operatorname{cosec}^2 \theta \).

Question 7. Prove that \( \frac{\sin \theta \tan \theta}{1-\cos \theta} = 1 + \sec \theta \).
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \frac{\sin \theta \tan \theta}{1-\cos \theta} \)
First, express \( \tan \theta \) in terms of \( \sin \theta \) and \( \cos \theta \). Remember that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
\( = \frac{\sin \theta \left(\frac{\sin \theta}{\cos \theta}\right)}{1-\cos \theta} \)
Multiply the terms in the numerator.
\( = \frac{\frac{\sin^2 \theta}{\cos \theta}}{1-\cos \theta} \)
Rewrite this as a single fraction.
\( = \frac{\sin^2 \theta}{\cos \theta (1-\cos \theta)} \)
Now, use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), which means \( \sin^2 \theta = 1 - \cos^2 \theta \).
\( = \frac{1-\cos^2 \theta}{\cos \theta (1-\cos \theta)} \)
Use the difference of squares formula in the numerator: \( a^2 - b^2 = (a-b)(a+b) \). So, \( 1-\cos^2 \theta = (1-\cos \theta)(1+\cos \theta) \).
\( = \frac{(1-\cos \theta)(1+\cos \theta)}{\cos \theta (1-\cos \theta)} \)
Cancel out the common factor \( (1-\cos \theta) \) from the numerator and denominator (assuming \( 1-\cos \theta \neq 0 \)).
\( = \frac{1+\cos \theta}{\cos \theta} \)
Separate the terms in the numerator over the common denominator.
\( = \frac{1}{\cos \theta} + \frac{\cos \theta}{\cos \theta} \)
We know that \( \frac{1}{\cos \theta} = \sec \theta \) and \( \frac{\cos \theta}{\cos \theta} = 1 \).
\( = \sec \theta + 1 \)
\( = 1 + \sec \theta \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This identity is a good example of simplifying by converting to sine and cosine and applying key identities.
In simple words: We change \( \tan \theta \) to \( \sin \theta / \cos \theta \), then use the rule \( \sin^2 \theta = 1-\cos^2 \theta \). We factor the top part and cancel terms. Finally, we split the fraction to get \( 1/\cos \theta \) plus 1, which means \( \sec \theta + 1 \).

🎯 Exam Tip: When a fraction has a \( (1-\cos \theta) \) or \( (1-\sin \theta) \) in the denominator, try to use \( \sin^2 \theta = 1-\cos^2 \theta \) (or vice versa) in the numerator to create a factor that can be cancelled. Converting everything to \( \sin \theta \) and \( \cos \theta \) is always a safe first step.

Question 8. Prove the identify: \( \frac{\sec A-1}{\sec A+1} = \frac{1-\cos A}{1+\cos A} \)
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \frac{\sec A-1}{\sec A+1} \)
First, express \( \sec A \) in terms of \( \cos A \). Remember that \( \sec A = \frac{1}{\cos A} \).
\( = \frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1} \)
Find a common denominator for the numerator and the denominator of the main fraction.
\( = \frac{\frac{1-\cos A}{\cos A}}{\frac{1+\cos A}{\cos A}} \)
Now, we can multiply the numerator by the reciprocal of the denominator.
\( = \frac{1-\cos A}{\cos A} \times \frac{\cos A}{1+\cos A} \)
Cancel out the common factor \( \cos A \).
\( = \frac{1-\cos A}{1+\cos A} \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This demonstrates a straightforward method of converting to cosine and simplifying.
In simple words: To prove this, we change \( \sec A \) to \( 1/\cos A \) on the left side. Then, we simplify the complex fraction by finding a common denominator in both the top and bottom parts. Finally, we cancel out \( \cos A \) from both to get the right side.

🎯 Exam Tip: When an identity involves reciprocal functions like \( \sec \theta \) or \( \operatorname{cosec} \theta \), it's often easiest to convert them into their sine or cosine equivalents first. This typically simplifies the expression into more manageable fractions.

Question 9. Prove that identity: \( \frac{\sin A}{1+\cos A} = \operatorname{cosec} A - \cot A \)
Answer:
Let's start from the Right Hand Side (R.H.S.) of the identity and transform it to match the Left Hand Side (L.H.S.).
\( \text{R.H.S.} = \operatorname{cosec} A - \cot A \)
Express \( \operatorname{cosec} A \) and \( \cot A \) in terms of \( \sin A \) and \( \cos A \).
Remember that \( \operatorname{cosec} A = \frac{1}{\sin A} \) and \( \cot A = \frac{\cos A}{\sin A} \).
\( = \frac{1}{\sin A} - \frac{\cos A}{\sin A} \)
Since the denominators are already the same, combine the numerators.
\( = \frac{1-\cos A}{\sin A} \)
To get a term of \( (1+\cos A) \) in the denominator, we can multiply both the numerator and the denominator by \( (1+\cos A) \). This is a useful step to create \( \sin^2 A \) in the numerator.
\( = \frac{(1-\cos A)(1+\cos A)}{\sin A (1+\cos A)} \)
Use the difference of squares formula in the numerator: \( (a-b)(a+b) = a^2-b^2 \). So, \( (1-\cos A)(1+\cos A) = 1-\cos^2 A \).
\( = \frac{1-\cos^2 A}{\sin A (1+\cos A)} \)
Using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \), we know that \( 1-\cos^2 A = \sin^2 A \).
\( = \frac{\sin^2 A}{\sin A (1+\cos A)} \)
Cancel out one \( \sin A \) from the numerator and denominator (assuming \( \sin A \neq 0 \)).
\( = \frac{\sin A}{1+\cos A} \)
This matches the Left Hand Side (L.H.S.).
Therefore, R.H.S. = L.H.S., and the identity is proven. This shows that working from the more complex side (often the one with multiple terms) can be an effective strategy.
In simple words: We start with the right side and change cosec and cot into sin and cos. Then we combine the fractions. To get the left side, we multiply the top and bottom by \( (1+\cos A) \) which helps us use the rule \( 1-\cos^2 A = \sin^2 A \) and simplify.

🎯 Exam Tip: When proving identities, if one side looks more complicated (e.g., involves subtraction or multiple terms), it's often easier to start from that side and simplify it. Remember to always use \( \sin \theta \) and \( \cos \theta \) for \( \operatorname{cosec} \theta \) and \( \cot \theta \) as a first step.

Question 10. Prove the identity: \( \frac{\sin A}{1+\cos A} + \frac{1+\cos A}{\sin A} = 2 \operatorname{cosec} A \).
Answer:
This question is identical to Question 5. Let's prove it again for clarity.
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \frac{\sin A}{1+\cos A} + \frac{1+\cos A}{\sin A} \)
To add these two fractions, we find a common denominator, which is \( \sin A (1+\cos A) \).
\( = \frac{\sin A \times \sin A + (1+\cos A) \times (1+\cos A)}{\sin A (1+\cos A)} \)
\( = \frac{\sin^2 A + (1+\cos A)^2}{\sin A (1+\cos A)} \)
Expand \( (1+\cos A)^2 \) using the algebraic identity \( (a+b)^2 = a^2+2ab+b^2 \).
\( = \frac{\sin^2 A + (1 + 2\cos A + \cos^2 A)}{\sin A (1+\cos A)} \)
Rearrange the terms in the numerator to group \( \sin^2 A \) and \( \cos^2 A \).
\( = \frac{(\sin^2 A + \cos^2 A) + 1 + 2\cos A}{\sin A (1+\cos A)} \)
Using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \).
\( = \frac{1 + 1 + 2\cos A}{\sin A (1+\cos A)} \)
Simplify the numerator.
\( = \frac{2 + 2\cos A}{\sin A (1+\cos A)} \)
Factor out 2 from the numerator.
\( = \frac{2(1+\cos A)}{\sin A (1+\cos A)} \)
Cancel out the common factor \( (1+\cos A) \) from the numerator and denominator (assuming \( 1+\cos A \neq 0 \)).
\( = \frac{2}{\sin A} \)
Since \( \operatorname{cosec} A = \frac{1}{\sin A} \).
\( = 2 \operatorname{cosec} A \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This shows a direct way to combine fractions and apply basic identities.
In simple words: We combine the two fractions on the left by finding a common denominator. Then, we expand and use the identity \( \sin^2 A + \cos^2 A = 1 \) to simplify the top part. After factoring, we cancel terms to get \( 2/\sin A \), which is \( 2 \operatorname{cosec} A \).

🎯 Exam Tip: Always look for common denominators when adding fractions. The identity \( \sin^2 A + \cos^2 A = 1 \) is fundamental and frequently used to simplify expressions after expansion.

Question 11. Prove that: \( (\operatorname{cosec} A - \sin A) ( \sec A - \cos A) \sec^2 A = \tan A \).
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = (\operatorname{cosec} A - \sin A) (\sec A - \cos A) \sec^2 A \)
First, express \( \operatorname{cosec} A \) and \( \sec A \) in terms of \( \sin A \) and \( \cos A \).
\( \operatorname{cosec} A = \frac{1}{\sin A} \) and \( \sec A = \frac{1}{\cos A} \).
Substitute these into the expression:
\( = \left(\frac{1}{\sin A} - \sin A\right) \left(\frac{1}{\cos A} - \cos A\right) \left(\frac{1}{\cos A}\right)^2 \)
Find common denominators for the terms inside the parentheses.
\( = \left(\frac{1 - \sin^2 A}{\sin A}\right) \left(\frac{1 - \cos^2 A}{\cos A}\right) \left(\frac{1}{\cos^2 A}\right) \)
Using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \), we know that \( 1 - \sin^2 A = \cos^2 A \) and \( 1 - \cos^2 A = \sin^2 A \).
Substitute these identities:
\( = \left(\frac{\cos^2 A}{\sin A}\right) \left(\frac{\sin^2 A}{\cos A}\right) \left(\frac{1}{\cos^2 A}\right) \)
Now, multiply these three fractions together.
\( = \frac{\cos^2 A \times \sin^2 A \times 1}{\sin A \times \cos A \times \cos^2 A} \)
Cancel out common terms. One \( \sin A \) from the numerator cancels one from the denominator. One \( \cos^2 A \) from the numerator cancels one from the denominator.
\( = \frac{\sin A}{\cos A} \)
We know that \( \frac{\sin A}{\cos A} = \tan A \).
\( = \tan A \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This highlights the power of converting all terms to sine and cosine.
In simple words: We start with the left side and change all terms to sin and cos. Then we combine the fractions inside the brackets. Using the rule \( \sin^2 A + \cos^2 A = 1 \), we simplify. Finally, after canceling common parts, we are left with \( \sin A / \cos A \), which is \( \tan A \).

🎯 Exam Tip: When an identity involves multiple trigonometric functions, a good general strategy is to convert all terms to \( \sin \) and \( \cos \). This often simplifies the problem into algebraic manipulation of fractions.

Question 12. Prove that \( \frac{\tan^2 \theta}{(\sec \theta-1)^2} = \frac{1+\cos \theta}{1-\cos \theta} \).
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \frac{\tan^2 \theta}{(\sec \theta-1)^2} \)
First, express \( \tan^2 \theta \) and \( \sec \theta \) in terms of \( \sin \theta \) and \( \cos \theta \).
\( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \).
Substitute these into the expression:
\( = \frac{\frac{\sin^2 \theta}{\cos^2 \theta}}{\left(\frac{1}{\cos \theta}-1\right)^2} \)
Simplify the denominator by finding a common denominator inside the parenthesis.
\( = \frac{\frac{\sin^2 \theta}{\cos^2 \theta}}{\left(\frac{1-\cos \theta}{\cos \theta}\right)^2} \)
Square the term in the denominator.
\( = \frac{\frac{\sin^2 \theta}{\cos^2 \theta}}{\frac{(1-\cos \theta)^2}{\cos^2 \theta}} \)
Now, multiply the numerator by the reciprocal of the denominator.
\( = \frac{\sin^2 \theta}{\cos^2 \theta} \times \frac{\cos^2 \theta}{(1-\cos \theta)^2} \)
Cancel out the common factor \( \cos^2 \theta \).
\( = \frac{\sin^2 \theta}{(1-\cos \theta)^2} \)
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we know that \( \sin^2 \theta = 1-\cos^2 \theta \).
\( = \frac{1-\cos^2 \theta}{(1-\cos \theta)^2} \)
Use the difference of squares formula in the numerator: \( a^2 - b^2 = (a-b)(a+b) \). So, \( 1-\cos^2 \theta = (1-\cos \theta)(1+\cos \theta) \).
\( = \frac{(1-\cos \theta)(1+\cos \theta)}{(1-\cos \theta)(1-\cos \theta)} \)
Cancel out one \( (1-\cos \theta) \) from the numerator and denominator (assuming \( 1-\cos \theta \neq 0 \)).
\( = \frac{1+\cos \theta}{1-\cos \theta} \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This shows a step-by-step method to simplify complex trigonometric fractions.
In simple words: We change \( \tan^2 \theta \) and \( \sec \theta \) to their sin/cos forms. We simplify the bottom part by making it a single fraction and squaring it. Then we cancel out \( \cos^2 \theta \) terms. After using \( \sin^2 \theta = 1-\cos^2 \theta \) and factoring, we cancel another common term to get the right side.

🎯 Exam Tip: When you see squares of functions, think about Pythagorean identities and the difference of squares. Converting everything to sine and cosine is a robust strategy for complex fractions.

Question 13. Show that \( \sqrt{\frac{1-\cos A}{1+\cos A}} = \frac{\sin A}{1+\cos A} \).
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \sqrt{\frac{1-\cos A}{1+\cos A}} \)
To remove the square root from the denominator, we multiply the numerator and denominator inside the square root by \( (1+\cos A) \). This is called rationalising.
\( = \sqrt{\frac{(1-\cos A)(1+\cos A)}{(1+\cos A)(1+\cos A)}} \)
Using the identity \( (a-b)(a+b) = a^2 - b^2 \) in the numerator and simplifying the denominator:
\( = \sqrt{\frac{1^2 - \cos^2 A}{(1+\cos A)^2}} \)
We use the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \), which means \( 1 - \cos^2 A = \sin^2 A \).
\( = \sqrt{\frac{\sin^2 A}{(1+\cos A)^2}} \)
Now, take the square root of both the numerator and the denominator.
\( = \frac{\sqrt{\sin^2 A}}{\sqrt{(1+\cos A)^2}} \)
\( = \frac{\sin A}{1+\cos A} \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This method is effective for expressions with square roots.
In simple words: We begin with the left side. Inside the square root, we multiply the top and bottom by \( (1+\cos A) \). This lets us use the rule \( 1-\cos^2 A = \sin^2 A \). After simplifying, we take the square root of the top and bottom parts to get the right side.

🎯 Exam Tip: When dealing with square roots in trigonometric identities, a common strategy is to multiply the numerator and denominator by the conjugate of the expression under the root to create a perfect square or use a Pythagorean identity.

Question 14. Prove the identity : \( (\sin \theta + \cos \theta) (\tan \theta + \cot \theta) = \sec \theta + \operatorname{cosec} \theta \).
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = (\sin \theta + \cos \theta) (\tan \theta + \cot \theta) \)
First, express \( \tan \theta \) and \( \cot \theta \) in terms of \( \sin \theta \) and \( \cos \theta \).
\( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).
Substitute these into the expression:
\( = (\sin \theta + \cos \theta) \left(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\right) \)
Find a common denominator for the terms inside the second parenthesis.
\( = (\sin \theta + \cos \theta) \left(\frac{\sin \theta \times \sin \theta + \cos \theta \times \cos \theta}{\cos \theta \sin \theta}\right) \)
\( = (\sin \theta + \cos \theta) \left(\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}\right) \)
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( = (\sin \theta + \cos \theta) \left(\frac{1}{\sin \theta \cos \theta}\right) \)
Multiply the terms.
\( = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \)
Separate the terms in the numerator over the common denominator.
\( = \frac{\sin \theta}{\sin \theta \cos \theta} + \frac{\cos \theta}{\sin \theta \cos \theta} \)
Cancel out common terms in each fraction.
\( = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \)
We know that \( \frac{1}{\cos \theta} = \sec \theta \) and \( \frac{1}{\sin \theta} = \operatorname{cosec} \theta \).
\( = \sec \theta + \operatorname{cosec} \theta \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This identity demonstrates how to expand and simplify terms efficiently.
In simple words: We start with the left side. We change tan and cot into sin and cos. Then we add the fractions inside the second bracket by finding a common bottom part. Using the rule \( \sin^2 \theta + \cos^2 \theta = 1 \), we simplify, then split the fraction into two parts. This gives us \( 1/\cos \theta \) and \( 1/\sin \theta \), which are \( \sec \theta \) and \( \operatorname{cosec} \theta \).

🎯 Exam Tip: When you have terms like \( (\tan \theta + \cot \theta) \), converting them to \( \sin \theta \) and \( \cos \theta \) and finding a common denominator usually leads to a simplification involving \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 15. Prove that \( \frac{\sin \theta}{(1-\cot \theta)} + \frac{\cos \theta}{(1 - \tan \theta)} = \cos \theta + \sin \theta \).
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \frac{\sin \theta}{1-\cot \theta} + \frac{\cos \theta}{1 - \tan \theta} \)
First, express \( \cot \theta \) and \( \tan \theta \) in terms of \( \sin \theta \) and \( \cos \theta \).
\( \cot \theta = \frac{\cos \theta}{\sin \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
Substitute these into the expression:
\( = \frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\cos \theta}{1 - \frac{\sin \theta}{\cos \theta}} \)
Simplify the denominators by finding common fractions.
\( = \frac{\sin \theta}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\cos \theta}{\frac{\cos \theta - \sin \theta}{\cos \theta}} \)
Now, invert the denominators and multiply.
\( = \sin \theta \times \frac{\sin \theta}{\sin \theta - \cos \theta} + \cos \theta \times \frac{\cos \theta}{\cos \theta - \sin \theta} \)
\( = \frac{\sin^2 \theta}{\sin \theta - \cos \theta} + \frac{\cos^2 \theta}{\cos \theta - \sin \theta} \)
Notice that \( (\cos \theta - \sin \theta) = -(\sin \theta - \cos \theta) \). We can make the denominators the same.
\( = \frac{\sin^2 \theta}{\sin \theta - \cos \theta} + \frac{\cos^2 \theta}{-(\sin \theta - \cos \theta)} \)
\( = \frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\cos^2 \theta}{\sin \theta - \cos \theta} \)
Now that the denominators are the same, combine the numerators.
\( = \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta - \cos \theta} \)
Use the difference of squares formula in the numerator: \( a^2 - b^2 = (a-b)(a+b) \). So, \( \sin^2 \theta - \cos^2 \theta = (\sin \theta - \cos \theta)(\sin \theta + \cos \theta) \).
\( = \frac{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{\sin \theta - \cos \theta} \)
Cancel out the common factor \( (\sin \theta - \cos \theta) \) from the numerator and denominator (assuming \( \sin \theta - \cos \theta \neq 0 \)).
\( = \sin \theta + \cos \theta \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This illustrates how to handle fractions with cotangent and tangent.
In simple words: We change cot and tan into sin and cos. Then we simplify the fractions by taking a common denominator for the parts inside the main fraction. After bringing terms up, we use the \( a^2-b^2 \) rule and cancel the common part \( (\sin \theta - \cos \theta) \) to get the final answer.

🎯 Exam Tip: When you see \( (1-\cot \theta) \) and \( (1-\tan \theta) \) in denominators of fractions to be added, always convert them to \( \sin/\cos \) form. Be careful with sign changes like \( (a-b) = -(b-a) \) to get common denominators.

Question 16. Prove that \( \frac{\cos A}{1+\sin A} + \tan A = \sec A \).
Answer:
Let's start from the Left Hand Side (L.H.S.) of the identity.
\( \text{L.H.S.} = \frac{\cos A}{1+\sin A} + \tan A \)
First, express \( \tan A \) in terms of \( \sin A \) and \( \cos A \). Remember that \( \tan A = \frac{\sin A}{\cos A} \).
\( = \frac{\cos A}{1+\sin A} + \frac{\sin A}{\cos A} \)
To add these two fractions, we find a common denominator, which is \( \cos A (1+\sin A) \).
\( = \frac{\cos A \times \cos A + \sin A \times (1+\sin A)}{\cos A (1+\sin A)} \)
\( = \frac{\cos^2 A + \sin A + \sin^2 A}{\cos A (1+\sin A)} \)
Rearrange the terms in the numerator to group \( \sin^2 A \) and \( \cos^2 A \).
\( = \frac{(\cos^2 A + \sin^2 A) + \sin A}{\cos A (1+\sin A)} \)
Using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \).
\( = \frac{1 + \sin A}{\cos A (1+\sin A)} \)
Cancel out the common factor \( (1+\sin A) \) from the numerator and denominator (assuming \( 1+\sin A \neq 0 \)).
\( = \frac{1}{\cos A} \)
We know that \( \frac{1}{\cos A} = \sec A \).
\( = \sec A \)
This matches the Right Hand Side (R.H.S.).
Therefore, L.H.S. = R.H.S., and the identity is proven. This is a common identity that demonstrates combining fractions and applying Pythagorean identities.
In simple words: We start with the left side and change \( \tan A \) to \( \sin A / \cos A \). Then, we add the two fractions by making their bottom parts the same. After using the rule \( \sin^2 A + \cos^2 A = 1 \), we simplify by canceling \( (1+\sin A) \) to get \( 1/\cos A \), which is \( \sec A \).

🎯 Exam Tip: For identities that combine different trigonometric functions, converting all terms to \( \sin \) and \( \cos \) is often the most direct path. Always look for opportunities to use \( \sin^2 A + \cos^2 A = 1 \) after finding a common denominator.