GSEB Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1

Question 1. In \( \Delta ABC \), right angled at B, AB = 24 cm, BC = 7 cm, determine
(i) sin A, cos A
(ii) sin C, cos C
Answer: First, we use the Pythagorean theorem to find the length of the hypotenuse AC.
In right \( \Delta ABC \):
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = 24^2 + 7^2 \)
\( AC^2 = 576 + 49 \)
\( AC^2 = 625 \)
\( AC = \sqrt{625} \)
\( AC = 25 \) cm

(i) For angle A:
\( \sin A = \frac{BC}{AC} = \frac{7}{25} \)
\( \cos A = \frac{AB}{AC} = \frac{24}{25} \)
(ii) For angle C:
\( \sin C = \frac{AB}{AC} = \frac{24}{25} \)
\( \cos C = \frac{BC}{AC} = \frac{7}{25} \)
In simple words: We have a right-angled triangle. First, we compute the length of the diagonal side (hypotenuse) using the Pythagorean theorem. Then, we determine the sine and cosine ratios for angle A and angle C by dividing the appropriate side lengths.

Exam Tip: Always remember SOH CAH TOA to correctly identify the opposite, adjacent, and hypotenuse sides relative to the angle you are considering.

Question 2. In Fig. find tan P – cot R.
Answer: We are given a right-angled triangle PQR, with a right angle at Q. We first calculate the length of side QR using the Pythagorean theorem.
In \( \Delta PQR \):
\( PR^2 = PQ^2 + QR^2 \)
\( 13^2 = 12^2 + QR^2 \)
\( 169 = 144 + QR^2 \)
\( QR^2 = 169 - 144 \)
\( QR^2 = 25 \)
\( QR = \sqrt{25} \)
\( QR = 5 \) cm

Now, we determine the values of \( \tan P \) and \( \cot R \):
\( \tan P = \frac{\text{Opposite side to P}}{\text{Adjacent side to P}} = \frac{QR}{PQ} = \frac{5}{12} \)
\( \cot R = \frac{\text{Adjacent side to R}}{\text{Opposite side to R}} = \frac{QR}{PQ} = \frac{5}{12} \)
Finally, we find the difference:
\( \tan P - \cot R = \frac{5}{12} - \frac{5}{12} = 0 \)
In simple words: We have a right triangle with two sides given. First, we find the third side using the Pythagorean theorem. Then, we calculate the tangent of angle P and the cotangent of angle R. Since both values happen to be the same, their difference is zero.

Exam Tip: Remember that `tan` and `cot` are reciprocal ratios. In a right triangle, `tan P` and `cot R` might be equal if P and R are complementary angles (i.e., `P + R = 90^\circ`).

Question 3. If \( \sin A = \frac{3}{4} \) calculate cos A and tan A.
Answer: Given \( \sin A = \frac{3}{4} \). We know that \( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \).
So, let Perpendicular (BC) \( = 3k \) and Hypotenuse (AC) \( = 4k \) for some positive number k.
In right \( \Delta ABC \) (right angled at B):
By Pythagorean theorem:
\( AC^2 = AB^2 + BC^2 \)
\( (4k)^2 = AB^2 + (3k)^2 \)
\( 16k^2 = AB^2 + 9k^2 \)
\( AB^2 = 16k^2 - 9k^2 \)
\( AB^2 = 7k^2 \)
\( AB = \sqrt{7k^2} = \sqrt{7}k \)

Now, we calculate \( \cos A \) and \( \tan A \):
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4} \)
\( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}} \)
In simple words: Given the sine of angle A, we first define the opposite and hypotenuse sides using a variable 'k'. Then, we apply the Pythagorean theorem to find the remaining side. Finally, we calculate the cosine and tangent of angle A using these side lengths.

Exam Tip: When given a trigonometric ratio, always visualize a right-angled triangle and use the Pythagorean theorem to find the unknown side before calculating other ratios.

Question 4. Given 15 cot A = 8 find sin A and sec A.
Answer: Given \( 15 \cot A = 8 \).
So, \( \cot A = \frac{8}{15} \). We know that \( \cot A = \frac{\text{Base}}{\text{Perpendicular}} \).
Let Base (AB) \( = 8k \) and Perpendicular (BC) \( = 15k \) for some positive number k.
In right \( \Delta ABC \) (right angled at B):
By Pythagorean theorem:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (8k)^2 + (15k)^2 \)
\( AC^2 = 64k^2 + 225k^2 \)
\( AC^2 = 289k^2 \)
\( AC = \sqrt{289k^2} = 17k \)

Now, we calculate \( \sin A \) and \( \sec A \):
\( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17} \)
\( \sec A = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8} \)
In simple words: From the given information, we deduce the cotangent of angle A. This allows us to represent the adjacent and opposite sides of the triangle. We then apply the Pythagorean theorem to find the hypotenuse AC. Finally, we determine the values of sin A and sec A using these calculated side lengths.

Exam Tip: Always simplify the given ratio (e.g., `15 cot A = 8` to `cot A = 8/15`) before setting up the sides of the triangle.

Question 5. Given \( \sec \theta = \frac{13}{12} \) calculate all other trigonometric ratios.
Answer: Given \( \sec \theta = \frac{13}{12} \). We know that \( \sec \theta = \frac{\text{Hypotenuse}}{\text{Base}} \).
So, let Hypotenuse (AC) \( = 13k \) and Base (AB) \( = 12k \) for some positive number k.
In right \( \Delta ABC \) (right angled at B):
By Pythagorean theorem:
\( AC^2 = AB^2 + BC^2 \)
\( (13k)^2 = (12k)^2 + BC^2 \)
\( 169k^2 = 144k^2 + BC^2 \)
\( BC^2 = 169k^2 - 144k^2 \)
\( BC^2 = 25k^2 \)
\( BC = \sqrt{25k^2} = 5k \)

Now, we calculate all other trigonometric ratios:
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13} \)
\( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13} \)
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12} \)
\( \text{cosec } \theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5} \)
\( \cot \theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{AB}{BC} = \frac{12k}{5k} = \frac{12}{5} \)
In simple words: We are given the secant of angle theta. From this, we can set up the hypotenuse and adjacent side of a right triangle. Using the Pythagorean theorem, we compute the length of the opposite side. With all sides known, we determine the values of the other five trigonometric ratios.

Exam Tip: List all six trigonometric ratios and their definitions (SOH CAH TOA, and their reciprocals) to ensure you calculate all required values accurately.

Question 6. If \( \angle A \) and \( \angle B \) are acute angles such that \( \cos A = \cos B \), then show that \( \angle A = \angle B \).
Answer: To demonstrate that \( \angle A = \angle B \) when their cosines are equal, we consider two separate right-angled triangles. Let these triangles be LMN (right-angled at M) and PQR (right-angled at Q). We consider \( \angle A \) in \( \Delta LMN \) and \( \angle B \) in \( \Delta PQR \).

Given \( \cos A = \cos B \).
In \( \Delta LMN \), \( \cos A = \frac{LM}{LN} \)
In \( \Delta PQR \), \( \cos B = \frac{PQ}{PR} \)
So, \( \frac{LM}{LN} = \frac{PQ}{PR} \). Let this common ratio be k.
\( \implies \frac{LM}{PQ} = \frac{LN}{PR} = k \)
\( \implies LM = k \cdot PQ \) and \( LN = k \cdot PR \)
Now, in \( \Delta LMN \), by Pythagorean theorem:
\( LN^2 = LM^2 + MN^2 \)
\( MN^2 = LN^2 - LM^2 \)
\( MN^2 = (k \cdot PR)^2 - (k \cdot PQ)^2 \)
\( MN^2 = k^2 PR^2 - k^2 PQ^2 \)
\( MN^2 = k^2 (PR^2 - PQ^2) \)
\( \implies MN = k \sqrt{PR^2 - PQ^2} \) (1)
In \( \Delta PQR \), by Pythagorean theorem:
\( PR^2 = PQ^2 + QR^2 \)
\( QR^2 = PR^2 - PQ^2 \)
\( \implies QR = \sqrt{PR^2 - PQ^2} \) (2)
Divide equation (1) by equation (2):
\( \frac{MN}{QR} = \frac{k \sqrt{PR^2 - PQ^2}}{\sqrt{PR^2 - PQ^2}} = k \)
So, we have \( \frac{LM}{PQ} = \frac{LN}{PR} = \frac{MN}{QR} = k \).
Since the ratio of corresponding sides is equal, the two triangles \( \Delta LMN \) and \( \Delta PQR \) are similar by SSS similarity criterion.
\( \implies \Delta LMN \sim \Delta PQR \)
For similar triangles, their corresponding angles are equal.
Since \( \angle A \) corresponds to \( \angle B \), we have \( \angle A = \angle B \).
In simple words: To show that angles A and B are equal when their cosines are equal, we use two right triangles. We set up the side ratios based on the given cosine equality. Using the Pythagorean theorem, we find that all corresponding sides of the two triangles are proportional. This means the triangles are similar. Because similar triangles have equal corresponding angles, it proves that angle A is equal to angle B.

Exam Tip: This is a standard proof. Remember to consider two separate right-angled triangles to establish similarity, which then proves the equality of angles.

Question 7. If \( \cot \theta = \frac{7}{8} \) evaluate:
(i) \( \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} \)
(ii) \( \cot^2 \theta \)
Answer: Given \( \cot \theta = \frac{7}{8} \). We know that \( \cot \theta = \frac{\text{Base}}{\text{Perpendicular}} \).
Let Base (AB) \( = 7k \) and Perpendicular (BC) \( = 8k \) for some positive number k.
In right \( \Delta ABC \) (right angled at B):
By Pythagorean theorem:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (7k)^2 + (8k)^2 \)
\( AC^2 = 49k^2 + 64k^2 \)
\( AC^2 = 113k^2 \)
\( AC = \sqrt{113k^2} = \sqrt{113}k \)

Now, we calculate \( \sin \theta \) and \( \cos \theta \):
\( \sin \theta = \frac{BC}{AC} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}} \)
\( \cos \theta = \frac{AB}{AC} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}} \)
(i) Evaluate \( \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} \):
Using the identity \( (a+b)(a-b) = a^2 - b^2 \):
\( \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} = \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} \)
Substitute the values of \( \sin \theta \) and \( \cos \theta \):
\( = \frac{1 - \left(\frac{8}{\sqrt{113}}\right)^2}{1 - \left(\frac{7}{\sqrt{113}}\right)^2} \)
\( = \frac{1 - \frac{64}{113}}{1 - \frac{49}{113}} \)
\( = \frac{\frac{113-64}{113}}{\frac{113-49}{113}} \)
\( = \frac{49/113}{64/113} = \frac{49}{64} \)
Alternatively, since \( \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta \), this simplifies directly to part (ii).
(ii) Evaluate \( \cot^2 \theta \):
\( \cot^2 \theta = (\cot \theta)^2 = \left(\frac{7}{8}\right)^2 = \frac{49}{64} \)
In simple words: Given the cotangent of angle theta, we first determine the lengths of the sides of the right-angled triangle. Using the Pythagorean theorem, we calculate the hypotenuse. Then, we find the sine and cosine values for theta. For part (i), we simplify the given expression using an algebraic identity, then substitute the values. For part (ii), we directly square the provided value of cot theta.

Exam Tip: Recognize the identities \( (1+\sin \theta)(1-\sin \theta) = 1-\sin^2 \theta = \cos^2 \theta \) and \( (1+\cos \theta)(1-\cos \theta) = 1-\cos^2 \theta = \sin^2 \theta \). This will quickly simplify part (i) to \( \cot^2 \theta \).

Question 8. If \( 3 \cot A = 4 \), check whether \( \frac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A \) or not.
Answer: Given \( 3 \cot A = 4 \).
So, \( \cot A = \frac{4}{3} \). We know that \( \cot A = \frac{\text{Base}}{\text{Perpendicular}} \).
Let Base (AB) \( = 4k \) and Perpendicular (BC) \( = 3k \) for some positive number k.
In right \( \Delta ABC \) (right angled at B):
By Pythagorean theorem:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (4k)^2 + (3k)^2 \)
\( AC^2 = 16k^2 + 9k^2 \)
\( AC^2 = 25k^2 \)
\( AC = \sqrt{25k^2} = 5k \)

Now, we calculate \( \tan A \), \( \sin A \), and \( \cos A \):
\( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{3k}{4k} = \frac{3}{4} \)
\( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5} \)
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5} \)
Evaluate the Left Hand Side (LHS):
\( LHS = \frac{1-\tan^2 A}{1+\tan^2 A} = \frac{1 - \left(\frac{3}{4}\right)^2}{1 + \left(\frac{3}{4}\right)^2} \)
\( = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} \)
\( = \frac{\frac{16-9}{16}}{\frac{16+9}{16}} \)
\( = \frac{7/16}{25/16} = \frac{7}{25} \)
Evaluate the Right Hand Side (RHS):
\( RHS = \cos^2 A - \sin^2 A = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 \)
\( = \frac{16}{25} - \frac{9}{25} = \frac{16-9}{25} = \frac{7}{25} \)
Since LHS = RHS, the given statement is true.
In simple words: Given the value of cot A, we first find the individual trigonometric ratios tan A, sin A, and cos A using a right triangle and the Pythagorean theorem. Then, we evaluate both sides of the given equation by substituting these values. By comparing the results, we confirm that the left side equals the right side, meaning the statement is true.

Exam Tip: Be careful with calculations involving squares of fractions. Ensure you find a common denominator correctly when adding or subtracting fractions.

Question 9. In \( \Delta ABC \), right angled at B, if \( \tan A = \frac{1}{\sqrt{3}} \), find the value of
(i) \( \sin A \cos C + \cos A \sin C \)
(ii) \( \cos A \cos C – \sin A \sin C \)
Answer: Given \( \tan A = \frac{1}{\sqrt{3}} \). We know that \( \tan A = \frac{\text{Perpendicular}}{\text{Base}} \).
So, let Perpendicular (BC) \( = k \) and Base (AB) \( = \sqrt{3}k \) for some positive number k.
In right \( \Delta ABC \) (right angled at B):
By Pythagorean theorem:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (\sqrt{3}k)^2 + (k)^2 \)
\( AC^2 = 3k^2 + k^2 \)
\( AC^2 = 4k^2 \)
\( AC = \sqrt{4k^2} = 2k \)

Now, we calculate \( \sin A, \cos A, \sin C \), and \( \cos C \):
\( \sin A = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2} \)
\( \cos A = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \)
\( \sin C = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \)
\( \cos C = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2} \)
(i) Evaluate \( \sin A \cos C + \cos A \sin C \):
\( = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right) \)
\( = \frac{1}{4} + \frac{3}{4} \)
\( = \frac{1+3}{4} = \frac{4}{4} = 1 \)
(ii) Evaluate \( \cos A \cos C – \sin A \sin C \):
\( = \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{1}{2}\right) - \left(\frac{1}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right) \)
\( = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 \)
In simple words: Given the tangent of angle A in a right-angled triangle, we first find the lengths of all sides using the Pythagorean theorem. Then, we calculate the sine and cosine for both angle A and angle C. For part (i), we substitute these values into the expression, which simplifies to 1. For part (ii), we substitute the values into the second expression, which simplifies to 0. These are also related to trigonometric identities for sum of angles.

Exam Tip: Recognize that (i) is the identity for \( \sin(A+C) \) and (ii) is for \( \cos(A+C) \). In a right-angled triangle, if B is \( 90^\circ \), then \( A+C = 90^\circ \). So, \( \sin(A+C) = \sin(90^\circ) = 1 \) and \( \cos(A+C) = \cos(90^\circ) = 0 \).

Question 10. In \( \Delta PQR \), right-angled at Q, \( PR + QR = 25 \) cm and \( PQ = 5 \) cm. Determine the value of \( \sin P \), \( \cos P \) and \( \tan P \).
Answer: We are given \( \Delta PQR \) is right-angled at Q.
Given: \( PR + QR = 25 \) cm
\( \implies PR = 25 - QR \) (1)
Also, given \( PQ = 5 \) cm.
By Pythagorean theorem:
\( PR^2 = PQ^2 + QR^2 \)
Substitute \( PR \) from equation (1):
\( (25 - QR)^2 = 5^2 + QR^2 \)
\( 625 - 50QR + QR^2 = 25 + QR^2 \)
Subtract \( QR^2 \) from both sides:
\( 625 - 50QR = 25 \)
\( 50QR = 625 - 25 \)
\( 50QR = 600 \)
\( QR = \frac{600}{50} \)
\( QR = 12 \) cm
Now substitute \( QR = 12 \) cm into equation (1) to find \( PR \):
\( PR = 25 - 12 \)
\( PR = 13 \) cm

With all three side lengths (PQ = 5 cm, QR = 12 cm, PR = 13 cm), we can now determine \( \sin P, \cos P \), and \( \tan P \):
\( \sin P = \frac{\text{Opposite side to P}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{12}{13} \)
\( \cos P = \frac{\text{Adjacent side to P}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{5}{13} \)
\( \tan P = \frac{\text{Opposite side to P}}{\text{Adjacent side to P}} = \frac{QR}{PQ} = \frac{12}{5} \)
In simple words: We are given a right triangle with one side length and the sum of the other two sides. We use substitution along with the Pythagorean theorem to find the lengths of the unknown sides. Once all three side lengths are known, we calculate the sine, cosine, and tangent of angle P by using the definitions of these ratios.

Exam Tip: When given the sum or difference of two sides, express one side in terms of the other. This lets you use the Pythagorean theorem to find the exact lengths.

Question 11. State whether the following are true or false. Justify your answer.
(i) The value of \( \tan A \) is always less than 1
(ii) \( \sec A = \frac{12}{5} \) for some value of A
(iii) \( \cos A \) is the abbreviation used for the cosecant of angle A
(iv) \( \cot A \) is the product of cot and A
(v) \( \sin \theta = \frac{4}{3} \) for some angle.
Answer:
(i) False.
The value of the perpendicular (opposite side) can be greater than the base (adjacent side) in a right triangle, making \( \tan A \) greater than 1. For example, in a right triangle, if the opposite side is 3 and the adjacent side is 1, then \( \tan A = 3 \), which is not less than 1. (e.g., \( \tan 60^\circ = \sqrt{3} \approx 1.732 > 1 \)).
In simple words: False. The tangent of an angle can be larger than 1 if the side opposite the angle is longer than the side next to it.
(ii) True.
\( \sec A \) is defined as the ratio of the Hypotenuse to the Base (adjacent side). The hypotenuse is always the longest side in a right-angled triangle, meaning it is always larger than the adjacent side. Therefore, the value of \( \sec A \) is always greater than 1 (or less than -1). Since \( \frac{12}{5} = 2.4 \), which is greater than 1, it is a possible value for \( \sec A \).
In simple words: True. Secant is hypotenuse divided by the adjacent side. The hypotenuse is always the longest side in a right triangle, so the secant value will always be more than 1 (or less than -1), making 12/5 a valid ratio.
(iii) False.
\( \cos A \) is the short form for the cosine of angle A. The abbreviation for cosecant is \( \text{cosec } A \) or \( \text{csc } A \).
In simple words: False. Cos A means cosine, not cosecant. Cosecant has its own abbreviation, cosec A.
(iv) False.
\( \cot A \) represents a single trigonometric ratio, the cotangent of angle A. \( \cot \) on its own is not a standalone mathematical operation and has no meaning without an angle. It is not a product.
In simple words: False. Cot A is a single mathematical term for the cotangent of angle A, not 'cot' multiplied by 'A'. 'Cot' alone does not have meaning.
(v) False.
The sine function's value (\( \sin \theta \)) always ranges between -1 and 1, inclusive. Since \( \frac{4}{3} \) is approximately 1.33, which is greater than 1, it is not a possible value for \( \sin \theta \).
In simple words: False. The sine of any angle can only be between -1 and 1. Since 4/3 is larger than 1, it cannot be a valid sine value for any angle.

Exam Tip: For True/False questions related to trigonometric ratios, remember the ranges of sine, cosine, secant, cosecant, and tangent. Also, know the correct abbreviations for each ratio.