OP Malhotra Class 11 Maths Solutions Chapter 4 Trigonometrical Functions Chapter Test

Question 1. If \( \cot A = \frac{3}{4} \), find the value of \( 3 \cos A + 5 \sin A \), where A lies in the first quadrant.
Answer: Given that \( \cot A = \frac{3}{4} \).
We know the identity: \( \csc^2 A = 1 + \cot^2 A \).
So, \( \csc A = \pm \sqrt{1+\cot^2 A} \)
\( = \pm \sqrt{1+\left(\frac{3}{4}\right)^2} \)
\( = \pm \sqrt{1+\frac{9}{16}} \)
\( = \pm \sqrt{\frac{16+9}{16}} \)
\( = \pm \sqrt{\frac{25}{16}} \)
\( = \pm \frac{5}{4} \)
Since angle A is in the first quadrant, all trigonometric ratios are positive. Therefore, \( \csc A = \frac{5}{4} \).
We know that \( \sin A = \frac{1}{\csc A} \), so \( \sin A = \frac{1}{\frac{5}{4}} = \frac{4}{5} \).
Now, we can find \( \cos A \) using the relation \( \cot A = \frac{\cos A}{\sin A} \).
So, \( \cos A = \cot A \cdot \sin A \)
\( = \frac{3}{4} \times \frac{4}{5} \)
\( = \frac{3}{5} \)
Finally, we need to find the value of \( 3 \cos A + 5 \sin A \):
\( 3 \cos A + 5 \sin A = 3\left(\frac{3}{5}\right) + 5\left(\frac{4}{5}\right) \)
\( = \frac{9}{5} + 4 \)
\( = \frac{9}{5} + \frac{20}{5} \)
\( = \frac{29}{5} \)
In simple words: First, we use the given cotangent value to find cosecant, remembering that all values are positive in the first quadrant. Then, we use cosecant to find sine, and cotangent with sine to find cosine. Finally, we plug these values into the expression to get the answer.

🎯 Exam Tip: When given a trigonometric ratio and a quadrant, always determine the signs of the other ratios carefully before calculating their values. Identities like \( \csc^2 A = 1 + \cot^2 A \) and \( \sin A = \frac{1}{\csc A} \) are fundamental.

Question 2. If \( \cos 120^\circ = -\frac{1}{2} \) find the values of \( \sin 120^\circ \) and \( \tan 120^\circ \).
Answer: Given \( \cos 120^\circ = -\frac{1}{2} \).
The angle \( 120^\circ \) lies in the second quadrant.
In the second quadrant, \( \sin \theta \) is positive, and \( \tan \theta \) is negative.
We use the trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
So, \( \sin^2 120^\circ = 1 - \cos^2 120^\circ \)
\( \sin 120^\circ = \pm \sqrt{1 - \left(-\frac{1}{2}\right)^2} \)
\( = \pm \sqrt{1 - \frac{1}{4}} \)
\( = \pm \sqrt{\frac{4-1}{4}} \)
\( = \pm \sqrt{\frac{3}{4}} \)
\( = \pm \frac{\sqrt{3}}{2} \)
Since \( 120^\circ \) is in the second quadrant, \( \sin 120^\circ \) must be positive.
Therefore, \( \sin 120^\circ = \frac{\sqrt{3}}{2} \).
Next, we find \( \tan 120^\circ \) using the relation \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
\( \tan 120^\circ = \frac{\sin 120^\circ}{\cos 120^\circ} \)
\( = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} \)
\( = -\frac{\sqrt{3}}{2} \times \frac{2}{1} \)
\( = -\sqrt{3} \)
In simple words: First, we use the cosine value and the fact that \( \sin^2 \theta + \cos^2 \theta = 1 \) to find the sine value. Since \( 120^\circ \) is in the second quadrant, the sine will be positive. Then, we divide the sine by the cosine to get the tangent value.

🎯 Exam Tip: Always remember the signs of trigonometric functions in each quadrant. This helps correctly determine the values when taking square roots, like for sine or cosine.

Question 3. Prove that \( \sec (-1680^\circ) \cdot \sin 330^\circ = 1 \).
Answer: We need to prove that the Left Hand Side (L.H.S.) equals 1.
L.H.S. \( = \sec (-1680^\circ) \sin 330^\circ \)
First, let's simplify \( \sec (-1680^\circ) \). We know \( \sec (-\theta) = \sec \theta \).
\( \sec (-1680^\circ) = \sec (1680^\circ) \)
To reduce the angle \( 1680^\circ \), we divide it by \( 360^\circ \): \( 1680 \div 360 = 4 \) with a remainder of \( 240 \).
So, \( 1680^\circ = 4 \times 360^\circ + 240^\circ \).
Therefore, \( \sec (1680^\circ) = \sec (240^\circ) \) (since \( \sec (n \cdot 360^\circ + \theta) = \sec \theta \)).
Now, \( 240^\circ \) is in the third quadrant. We can write \( 240^\circ \) as \( 180^\circ + 60^\circ \).
\( \sec (240^\circ) = \sec (180^\circ + 60^\circ) \).
In the third quadrant, \( \sec \theta \) is negative. So, \( \sec (180^\circ + 60^\circ) = -\sec 60^\circ \).
We know \( \sec 60^\circ = 2 \). Thus, \( \sec (240^\circ) = -2 \).
Next, let's simplify \( \sin 330^\circ \).
The angle \( 330^\circ \) is in the fourth quadrant. We can write \( 330^\circ \) as \( 360^\circ - 30^\circ \).
\( \sin 330^\circ = \sin (360^\circ - 30^\circ) \).
In the fourth quadrant, \( \sin \theta \) is negative. So, \( \sin (360^\circ - 30^\circ) = -\sin 30^\circ \).
We know \( \sin 30^\circ = \frac{1}{2} \). Thus, \( \sin 330^\circ = -\frac{1}{2} \).
Now, substitute these simplified values back into the L.H.S. expression:
L.H.S. \( = (-2) \times \left(-\frac{1}{2}\right) \)
\( = 1 \)
This is equal to the Right Hand Side (R.H.S.).
In simple words: We break down each trigonometric term. For secant, we first handle the negative angle, then reduce it to a smaller angle using \( 360^\circ \) cycles. For sine, we reduce the angle using \( 360^\circ \). After finding the simple values for each, we multiply them to show the result is 1.

🎯 Exam Tip: When simplifying trigonometric expressions with large or negative angles, always use periodicity and quadrant rules to reduce them to standard angles between \( 0^\circ \) and \( 90^\circ \).

Question 4. If A, B, C and D are the angles of a cyclic quadrilateral, show that \( \cos A + \cos B + \cos C + \cos D = 0 \).
Answer: For a cyclic quadrilateral, the sum of opposite angles is \( 180^\circ \) or \( \pi \) radians.
Therefore, we have two main properties:
1. \( A + C = 180^\circ \implies C = 180^\circ - A \)
2. \( B + D = 180^\circ \implies D = 180^\circ - B \)
Now, let's substitute these into the expression we need to prove:
L.H.S. \( = \cos A + \cos B + \cos C + \cos D \)
\( = \cos A + \cos B + \cos (180^\circ - A) + \cos (180^\circ - B) \)
We know the identity \( \cos (180^\circ - \theta) = -\cos \theta \).
Using this identity for the third and fourth terms:
\( \cos (180^\circ - A) = -\cos A \)
\( \cos (180^\circ - B) = -\cos B \)
Substitute these back into the L.H.S.:
L.H.S. \( = \cos A + \cos B + (-\cos A) + (-\cos B) \)
\( = \cos A + \cos B - \cos A - \cos B \)
\( = 0 \)
This equals the Right Hand Side (R.H.S.).
In simple words: For a cyclic quadrilateral, opposite angles add up to \( 180^\circ \). This means we can replace C and D with \( (180^\circ - A) \) and \( (180^\circ - B) \). Since \( \cos (180^\circ - \theta) \) is \( -\cos \theta \), the terms will cancel each other out, giving zero.

🎯 Exam Tip: Remember the fundamental property of cyclic quadrilaterals: opposite angles sum to \( 180^\circ \). This identity is key to simplifying expressions involving their angles.

Question 5. If \( \tan 25^\circ = a \), prove that \( \frac{\tan 155^\circ - \tan 115^\circ}{1 + \tan 155^\circ \cdot \tan 115^\circ} = \frac{1-a^2}{2a} \).
Answer: Given \( \tan 25^\circ = a \).
We need to prove that L.H.S. \( = \frac{1-a^2}{2a} \).
Let's simplify the terms in the L.H.S.:
\( \tan 155^\circ = \tan (180^\circ - 25^\circ) \)
We know that \( \tan (180^\circ - \theta) = -\tan \theta \).
So, \( \tan 155^\circ = -\tan 25^\circ = -a \).
Next, \( \tan 115^\circ = \tan (90^\circ + 25^\circ) \)
We know that \( \tan (90^\circ + \theta) = -\cot \theta \).
So, \( \tan 115^\circ = -\cot 25^\circ \).
Since \( \cot 25^\circ = \frac{1}{\tan 25^\circ} = \frac{1}{a} \), we have \( \tan 115^\circ = -\frac{1}{a} \).
Now, substitute these values into the L.H.S. expression:
L.H.S. \( = \frac{\tan 155^\circ - \tan 115^\circ}{1 + \tan 155^\circ \cdot \tan 115^\circ} \)
\( = \frac{(-a) - \left(-\frac{1}{a}\right)}{1 + (-a) \cdot \left(-\frac{1}{a}\right)} \)
\( = \frac{-a + \frac{1}{a}}{1 + 1} \)
\( = \frac{\frac{-a^2+1}{a}}{2} \)
\( = \frac{1-a^2}{2a} \)
This equals the Right Hand Side (R.H.S.).
In simple words: We first rewrite \( \tan 155^\circ \) and \( \tan 115^\circ \) using \( 180^\circ - \theta \) and \( 90^\circ + \theta \) identities, replacing them with expressions involving \( \tan 25^\circ \) or \( \cot 25^\circ \). Then, we substitute \( a \) for \( \tan 25^\circ \) and simplify the whole fraction to get the required result.

🎯 Exam Tip: Mastering angle reduction formulas (like \( \tan(180^\circ - \theta) \) and \( \tan(90^\circ + \theta) \)) is crucial for simplifying complex trigonometric expressions effectively.

Question 6. If A, B, C be the angles of a triangle, prove that \( \frac{\sin (B+C)+\sin (C+A)+\sin (A+B)}{\sin (\pi+A)+\sin (3 \pi+B)+\sin (5 \pi+C)} = 1 \).
Answer: Given that A, B, C are the angles of a triangle.
This means their sum is \( 180^\circ \) or \( \pi \) radians: \( A + B + C = \pi \).
From this, we can write:
\( B+C = \pi - A \)
\( C+A = \pi - B \)
\( A+B = \pi - C \)
Now, let's simplify the numerator of the given expression:
Numerator \( = \sin (B+C) + \sin (C+A) + \sin (A+B) \)
\( = \sin (\pi - A) + \sin (\pi - B) + \sin (\pi - C) \)
Using the identity \( \sin (\pi - \theta) = \sin \theta \):
Numerator \( = \sin A + \sin B + \sin C \).
Next, let's simplify the denominator of the given expression:
Denominator \( = \sin (\pi+A) + \sin (3\pi+B) + \sin (5\pi+C) \)
For each term, we use the identity \( \sin (n\pi + \theta) = (-1)^n \sin \theta \).
For \( \sin (\pi+A) \): here \( n=1 \), so \( \sin (\pi+A) = (-1)^1 \sin A = -\sin A \).
For \( \sin (3\pi+B) \): here \( n=3 \), so \( \sin (3\pi+B) = (-1)^3 \sin B = -\sin B \).
For \( \sin (5\pi+C) \): here \( n=5 \), so \( \sin (5\pi+C) = (-1)^5 \sin C = -\sin C \).
So, Denominator \( = (-\sin A) + (-\sin B) + (-\sin C) \)
\( = -(\sin A + \sin B + \sin C) \).
Now, put the simplified numerator and denominator back into the original expression:
L.H.S. \( = \frac{\sin A + \sin B + \sin C}{-(\sin A + \sin B + \sin C)} \)
\( = -1 \)
In simple words: Since A, B, C are angles of a triangle, their sum is \( 180^\circ \). We use this to rewrite the terms in the numerator. For the denominator, we apply a rule about \( \sin(n\pi + \theta) \). After simplifying both the top and bottom parts, we find that the expression equals -1.

🎯 Exam Tip: Always remember that for angles A, B, C in a triangle, \( A+B+C = \pi \) radians. Also, recall the general identity for sine with multiples of \( \pi \): \( \sin (n\pi + \theta) = (-1)^n \sin \theta \).

Question 7. Prove that \( \frac{\cos \theta}{1-\sin \theta}+\frac{1-\sin \theta}{\cos \theta} = 2 \sec \theta \).
Answer: We will simplify the Left Hand Side (L.H.S.) of the equation.
L.H.S. \( = \frac{\cos \theta}{1-\sin \theta}+\frac{1-\sin \theta}{\cos \theta} \)
To add these fractions, we find a common denominator, which is \( (1-\sin \theta)\cos \theta \).
\( = \frac{\cos \theta \cdot \cos \theta + (1-\sin \theta) \cdot (1-\sin \theta)}{(1-\sin \theta)\cos \theta} \)
\( = \frac{\cos^2 \theta + (1-\sin \theta)^2}{(1-\sin \theta)\cos \theta} \)
Now, expand \( (1-\sin \theta)^2 \). It becomes \( 1^2 - 2(1)(\sin \theta) + (\sin \theta)^2 = 1 - 2\sin \theta + \sin^2 \theta \).
\( = \frac{\cos^2 \theta + 1 - 2\sin \theta + \sin^2 \theta}{(1-\sin \theta)\cos \theta} \)
We know the fundamental trigonometric identity: \( \sin^2 \theta + \cos^2 \theta = 1 \).
Substitute this into the numerator:
\( = \frac{1 + 1 - 2\sin \theta}{(1-\sin \theta)\cos \theta} \)
\( = \frac{2 - 2\sin \theta}{(1-\sin \theta)\cos \theta} \)
Factor out 2 from the numerator:
\( = \frac{2(1 - \sin \theta)}{(1-\sin \theta)\cos \theta} \)
Assuming \( 1-\sin \theta \neq 0 \) (i.e., \( \sin \theta \neq 1 \)), we can cancel the term \( (1 - \sin \theta) \) from both the numerator and the denominator.
\( = \frac{2}{\cos \theta} \)
We know that \( \frac{1}{\cos \theta} = \sec \theta \).
So, \( = 2 \sec \theta \)
This is equal to the Right Hand Side (R.H.S.).
In simple words: We combine the two fractions by finding a common bottom part. Then, we expand the squared term and use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). After simplifying, we factor out a 2 and cancel matching terms, which leaves us with \( 2/\cos \theta \), which is the same as \( 2 \sec \theta \).

🎯 Exam Tip: When proving trigonometric identities, look for opportunities to combine fractions, expand terms, and use fundamental identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) to simplify the expression towards the target.

Question 8. If \( \sec \theta = \sqrt{2} \) and \( \frac{3\pi}{2} < \theta < 2\pi \), find the value of \( \frac{1+\tan \theta+\csc \theta}{1+\cot \theta-\csc \theta} \).
Answer: Given \( \sec \theta = \sqrt{2} \) and that \( \theta \) lies in the interval \( \frac{3\pi}{2} < \theta < 2\pi \). This means \( \theta \) is in the fourth quadrant.
In the fourth quadrant, \( \cos \theta \) is positive, while \( \sin \theta \), \( \tan \theta \), \( \cot \theta \), and \( \csc \theta \) are all negative.
First, let's find the values of \( \tan \theta \), \( \csc \theta \), and \( \cot \theta \).
1. To find \( \tan \theta \):
We use the identity \( \tan^2 \theta = \sec^2 \theta - 1 \).
\( \tan \theta = \pm \sqrt{\sec^2 \theta - 1} \)
\( = \pm \sqrt{(\sqrt{2})^2 - 1} \)
\( = \pm \sqrt{2 - 1} \)
\( = \pm \sqrt{1} = \pm 1 \)
Since \( \theta \) is in the fourth quadrant, \( \tan \theta \) is negative. So, \( \tan \theta = -1 \).
2. To find \( \cos \theta \) and then \( \sin \theta \):
\( \cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{2}} \).
We know \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), so \( \sin \theta = \tan \theta \cdot \cos \theta \).
\( = (-1) \cdot \left(\frac{1}{\sqrt{2}}\right) \)
\( = -\frac{1}{\sqrt{2}} \).
3. To find \( \csc \theta \):
\( \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{1}{\sqrt{2}}} = -\sqrt{2} \).
4. To find \( \cot \theta \):
\( \cot \theta = \frac{1}{\tan \theta} = \frac{1}{-1} = -1 \).
Now, substitute these values into the given expression:
\( \frac{1+\tan \theta+\csc \theta}{1+\cot \theta-\csc \theta} \)
\( = \frac{1 + (-1) + (-\sqrt{2})}{1 + (-1) - (-\sqrt{2})} \)
\( = \frac{1 - 1 - \sqrt{2}}{1 - 1 + \sqrt{2}} \)
\( = \frac{-\sqrt{2}}{\sqrt{2}} \)
\( = -1 \)
In simple words: First, we determine the quadrant of the angle and the signs of the trigonometric functions in that quadrant. Then, using the given secant value and identities, we calculate the values of tangent, sine, cosecant, and cotangent. Finally, we substitute these calculated values into the expression and simplify to get the answer.

🎯 Exam Tip: Always pay close attention to the given quadrant for \( \theta \), as it dictates the positive or negative sign for each trigonometric ratio when derived from identities.