CBSE Class 11 Mathematics Trigonometric Functions Worksheet Set 08

CBSE Class 11 Mathematics Worksheet - Trigonometric Functions (8). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. If \( \tan(\pi + \theta) = n \tan(\pi - \theta) \). Show that \( (n + 1) \sin(2\theta) = (n - 1) \sin(2\pi) \)?
Answer: We have, \( \tan(\pi + \theta) = n \tan(\pi - \theta) \)
\( \implies \frac{\tan(\pi+\theta)}{\tan(\pi-\theta)} = \frac{n}{1} \)
\( \implies \frac{\tan(\pi+\theta)+\tan(\pi-\theta)}{\tan(\pi+\theta)-\tan(\pi-\theta)} = \frac{n+1}{n-1} \) ............ apply componendo & dividendo
\( \implies \frac{\frac{\sin(\pi+\theta)}{\cos(\pi+\theta)} + \frac{\sin(\pi-\theta)}{\cos(\pi-\theta)}}{\frac{\sin(\pi+\theta)}{\cos(\pi+\theta)} - \frac{\sin(\pi-\theta)}{\cos(\pi-\theta)}} = \frac{n+1}{n-1} \)
\( \implies \frac{\sin(\pi+\theta)\cos(\pi-\theta) + \sin(\pi-\theta)\cos(\pi+\theta)}{\sin(\pi+\theta)\cos(\pi-\theta) - \cos(\pi+\theta)\sin(\pi-\theta)} = \frac{n+1}{n-1} \)
\( \implies \frac{\sin(\pi+\theta + \pi-\theta)}{\sin(\pi+\theta - \pi+\theta)} = \frac{n+1}{n-1} \)
\( \implies \frac{\sin(2\pi)}{\sin(2\theta)} = \frac{n+1}{n-1} \)
\( \implies \sin(2\pi) \cdot (n - 1) = \sin(2\theta) \cdot (n + 1) \) (proved)

Question. If \( \cos(\alpha - \beta) + \cos(\beta - \gamma) + \cos(\gamma - \alpha) = -\frac{3}{2} \), show that \( \sin \alpha + \sin \beta + \sin \gamma = 0 \) and \( \cos \alpha + \cos \beta + \cos \gamma = 0 \)?
Answer: We have, \( \cos(\alpha - \beta) + \cos(\beta - \gamma) + \cos(\gamma - \alpha) = -\frac{3}{2} \)
\( \implies \cos \alpha \cos \beta + \sin \alpha \sin \beta + \cos \beta \cos \gamma + \sin \beta \sin \gamma + \cos \gamma \cos \alpha + \sin \gamma \sin \alpha = -\frac{3}{2} \)
\( \implies 2\cos \alpha \cos \beta + 2\sin \alpha \sin \beta + 2\cos \beta \cos \gamma + 2\sin \beta \sin \gamma + 2\cos \gamma \cos \alpha + 2\sin \gamma \sin \alpha + 1 + 1 + 1 = 0 \)
\( \implies 2\cos \alpha \cos \beta + 2\sin \alpha \sin \beta + 2\cos \beta \cos \gamma + 2\sin \beta \sin \gamma + 2\cos \gamma \cos \alpha + 2\sin \gamma \sin \alpha + \sin^2 \alpha + \cos^2 \alpha + \sin^2 \beta + \cos^2 \beta + \sin^2 \gamma + \cos^2 \gamma = 0 \)
\( \implies (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + 2\sin \alpha \sin \beta + 2\sin \beta \sin \gamma + 2\sin \gamma \sin \alpha) + (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2\cos \alpha \cos \beta + 2\cos \beta \cos \gamma + 2\cos \gamma \cos \alpha) = 0 \)
\( \implies (\sin \alpha + \sin \beta + \sin \gamma)^2 + (\cos \alpha + \cos \beta + \cos \gamma)^2 = 0 \)
This is possible only when \( \sin \alpha + \sin \beta + \sin \gamma = 0 \) and \( \cos \alpha + \cos \beta + \cos \gamma = 0 \) ans.

Question. If \( \alpha \) & \( \beta \) are the solutions of the equation \( a \tan \theta + b \sec \theta = c \), then show that \( \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2} \)?
Answer: We have, \( a \tan \theta + b \sec \theta = c \)      .........(1)
\( \implies c - a \tan \theta = b \sec \theta \)
Squaring both sides
\( \implies c^2 + a^2 \tan^2 \theta - 2ac \tan \theta = b^2 \sec^2 \theta \)
\( \implies c^2 + a^2 \tan^2 \theta - 2ac \tan \theta = b^2 (1 + \tan^2 \theta) \)
\( \implies c^2 + a^2 \tan^2 \theta - 2ac \tan \theta = b^2 + b^2 \tan^2 \theta \)
\( \implies a^2 \tan^2 \theta - b^2 \tan^2 \theta - 2ac \tan \theta + c^2 - b^2 = 0 \)
\( \implies (a^2 - b^2) \tan^2 \theta - 2ac \tan \theta + (c^2 - b^2) = 0 \)      .........(2)
We are given that \( \alpha \) & \( \beta \) are the solutions of the equation (1) \( \therefore \tan \alpha \) & \( \tan \beta \) are the roots of equation (2)
Now, sum of roots = \( \tan \alpha + \tan \beta = \frac{2ac}{a^2 - b^2} \) and
product of roots = \( \tan \alpha \tan \beta = \frac{c^2 - b^2}{a^2 - b^2} \)
Now, \( \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \)
\( = \frac{\frac{2ac}{a^2 - b^2}}{1 - \frac{c^2 - b^2}{a^2 - b^2}} \)
\( \implies \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2} \) (proved)

Question. If \( \alpha \) & \( \beta \) are the solutions of the equation \( a \cos \theta + b \sin \theta = c \), then show that
1. \( \cos(\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2} \)
2. \( \cos(\alpha - \beta) = \frac{2c^2 - (a^2 + b^2)}{a^2 + b^2} \)
Answer: We have, \( a \cos \theta + b \sin \theta = c \)      ............. (1)
\( \implies c - a \cos \theta = b \sin \theta \)
Squaring both sides
\( \implies (c - a \cos \theta)^2 = b^2 \sin^2 \theta \)
\( \implies c^2 + a^2 \cos^2 \theta - 2ac \cos \theta = b^2 (1 - \cos^2 \theta) \)
\( \implies c^2 + a^2 \cos^2 \theta - 2ac \cos \theta = b^2 - b^2 \cos^2 \theta \)
\( \implies (a^2 + b^2) \cos^2 \theta - 2ac \cos \theta + (c^2 - b^2) = 0 \)      ............. (2)
We are given that \( \alpha \) & \( \beta \) are the solutions of the equation (1) \( \therefore \cos \alpha \) & \( \cos \beta \) are the roots of equation (2)
\( \therefore \) product of roots = \( \cos \alpha \cos \beta = \frac{c^2 - b^2}{a^2 + b^2} \)
Consider again, \( a \cos \theta + b \sin \theta = c \)
\( \implies c - b \sin \theta = a \cos \theta \)
Squaring,
\( \implies (c - b \sin \theta)^2 = a^2 \cos^2 \theta \)
\( \implies c^2 + b^2 \sin^2 \theta - 2bc \sin \theta = a^2 (1 - \sin^2 \theta) \)
\( \implies (a^2 + b^2) \sin^2 \theta - 2bc \sin \theta + (c^2 - a^2) = 0 \)      ............. (3)
\( \sin \alpha \) and \( \sin \beta \) are the solutions/roots of equation (3)
\( \therefore \) product of roots = \( \sin \alpha \sin \beta = \frac{c^2 - a^2}{a^2 + b^2} \)
Now, \( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)
\( = \frac{c^2 - b^2}{a^2 + b^2} - \frac{c^2 - a^2}{a^2 + b^2} \)
\( \implies \cos(\alpha + \beta) = \frac{c^2 - b^2 - c^2 + a^2}{a^2 + b^2} = \frac{a^2 - b^2}{a^2 + b^2} \) (proved) and
\( \implies \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \)
\( = \frac{c^2 - b^2}{a^2 + b^2} + \frac{c^2 - a^2}{a^2 + b^2} \)
\( = \frac{2c^2 - (a^2 + b^2)}{a^2 + b^2} \) (proved)

Question. If \( 3 \tan(\theta - 15^\circ) = \tan(\theta + 15^\circ) \). Find the value of \( \theta \)?
Answer: We have, \( 3 \tan(\theta - 15^\circ) = \tan(\theta + 15^\circ) \)
\( \implies \frac{\tan(\theta + 15^\circ)}{\tan(\theta - 15^\circ)} = \frac{3}{1} \)
Apply componendo & dividendo \( \left(\frac{n+d}{n-d}\right) \)
\( \implies \frac{\tan(\theta + 15^\circ) + \tan(\theta - 15^\circ)}{\tan(\theta + 15^\circ) - \tan(\theta - 15^\circ)} = \frac{3+1}{3-1} = 2 \)
Converting into \( \sin \theta \) & \( \cos \theta \), we get (after taking L.C.M)
\( \implies \frac{\sin(\theta + 15^\circ)\cos(\theta - 15^\circ) + \sin(\theta - 15^\circ)\cos(\theta + 15^\circ)}{\sin(\theta + 15^\circ)\cos(\theta - 15^\circ) - \sin(\theta - 15^\circ)\cos(\theta + 15^\circ)} = 2 \)
\( \implies \frac{\sin(\theta + 15^\circ + \theta - 15^\circ)}{\sin(\theta + 15^\circ - \theta + 15^\circ)} = 2 \)      ............ \( \{ \sin a \cos b + \cos a \sin b = \sin(a+b) \} \)
\( \implies \frac{\sin(2\theta)}{\sin(30^\circ)} = 2 \)
\( \implies \sin(2\theta) = 2 \sin(30^\circ) \)
\( \implies \sin(2\theta) = 1 \)
\( \implies \sin(2\theta) = \sin\left(\frac{\pi}{2}\right) \)
\( \implies 2\theta = \frac{\pi}{2} \)
\( \implies \theta = \frac{\pi}{4} \) ans.