Trigonometric Identities Question 32
Question: If angle $ \theta $ be divided into two parts such that the tangent of one part is $ k $ times the tangent of the other and $ \varphi $ is their difference, then $ \sin \theta = $
Options:
A) $ \frac{k+1}{k-1}\sin \varphi $
B) $ \frac{k-1}{k+1}\sin \varphi $
C) $ \frac{2k-1}{2k+1}\sin \varphi $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ A+B=\theta $ and $ A-B=\varphi $ . Then $ \tan A=k\tan B $ or $ \frac{k}{1}=\frac{\tan A}{\tan B}=\frac{\sin A\cos B}{\cos A\sin B} $ Applying componendo and dividendo
$ \Rightarrow \frac{k+1}{k-1}=\frac{\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B} $ $ =\frac{\sin (A+B)}{\sin (A-B)}=\frac{\sin \theta }{\sin \varphi }\Rightarrow \sin \theta =\frac{k+1}{k-1}\sin \varphi $ .
BETA
