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
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 \[\Rightarrow \frac{k+1}{k-1}=\frac{\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B}\] \[=\frac{\sin (A+B)}{\sin (A-B)}=\frac{\sin \theta }{\sin \varphi }\Rightarrow \sin \theta =\frac{k+1}{k-1}\sin \varphi \].You need to login to perform this action.
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