Q. If angle $\theta\$be divided into two parts such that the tangent of one part is $k$ times the tangent of the other and $\phi$ is their difference, then sin $\theta$ is equal to
A) $\frac{k + 1}{k - 1}\sin\phi$
B) $\frac{k - 1}{k + 1}\sin\phi$
C) $\frac{2k - 1}{2k + 1}\sin\phi$
D) None of these
Explanation
(a)
Let $A + B = \theta\ and\ A - B = \phi$
Then,$\tan{A = k\tan{B\ \Rightarrow \frac{\tan A}{\tan B} = \frac{k}{1}}}$
$\Rightarrow \ \frac{k}{1} = \frac{\sin{A\cos B}}{\cos{A\sin B}}$
Applying componendo and dividend rule, we get
$\Rightarrow \frac{k + 1}{k - 1} = \frac{\sin{A\cos{B + \cos{A\sin B}}}}{\sin{A\cos{B - \cos{A\sin B}}}}$
$= \frac{\sin{(\ A + B)}}{\sin{(A - B)}} = \frac{\sin\theta}{\sin\phi}$
$\Rightarrow \sin{\theta = \frac{k + 1}{k - 1}\sin\phi}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.