Question

If an angle $\alpha$ is divided into two parts $A$ and $B$ such that $A-B=x$ and $\tan A:\tan B=K:1,$ then the value of $sinx$ is

• A. $\frac{K+1}{K-1}\sin \alpha$
• B. $\frac{K}{K+1}\sin \alpha$
• C. $\frac{K-1}{K+1}\sin \alpha$
• D. none

Answer 1

Answer: (C)

$\frac{K-1}{K+1}\sin \alpha$

$A+B=\alpha$

$A-B=x$

$\frac{\tan A}{\tan B}=\frac{K}{1}$

$\frac{\sin A}{\cos A}\frac{\cos B}{\sin B}=\frac{K}{1}$

$\frac{\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B}=\frac{K+1}{K-1}$

$\frac{\sin (A+B)}{\sin (A-B)}=\frac{K+1}{K-1}$

$\frac{\sin \alpha }{\sin x}\frac{K+1}{K-1}$

$\sin x=\frac{K-1}{K+1}\sin \alpha$.