Question

If an angle $\theta$ is divided into 2 parts A and B such that $A-B=k$ and $A+B=\theta$ and tan $A:tanB=k:1$, then the value of sin k is :

• A. $\frac{k+1}{k-1}sin\theta$
• B. $\frac{k}{k+1}sin\theta$
• C. $\frac{k-1}{k+1}sin\theta$
• D. None of these

Answer 1

Answer: (C)

Given an angle $\theta$ which is divided into two parts A and B such that $A-B=k$ and $A+B=\theta$ ,
and tan A : tan B = k : 1, i.e. $\frac{tanA}{tanB}=\frac{k}{1}$
$\Rightarrow \frac{tanA+tanB}{tanA-tanB}=\frac{k+1}{k-1}$
(by componendo and dividendo)
$\Rightarrow \frac{sin\left(A+B\right)}{sin\left(A-B\right)}=\frac{k+1}{k-1}\Rightarrow \frac{sin\theta }{sink}=\frac{k+1}{k-1}$
$\Rightarrow sink=\frac{k+1}{k-1}sin\theta$