Question

If an angle $\theta$ is divided into 2 parts A and B such that $A - B = k$ and $A + B = \theta$ and tan $A : tan \,B = k :\, 1$, 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)

$\frac{k-1}{k+1} sin\,\theta$

Answer 2

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. $\frac{tan\,A}{tan\,B} = \frac{k}{1}$ $\Rightarrow \frac{tan\,A+tan\,B}{tan\,A-tan\,B} = \frac{k+1}{k-1}$ (by componendo and dividendo) $\Rightarrow\quad \frac{sin \left(A + B\right)}{sin \left(A - B\right)} = \frac{k+1}{k-1} \quad\Rightarrow \frac{sin\,\theta}{sin\,k} = \frac{k+1}{k-1}$ $\Rightarrow \quad sin\,k = \frac{k+1}{k-1} sin\,\theta$