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Q.
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 :
Trigonometric Functions
Solution:
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 $