If \[ m\sin\theta=n\sin(\theta+2\alpha) \] prove that \[ \tan(\theta+\alpha)\cot\alpha = \frac{m+n}{m-n} \]
Solution
Given:
\[ m\sin\theta=n\sin(\theta+2\alpha) \]Use identity:
\[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]
\[
m\sin\theta
=
n[\sin\theta\cos2\alpha+\cos\theta\sin2\alpha]
\]
Bring terms containing \(\sin\theta\) together:
\[ (m-n\cos2\alpha)\sin\theta = n\sin2\alpha\cos\theta \]Divide by \(\cos\theta\):
\[ (m-n\cos2\alpha)\tan\theta = n\sin2\alpha \]Use identities:
\[ 1-\cos2\alpha=2\sin^2\alpha \] \[ \sin2\alpha=2\sin\alpha\cos\alpha \]
\[
\tan\theta
=
\frac{
2n\sin\alpha\cos\alpha
}{
m-n\cos2\alpha
}
\]
Now use formula for \(\tan(\theta+\alpha)\):
\[ \tan(\theta+\alpha) = \frac{\tan\theta+\tan\alpha} {1-\tan\theta\tan\alpha} \]After simplification:
\[ \tan(\theta+\alpha) = \frac{m+n}{m-n}\tan\alpha \]
\[
\tan(\theta+\alpha)\cot\alpha
=
\frac{m+n}{m-n}
\]
Hence Proved.