If m sin θ = n sin (θ + 2α),

According to the question,

m sin θ = n sin (θ + 2α)

To prove:

tan (θ + α)cot α =(m + n)/(m – n)

Proof:

m sin θ = n sin (θ + 2α)

⇒ sin(θ + 2α) / sinθ = m/n

Applying componendo-dividendo rule, we have,

$\Rightarrow \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{m+n}{m-n}$

By transformation formula of T-ratios,

We know that,

sin A + sin B = 2 sin ((A+B)/2) cos ((A – B)/2)

And,

sin A – sin B = 2 cos ((A+B)/2) sin ((A – B)/2)

On applying the formula, we get,

$\frac{2 \sin \left(\frac{2 \theta+2 \alpha}{2}\right) \cos \left(\frac{\theta+2 \alpha-\theta}{2}\right)}{2 \cos \left(\frac{2 \theta+2 \alpha}{2}\right) \sin \left(\frac{\theta+2 \alpha-\theta}{2}\right)}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{m}-\mathrm{n}}$

$\Rightarrow \frac{\sin (\theta+\alpha) \cos (\alpha)}{\cos (\theta+\alpha) \sin (\alpha)}=\frac{m+n}{m-n}$

$\{\because \tan x=(\sin x) /(\cos x)\}$

$\Rightarrow \tan (\theta+\alpha) \cot \alpha=\frac{m+n}{m-n}$

Therefore, tan (θ + α) cot α = (m + n)/(m – n)

Hence Proved