If m =n ( +2 ), prove that ( + ) = m+n m-n .

Question

$\text{If}\ \text{m}\sin\theta=\text{n}\sin(\theta+2\alpha),$ prove that $\tan(\theta+\alpha)\cot\alpha=\frac{\text{m+n}}{\text{m}-\text{n}}.$

✓

Answer

Given that $\text{m}\sin\theta=\text{n}\sin(\theta+2\alpha),$
We need to prove that $\tan(\theta+\alpha)=\frac{\text{m+n}}{\text{m}-\text{n}}\tan\alpha$
$\text{m}\sin\theta=\text{n}\sin(\theta+2\alpha)$
$\Rightarrow\ \frac{\sin(\theta+2\alpha)}{\sin\theta}=\frac{\text{m}}{\text{n}}$
Using Componendo - Dividendo, we have,
$\Rightarrow\ \frac{\sin(\theta+2\alpha)+\sin\theta}{\sin(\theta+2\theta)-\sin\theta}=\frac{\text{m+n}}{\text{m}-\text{n}}...(1)$
We know that,
$\sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}$
and
$\sin\text{C}-\sin\text{D}=2\cos\frac{\text{C+D}}{2}\sin\frac{\text{C}-\text{D}}{2}$
Applying the above formula in equation (1), we have,
$\frac{2\sin\frac{\theta+2\theta+\theta}{2}\cos\frac{\theta+2\theta-\theta}{2}}{2\cos\frac{\theta+2\theta+\theta}{2}\sin\frac{\theta+2\theta-\theta}{2}}=\frac{\text{m+n}}{\text{m}-\text{n}}$
$\Rightarrow\ \frac{2\sin(\theta+\alpha)\cos\alpha}{2\cos(\theta+\alpha)\sin\alpha}=\frac{\text{m+n}}{\text{m}-\text{n}}$
$\Rightarrow\ \frac{\tan(\theta+\alpha)}{\tan\alpha}=\frac{\text{m+n}}{\text{m}-\text{n}}$
$\Rightarrow\ \tan(\theta+\alpha)=\frac{\text{m+n}}{\text{m}-\text{n}}\times\tan\alpha$
Hence proved.

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