# If M Sin θ = N Sin ( θ + 2 α ) , Prove that Tan ( θ + α ) Cot α = M + N M − N - Mathematics

Sum

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

#### Solution

Given:
$m \sin\theta = n \sin\left( \theta + 2\alpha \right)$

$\Rightarrow \frac{m}{n} = \frac{\sin\left( \theta + 2\alpha \right)}{\sin\theta}$
Applying componendo and dividendo, we get

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

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

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

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

$\therefore \tan\left( \theta + \alpha \right) \cot\alpha = \frac{m + n}{m - n}$
Concept: Transformation Formulae
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Exercise 8.2 | Q 19 | Page 19