Question

help teachersplsssyyyy​

Answer 1

Step-by-step explanation:

Given that:

$\frac{sin \theta}{cot\theta + cosec \theta} = 2 + \frac{sin \theta}{cot\theta - cosec \theta}$

Solution:

Take LHS

Multiply and divide by

$cot\theta - cosec \theta$

$\frac{sin \theta}{cot\theta + cosec \theta} \times \frac{cot\theta - cosec \theta}{cot\theta - cosec \theta} \\ \\ = \frac{sin \theta(cot\theta - cosec \theta)}{ {cot}^{2} \theta - {cosec}^{2} \theta} \\ \\ \because {cosec}^{2} \theta - {cot}^{2} \theta = 1 \\ \\ = \frac{sin \theta(cosec\theta - cot \theta)}{1} \\ \\ = sin \theta\bigg( \frac{1}{sin \theta} - \frac{cos \theta}{sin \theta} \bigg)$

$= sin \theta\bigg( \frac{1 - cos \theta}{sin \theta}\bigg ) \\ \\ = 1 - cos \theta \: \: \: \: ...eq1$

Now take RHS

$2 + \frac{sin \theta}{cot\theta - cosec \theta} \\ \\ = 2 + \frac{sin \theta}{cot\theta - cosec \theta} \times \frac{cot\theta + cosec \theta}{cot\theta + cosec \theta} \\ \\ = 2 - \frac{sin \theta(cot\theta + cosec \theta)}{ {cosec}^{2} \theta - {cot}^{2} \theta } \\ \\ = 2 - sin \theta(cot\theta + cosec \theta) \\ \\ = 2 - sin \theta\bigg(\frac{cos \theta}{sin \theta} + \frac{1}{sin \theta} \bigg) \\ \\ = 2 - \frac{sin \theta(1 + cos \theta)}{sin \theta} \\ \\ = 2 - 1 - cos \theta \\ \\ = 1 - cos \theta \: \: \: \: ...eq2$

From Eq1 and eq2

it is proved that

$\frac{sin \theta}{cot\theta + cosec \theta} = 2 + \frac{sin \theta}{cot\theta - cosec \theta}$

Hope it helps you