Question

$\frac{1}{cosec \theta + cot \theta } - \frac{1}{sin \theta} = \frac{1}{sin \theta} - \frac{1}{cosec \theta - cot \theta}$
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Answer 1

Step-by-step explanation:

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Answer 2

$\rm\frac{1}{cosec \theta + cot \theta } - \frac{1}{sin \theta} = \frac{1}{sin \theta} - \frac{1}{cosec \theta - cot \theta}$

Solution-

Here we do some important thing

$\rm \frac{1}{cosec \theta + cot \theta} + \frac{1}{cosec \theta - cot \theta} = \frac{2}{sin \theta} \\ \rm \: now \\ \rm \: lhs = \frac{cosec \theta - cot \theta + cosec \theta + cot \theta}{ {cosec}^{2} \theta - {cot}^{2} \theta } \\ \rm \implies \frac{2cosec \theta}{1} \rm \\ \implies \: lhs = \frac{2}{sin \theta}$

Here we take LHS on RHS and then solve the alternative method is to rationalise this is the most hardest method to that question here we go for RHS

$\rm \: rhs = \frac{2}{sin \theta}$

LHS=RHS

Hence proved