Question

How to solve this????

Answer 1

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Answer 2

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$\frac{1}{sec\: \theta-1} + \frac{1}{sec\: \theta+1} = 2 cot\: \theta × cosec\: \theta$

$\huge\bigstar\underline{Answer}\bigstar$

$\underline{LHS\: = }$

= $\frac{1}{sec\: \theta-1} + \frac{1}{sec\: \theta+1}$

$\implies \frac{sec\: \theta+1+sec\: \theta-1}{(sec\: \theta+1)(sec\: \theta-1)}$

$\implies \frac{2sec\: \theta}{sec^{2}\: \theta-1}$

We have,
$sec^{2}\: \theta-1 = tan^{2}\: \theta$

$\implies \frac{2sec\: \theta}{tan^{2}\: \theta}$

$\implies \frac{2sec\: \theta}{tan\: \theta × tan\: \theta}$

We know,
$sec\: \theta = \frac{1}{cos\: \theta}$

& $tan\: \theta = \frac{sin\: \theta}{cos\: \theta}$

$\implies 2\frac{\frac{1}{cos\: \theta}}{tan\: \theta× \frac{sin\: \theta}{cos\: \theta}}$

$\implies 2\frac{1}{tan\: \theta}× \frac{1}{sin\: \theta}$


We have,
$cot\: \theta= \frac{1}{tan\: \theta}$

& $cosec\: \theta= \frac{1}{sin\: \theta}$


$\implies 2cot\: \theta × cosec\: \theta$

$\underline{=\: RHS }$