Question

PLEASE help EMERGENCY
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Answer 1

this is the answer for the question you asked

Answer 2

Refer To Attachment

Points To remember.

Trigonometry identies

$\sin{}^{2}\theta + \cos {}^{2} \theta = 1 \\ \\ \sec {}^{2} \theta - \tan {}^{2} \theta = 1 \\ \\ cosec {}^{2} \theta - \cot {}^{2} \theta = 1$

$cosec\theta = \frac{1}{ \sin \theta } \\ \\ \sec \theta = \frac{1}{ \cos \theta} \\ \\ \cot\theta = \frac{ \cos\theta }{ \sin\theta}$

Using this You will Get LHS

$\frac{(1 - cos\theta)(1 - \cos\theta)}{1 - \cos {}^{2} \theta } \\ \\ 1 - \cos {}^{2} \theta$

Can Written as

a²-b²=(a+B)(a-b)

$\frac{(1 - \cos \theta)(1 - \cos\theta) }{ {1}^{2} - { \cos \theta}^{2} } \\ \\ \frac{\cancel{(1 - \cos \theta)}(1 - \cos\theta) }{\cancel{ (1 - \cos \theta)}(1 + \cos \theta)}$

Hence

Its Proved

LHS=RHS

$\boxed{\mathbf{\frac{(1-\cos\theta)}{(1+\cos\theta)}=\frac{(1-\cos\theta)}{(1+\cos\theta)}}}$