Question

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‌✌️ ✨Urgently ✨ ✌️ 
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Answer 1

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$\sf {\large {\underline {TRIGONOMETRIC \:PROVING}}}$

$\sf {\large {\underline {FORMULAE }}}$

A ). ${sin} ^{2}$ theta + ${cos} ^{2}$ theta = 1

B ). 1 - ${sin} ^{2}$ = ${cos} ^{2}$

C ). $\frac{1}{cos\:theta}$ = sec theta

D ). $\frac{Sin\:theta}{cos\:theta}$ = Tan theta .

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Answer 2

Answer :

L.H.S =

$\tt {\sqrt{ \frac{1 - \sin \theta}{1 + \sin \theta} }}$

$\tt{ = \sqrt{ \frac{1 - \sin \theta}{ 1 + \sin \theta} \times \frac{ 1 - \sin \theta}{1 - \sin \theta}} }$

$\tt {= \sqrt{ \frac{(1 - \sin \theta) {}^{2} }{1 {}^{2} - \sin {}^{2} \theta } }}$

$\tt{ = \sqrt{ \frac{(1 - \sin \theta) {}^{2} }{ \cos {}^{2} \theta}} }$

$\tt {= \frac{1 - \sin \theta}{ \cos \theta} }$

$\tt { = \frac{1}{ \cos \theta} - \frac{ \sin \theta}{ \cos \theta} }$

$\tt {= \sec \theta - \tan \theta}$

= R.H.S

Hence, Proved.

If any doubt, write in comment box.

Hope it helps! :)