Question

With step by step explaination ​

Answer 1

Answer:

$\sqrt{ \frac{1 - sinθ}{1 + sinθ} } = sec θ- tanθ$

$\sqrt{ \frac{1 - sinθ}{1 + sinθ} } \times \sqrt{ \frac{1 - sinθ}{1 - sinθ} } = secθ - tanθ$

$\sqrt{ \frac{(1 - sinθ)(1 - sinθ)}{(1 + sinθ)(1 - sinθ)} } =secθ - tanθ$

$\sqrt{ \frac{ {(1 - sinθ)}^{2} }{ {(1)}^{2} - {(sinθ)}^{2} } } =secθ - tanθ$

$\sqrt{ \frac{ {(1 - sinθ)}^{2} }{1 - {sinθ}^{2} } } = secθ - tanθ$

$\sqrt{ \frac{ {(1 - sinθ)}^{2} }{ {cosθ}^{2} } } = secθ- tanθ$

$\frac{1 - sinθ}{cosθ} = secθ - tanθ$

$\frac{1}{cosθ} - \frac{sinθ}{cosθ} = secθ- tanθ$

$secθ - tanθ = secθ- tanθ$

$L.H.S=R.H.S$

$Hence \: proved$

Answer 2

TO PROVE :-

$\sf \sqrt{ \frac{1 - \sin( \theta) }{1 + \sin( \theta) } } = \sec( \theta) - \tan( \theta)$

SOLUTION :-

$\sf \: Taking \: \: LHS.$

$\sf \sqrt{ \frac{1 - \sin( \theta) }{1 + \sin( \theta) } }$

$\sf \: Rationalization \: \: of \: \: denominator$

$= \sf \sqrt{ \frac{(1 - \sin( \theta) )}{(1 + \sin( \theta)) } \times \frac{(1 - \sin( \theta) )}{(1 - \sin( \theta)) }}$

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

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

$= \sf \frac{1 - \sin(\theta)}{ \cos( \theta) }$

$= \sf \: \frac{1}{ \cos( \theta) } - \frac{ \sin( \theta) }{ \cos( \theta) }$

$\sf \: = \sec( \theta) - \tan( \theta)$

• PROVED