Question

Prove:-

$( \secθ - \tanθ)^{2} = \frac{1 - \sinθ }{1 + \sinθ }.$
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Answer 1

Answer:

$\huge \fbox \pink {QUESTION}$

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

$\huge \fbox \pink {SOLUTION}$

$\bf L.H.S - ( \secθ - \tanθ)^{2} \\ \bf = ( \frac{1}{ \cosθ} - \frac{ \sinθ }{ \cosθ})^{2} \\ \bf = \frac{(1 - \sinθ)(1 - \sinθ) }{(1 - \sin ^{2}θ) } \\ \bf = \frac{ (1 - \sinθ)(1 - \sinθ)}{(1 - \sinθ)(1 + \sinθ)} \\ \bf = \frac{1 - \sinθ}{1 + \sinθ} = R.H.S$

Hence Proved !!