Question

Identity transformations of Trigonometric Expressions.
prove the following identities.
$\frac{1-sin2\alpha }{1+sin2\alpha }=co{t}^2(\frac{\pi }{4}+\alpha ).$

Answer 1

$\frac{1-sin2\alpha }{1+sin2\alpha }=co{t}^2(\pi /4+\alpha )$

L.H.S $\frac{1-sin2\alpha }{1+sin2\alpha }$

$\frac{(cos\alpha +sin\alpha {)}^2}{(cos\alpha +sin\alpha {)}^2}=\frac{(1-tan\alpha {)}^2}{(1+tan\alpha {)}^2}$

$=\frac{1}{{\left(\frac{1+tan\alpha }{1-tan\alpha }\right)}^2}=\frac{1}{{\left(\frac{tan\pi /4+tan\alpha }{1-tan\pi /4tan\alpha }\right)}^2}$

$=\frac{1}{\left(tan\right(\pi /4+\alpha ){)}^2}=co{t}^2(\pi /4+\alpha )$