Question

Cos2theta/(1+sin2theta)

Answer 1

$\huge\bigstar\mathfrak\blue{\underline{\underline{SOLUTION:}}}$

To prove:

$\frac{cos2 \theta}{1 + sin2 \theta}$

Proof:

L.H.S

$\frac{cos2 \theta}{1 + sin \theta} = \frac{ \frac{ \frac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta } }{ 1 + 2tan \theta} }{1 + {tan}^{2} \theta}$

[Using: sin2x=

$\frac{2tanx}{1 + {tan}^{2} x}$

and cos2x=

$\frac{1 - {tan}^{2}x }{1 + {tan}^{2}x }$]

$= > \frac{ \frac{ \frac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta } }{1 + {tan}^{2} \theta + 2tan \theta } }{ 1 + {tan}^{2} \theta } = \frac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta + 2tan \theta } \\ \\ = > \frac{(1 - tan \theta)(1 + tan \theta)}{(1 + tan \theta) {}^{2} } = \frac{(1 - tan \theta)}{(1 + tan \theta)} \\ \\ = > \frac{tan \frac{\pi}{4} - tan \theta }{tan \frac{\pi}{4} + tan \theta } = tan( \frac{\pi}{4} - \theta)$

R.H.S.

Hence proved ☺️