Question

P.T sec2x + tan2x = tan(45 + x)

Answer 1

This is correct answer.
check it.

Answer 2

Hi...☺

Here is your answer....✌

LHS

$= \sec2x + \tan2x \\ \\ = \frac{1}{ \cos2x} + \frac{ \sin2x}{ \cos2x } \\ \\ = \frac{1 + \sin2x }{ \cos2x } \\ \\ = \frac{ { { \cos}^{2}x + \sin }^{2} x+ 2 \sin x \cos x } { \cos {}^{2} x - { \sin }^{2} x} \\ \\ \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: | \sin2x= 2 \sin x \cos x |\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: | \cos2x = \cos {}^{2} x - { \sin }^{2} x| \\ \\ = \frac{(\cos x + \sin x ){}^{2} }{ (\cos x + \sin x)( \cos x - \sin x) } \\ \\ = \frac{\cos x + \sin x}{\cos x - \sin x} \\ \\ = \frac{1 + \tan x}{1 - \tan x} \: \: \: \: \: \: \:$

[ By dividing numerator and denominator by cos x ]

$= \frac{ \tan45 \degree - \tan x}{1 - \tan45\degree . \tan x } \: \: \: \: \: \: \: \: | \because \: \tan45\degree = 1 | \\ \\ = \tan(45\degree - x)$

= RHS [ Proved ]