Question

Prove that $\frac{1+sin2x}{cos2x}=tan(\frac{\pi }{4}+x)$

Answer 1

L.H.S $=\frac{1+sin2x}{cos2x.}$

$=\frac{1+\frac{2tanx}{1+ta{n}^{2}x}}{\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}}$ $[\because sin2x=\frac{2tanx}{1+ta{n}^{2}x}$ $cos2x=\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}]$

$=\frac{1+ta{n}^{2}x+2tanx}{1-ta{n}^{2}x}$

$=\frac{(1+tanx{)}^2}{(1-tanx)(1+tanx)}$

$=\frac{1+tanx}{1-tanx}$

$=\frac{tan\pi /4+tanx}{1-tan\frac{\pi }{4}tanx}$ $[\because tan\frac{\pi }{4}=1]$

$=tan(\frac{\pi }{4}+x)$

= R.H.S