Question

prove that 1 - sin2x / 1 + sin2x = tan²(π/4 - x)

Answer 1

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Answer 2

$\text { Proved that } \frac{1-\sin 2 x}{1+\sin 2 x}=\tan ^{2}\left(\frac{\pi}{4}-x\right)$

Solution:

Need to prove:

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

Lets solve Left hand side

$\frac{1-\sin 2 x}{1+\sin 2 x}$

$\text { As } \sin ^{2} x+\cos ^{2} x=1 \text { and } \sin 2 x=2 \sin x \cos x$

$=>\frac{1-\sin 2 x}{1+\sin 2 x}=\frac{\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x}{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}$

$=>\frac{1-\sin 2 x}{1+\sin 2 x}=\frac{(\cos x-\sin x)^{2}}{(\cos x+\sin x)^{2}}$

$\begin{aligned}&=>\frac{1-\sin 2 x}{1+\sin 2 x}=\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)^{2}\\\\&=>\frac{1-\sin 2 x}{1+\sin 2 x}=\left(\frac{\cos x\left(1-\frac{\sin x}{\cos x}\right)}{\cos x\left(1+\frac{\sin x}{\cos x}\right)}\right)^{2}\\\\&\Rightarrow \frac{1-\sin 2 x}{1+\sin 2 x}=\left(\frac{1-\tan x}{1+\tan x \times 1}\right)^{2}\end{aligned}$

$\text { As } \tan \left(\frac{\pi}{4}\right)=1$

$=>\frac{1-\sin 2 x}{1+\sin 2 x}=\left(\frac{\tan \left(\frac{\pi}{4}\right)-\tan x}{1+\tan x \times \tan \left(\frac{\pi}{4}\right)}\right)^{2}$

$\text { Using } \tan (\mathrm{A}-\mathrm{B})=\frac{\tan (\mathrm{A})-\tan (B)}{1+\tan (A) \times \tan (B)}$

$\begin{aligned}&=>\frac{1-\sin 2 x}{1+\sin 2 x}=\left(\tan \left(\frac{\pi}{4}-x\right)\right)^{2}\\\\&=>\frac{1-\sin 2 x}{1+\sin 2 x}=\tan ^{2}\left(\frac{\pi}{4}-x\right)\end{aligned}$

$\text { Hence proved that } \frac{1-\sin 2 x}{1+\sin 2 x}=\tan ^{2}\left(\frac{\pi}{4}-x\right)$

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