^2 (2 ^ -1 1+x 1-x )= _ _ _ _ , where -1 x<1 - Vidyadip

MCQ

$\sin ^2\left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)=$ $\_\_\_\_$ , where $-1 \leq x<1$

✓
$1-x^2$
B
$1+x^2$
C
$x^2-1$
D
$-x^2$

✓

Answer

Correct option: A.

$1-x^2$

(A) $\sin ^2\left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$
$=\sin ^2(2 \theta)$, where $\theta=\tan ^{-1} \sqrt{\frac{1+x}{1-x}}$
$=\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)^2$, where $\tan \theta=\sqrt{\frac{1+x}{1-x}}$
$=\left\{\frac{\frac{2 \sqrt{1+x}}{\sqrt{1-x}}}{1+\left(\frac{1+x}{1-x}\right)}\right\}^2=\frac{4(1+x)(1-x)}{(1-x+1+x)^2}=1-x^2$

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