If ^ -1 1+x^2 - 1-x^2 1+x^2 + 1-x^2 = , then x^2 =

MCQ

If $\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha,$ then $x^2 =$

✓
$\sin2\alpha$
B
$\sin\alpha$
C
$\cos2\alpha$
D
$\cos\alpha$

✓

Answer

Correct option: A.

$\sin2\alpha$

$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha$
$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}=\tan\alpha$
$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}=\tan\alpha$
$\frac{1+\text{x}^2-2\sqrt{1-\text{x}^2}\sqrt{1+\text{x}^2}+1-\text{x}^2}{1+\text{x}^2-1+\text{x}^2}=\tan\alpha$
$\frac{1-\sqrt{1-\text{x}^4}}{\text{x}^2}=\tan\alpha$
$1-\sqrt{1-\text{x}^4}=\text{x}^2\tan\alpha$
$\big(1-\text{x}^2\tan\alpha\big)^2=1-\text{x}^4$
$1-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=1-\text{x}^4$
$\text{x}^4-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=0$
$\text{x}^2\big(\text{x}^2-2\tan\alpha+\text{x}^2\tan^2\alpha\big)=0$
$\text{x}^2=\frac{2\tan\alpha}{1+\tan^2\alpha}$
$\text{x}^2=\frac{2\tan\alpha}{\sec^2\alpha}$
$\text{x}^2=2\tan\alpha\cos^2\alpha$
$\text{x}^2=2\sin\alpha\cos\alpha$
$=2\sin\alpha$

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