Question

sin[2tan inverse √1-x/1+x]= √1-x²​

Answer 1

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Answer 2

$\sin(2\tan^{1} \dfrac{\sqrt{1-x}}{\sqrt{1+x}})=\sqrt{1-x^{2}},$ proved.

Step-by-step explanation:

L.H.S=$\sin(2\tan^{1} \dfrac{\sqrt{1-x}}{\sqrt{1+x}})$

Put $x=\cos 2A$

$=\sin(2\tan^{1} \dfrac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}})$

$=\sin(2\tan^{-1} \sqrt{{\dfrac{2\sin^2 A}{2\cos^2 A}}} )$

$=\sin(2\tan^{-1} \sqrt{{\dfrac{2\sin^2 A}{2\cos^2 A}}})$

$=\sin(2\tan^{-1} \sqrt{{\tan^2 A}})$

$=\sin 2A$

∴ $\sin 2A=\sqrt{1-\cos^2 2A}$

$\sin 2A=\sqrt{1-x^{2}}$

=R.H.S, proved.

Hence, $\sin(2\tan^{1} \dfrac{\sqrt{1-x}}{\sqrt{1+x}})=\sqrt{1-x^{2}}$, proved.