Question:
Prove that:
$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x,|x| \leq \frac{1}{\sqrt{2}}$
Solution:
To Prove: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$
Formula Used: $\sin 2 A=2 \times \sin A \times \cos A$
Proof:
$\mathrm{LHS}=\sin ^{-1}\left(2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}\right) \cdots$(1)
Let $x=\sin A \ldots$ (2)
Substituting $(2)$ in $(1)$,
LHS $=\sin ^{-1}\left(2 \sin A \sqrt{1-\sin ^{2} A}\right)$
$=\sin ^{-1}(2 \times \sin A \times \cos A)$
$=\sin ^{-1}(\sin 2 A)$
$=2 A$
From $(2), A=\sin ^{-1} x$
$2 A=2 \sin ^{-1} x$
$=$ RHS
Therefore, LHS $=$ RHS
Hence proved.
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