Question

2sin^-1x=π-sin^-1(2x√1-x^2)​

Answer 1

Step-by-step explanation:

$\alpha = { \sin ^{ - 1} (x) } \\ \sin( \alpha ) = x \\ rhs \\ \sin ^{ - 1} (2x \sqrt{1 - {x}^{2} } ) \\ \sin ^{ - 1} (2 \sin( \alpha ) \sqrt{1 - \sin ^{2} ( \alpha ) } ) \\ \sin ^{ - 1} (2 \sin( \alpha ) \cos( \alpha ) ) \\ note \: 2 \sin( \alpha ) \cos( \alpha ) = \sin(2 \alpha ) \\ \sin ^{ - 1} ( \sin(2 \alpha ) ) = 2 \alpha \\ given \: \alpha = \sin ^{ - 1} (x) \\ 2 \alpha = 2 \sin {}^{ - 1} (x) \\ lhs = rhs = 2 \sin {}^{ - 1} (x) \\ hence \: proved \: (\pi \: is \: not \: there$