tan^{-1} [2 cos (2 sin^{-1} frac{1}{2})] = do | vedclass

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(B) We know that $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$.Substituting this into the expression: $2 \sin^{-1}(\frac{1}{2}) = 2(\frac{\pi}{6}) = \frac{

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) We know that $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$.Substituting this into the expression: $2 \sin^{-1}(\frac{1}{2}) = 2(\frac{\pi}{6}) = \frac{\pi}{3}$.Now, the expression becomes $	an^{-1} [2 \cos(\frac{\pi}{3})]$.Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, we have $	an^{-1} [2 	imes \frac{1}{2}] = 	an^{-1}(1)$.Since $	an(\frac{\pi}{4}) = 1$, the final value is $\frac{\pi}{4}$.

$\tan^{-1} [2 \cos (2 \sin^{-1} \frac{1}{2})] = \dots \dots \dots$

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