\(tan^{-1}\begin{Bmatrix}2cos\left(2sin^{-1}\frac{1}{2}\right)\end{Bma

1.	\(tan^{-1}\begin{Bmatrix}2cos\left(2sin^{-1}\frac{1}{2}\right)\end{Bmatrix}\)=?A. \(\frac{\pi}{3}\) B. \(\frac{\pi}{4}\) C. \(\frac{3\pi}{4}\) D. \(\frac{2\pi}{3}\)
Answer» Correct Answer is (B) \(\frac{\pi}{4}\) Let , x = \(tan^{-1}\begin{Bmatrix}2cos\left(2sin^{-1}\frac{1}{2}\right)\end{Bmatrix}\) ⇒ x= \(tan^{-1}\begin{Bmatrix}2cos\left(2cos(\frac{\pi}{6})\right)\end{Bmatrix}\) \((\because sin(\frac{\pi}{6})=\frac{1}{2})\) ⇒ x = tan^{-1\((2cos\frac{\pi}{3})\)} ⇒ x = tan^-1\((2(\frac{1}{2}))\)\(tan^{-1}1=\frac{\pi}{4}\) \((\because cos(\frac{\pi}{3})=\frac{1}{2} and \tan(\frac{\pi}{4}=1)\)

\(tan^{-1}\begin{Bmatrix}2cos\left(2sin^{-1}\frac{1}{2}\right)\end{Bmatrix}\)=?A. \(\frac{\pi}{3}\) B. \(\frac{\pi}{4}\) C. \(\frac{3\pi}{4}\) D. \(\frac{2\pi}{3}\)

Answer»

Correct Answer is (B) \(\frac{\pi}{4}\)

Let , x = \(tan^{-1}\begin{Bmatrix}2cos\left(2sin^{-1}\frac{1}{2}\right)\end{Bmatrix}\)

⇒ x= \(tan^{-1}\begin{Bmatrix}2cos\left(2cos(\frac{\pi}{6})\right)\end{Bmatrix}\)

\((\because sin(\frac{\pi}{6})=\frac{1}{2})\)

⇒ x = tan^{-1\((2cos\frac{\pi}{3})\)}

⇒ x = tan^-1\((2(\frac{1}{2}))\)\(tan^{-1}1=\frac{\pi}{4}\)

\((\because cos(\frac{\pi}{3})=\frac{1}{2} and \tan(\frac{\pi}{4}=1)\)