Question

The value of ${\sin }^{-1}\left\{\cot \left\{{\sin }^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}+{\cos }^{}\frac{\sqrt{12}}{4}+{\sec }^{-1}\sqrt{2}\right\}\right\}$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{6}$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $0$

Solution:-

$si{n}^{-1}\left\{cot\{si{n}^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}+co{s}^{-1}\frac{\sqrt{12}}{4}+se{c}^{-1}\sqrt{2}\}\right\}$

We know that

$sin\frac{\pi }{12}=\sqrt{\frac{2-\sqrt{3}}{4}}cos\frac{\pi }{6}=\frac{\sqrt{3}}{2}=\frac{\sqrt{12}}{4}$

$sec\frac{\pi }{4}=\sqrt{2}$

$si{n}^{-1}\left(cot\right(\frac{\pi }{12}+\frac{\pi }{6}+\frac{\pi }{4}\left)\right)$

$si{n}^{-1}\left\{cot\left(\frac{\pi }{2}\right)\right\}$

$si{n}^{-1}0$

$0$

C is correct.