Question

${\sin }^{-1}\left(\frac{-\sqrt{2}}{2}\right)+{\cos }^{-1}\left(\frac{-1}{2}\right)-{\tan }^{-1}\left(-\sqrt{3}\right)+{\cot }^{-1}\left(\frac{-1}{\sqrt{3}}\right)=$

• A. $\frac{7\pi }{12}$
• B. $\frac{17\pi }{12}$
• C. $\frac{5\pi }{12}$
• D. $\frac{\pi }{12}$

Answer 1

The correct option is B

$\frac{17\pi }{12}$

Explanation for the correct answer:

Simplifying the given expression:

$$\begin{gathered} {\sin }^{-1}\left(\frac{-\sqrt{2}}{2}\right)+{\cos }^{-1}\left(\frac{-1}{2}\right)-{\tan }^{-1}\left(-\sqrt{3}\right)+{\cot }^{-1}\left(\frac{-1}{\sqrt{3}}\right) \\ ={\sin }^{-1}\left(-\frac{1}{\sqrt{2}}\right)+{\cos }^{-1}\left(\frac{-1}{2}\right)-{\tan }^{-1}\left(-\sqrt{3}\right)+{\cot }^{-1}\left(\frac{-1}{\sqrt{3}}\right) \\ =-{\sin }^{-1}\left(\sin 45°\right)+{\cos }^{-1}\left(\cos 120°\right)-{\tan }^{-1}\left(\tan 120°\right)+{\cot }^{-1}\left(cot120°\right) \end{gathered}$$

$\because -\sin n45°=-\frac{1}{\sqrt{2}},\cos 120°=\frac{-1}{2},\tan 120°=-\sqrt{3},cot120°=\frac{-1}{\sqrt{3}}$, substituting we have,

$$\begin{gathered} =-{\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{4}\right)+{\cos }^{-1}\left(\cos \frac{2\pi }{3}\right)-{\tan }^{-1}\left(\tan \frac{2\pi }{3}\right)+{\cot }^{-1}\left(cot\frac{2\pi }{3}\right) \\ =\left(-\frac{\mathrm{\pi }}{4}\right)+\left(\frac{2\pi }{3}\right)-\left(\frac{2\pi }{3}-\mathrm{\pi }\right)+\left(\frac{2\pi }{3}\right) \\ =-\frac{\pi }{4}+\frac{2\pi }{3}+\frac{\pi }{3}+\frac{2\pi }{3} \\ =\frac{17\pi }{12} \end{gathered}$$

Therefore, the correct answer is option (B).