Question

Answer the question number 6
Q.6. Find the principal value of ${\sin }^{-1}\left(\frac{-\sqrt{3}}{2}\right)+{\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
Ans : $\frac{\mathrm{\pi }}{2}$

Answer 1

Dear student,
Here is the solution of your asked query:
$$\begin{gathered} \mathrm{Q}.6.{\sin }^{-1}\left(\frac{-\sqrt{3}}{2}\right)+{\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right) \\ =-{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)+\mathrm{\pi }-{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)\{\mathrm{Since},{\sin }^{-1}\left(-\mathrm{x}\right)=-{\sin }^{-1}\mathrm{x}\mathrm{and}{\cos }^{-1}\left(-\mathrm{x}\right)=\mathrm{\pi }-{\cos }^{-1}\mathrm{x}\mathrm{for}\mathrm{all}\mathrm{x}\in \left[-1,1\right]\} \\ =-{\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{3}\right)+\mathrm{\pi }-{\cos }^{-1}\left(\cos \frac{\mathrm{\pi }}{6}\right) \\ =-\frac{\mathrm{\pi }}{3}+\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\{\mathrm{since}{\sin }^{-1}\left(\mathrm{sin\theta }\right)=\mathrm{\theta }\mathrm{and}{\cos }^{-1}\left(\mathrm{cos\theta }\right)=\mathrm{\theta }\} \\ =\mathrm{\pi }-\left(\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}\right) \\ =\mathrm{\pi }-\frac{\mathrm{\pi }}{2} \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$
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