Question

prove that sin^-1 (-1/2) + ? - cos^-1 (?3/2) = cos^-1 (-1/2)
Q.2. Prove that ${\sin }^{-1}\left(-\frac{1}{2}\right)+{\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right)={\cos }^{-1}\left(-\frac{1}{2}\right)$

Answer 1

Dear student
$$\begin{gathered} {\sin }^{-1}\left(\frac{-1}{2}\right)+{\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right)={\cos }^{-1}\left(-\frac{1}{2}\right) \\ \mathrm{LHS}={\sin }^{-1}\left(\frac{-1}{2}\right)+{\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right) \\ =-{\sin }^{-1}\left(\frac{1}{2}\right)+\mathrm{\pi }-{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ =-{\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{6}\right)+\mathrm{\pi }-{\cos }^{-1}\left(\cos \frac{\mathrm{\pi }}{6}\right) \\ =-\frac{\mathrm{\pi }}{6}+\mathrm{\pi }-\frac{\mathrm{\pi }}{6} \\ =\frac{-\mathrm{\pi }+6\mathrm{\pi }-\mathrm{\pi }}{6} \\ =\frac{2\mathrm{\pi }}{3} \\ \mathrm{RHS}={\cos }^{-1}\left(-\frac{1}{2}\right) \\ =\mathrm{\pi }-{\cos }^{-1}\left(\frac{1}{2}\right) \\ =\mathrm{\pi }-{\cos }^{-1}\left(\cos \frac{\mathrm{\pi }}{3}\right) \\ =\mathrm{\pi }-\frac{\mathrm{\pi }}{3} \\ =\frac{2\mathrm{\pi }}{3} \\ \mathrm{SO},{\sin }^{-1}\left(\frac{-1}{2}\right)+{\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right)={\cos }^{-1}\left(-\frac{1}{2}\right) \end{gathered}$$
Regards