tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2)

We know that the range of principal values of tan^-1, cos^-1 and sin^-1 are $\left(\frac{-\mathrm{\pi }}{2},\frac{\pi }{2}\right) , \left(0,\pi \right) \mathrm{and} \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \mathrm{respectively},$ Let tan^-1(1) = ${\theta }_1$. Then, $$\begin{gathered} \tan {\theta }_1 = 1 = \tan \frac{\pi }{4} \Rightarrow {\theta }_1 = \frac{\pi }{4} \in \left[0, \pi \right]. \\ Let {\cos }^{-1}\left(\frac{-1}{2}\right)={\theta }_2. \mathrm{Then}, \\ \cos {\theta }_2=\frac{-1}{2} = -\cos \frac{\pi }{3}=\cos \left(\pi - \frac{\pi }{3}\right)=\cos \frac{2\pi }{3} \\ So {\theta }_2 =\frac{2\pi }{3}\in \left[0 , \pi \right]. \\ Let {\sin }^{-1}\left(\frac{-1}{2}\right)={\theta }_3. \mathrm{Then}, \\ \sin {\theta }_3=\frac{-1}{2} = -\sin \frac{\pi }{6}=\sin \left(-\frac{\pi }{6}\right) \\ So {\theta }_3 =-\frac{\pi }{6}\in \left[\frac{-\pi }{2} , \frac{\pi }{2}\right]. \\ So {\tan }^{-1}\left(1\right) +{\cos }^{-1}\left(\frac{-1}{2}\right)+{\sin }^{-1}\left(\frac{-1}{2}\right)= \frac{\pi }{4}+ \frac{2\pi }{3}-\frac{\pi }{6}=\frac{3\mathrm{\pi }}{4} \end{gathered}$$