Question: Find the principal value of \begin{align} tan^{-1} (1) + cos^{-1}\left(-\frac{1}{2}\right) + sin^{-1}\left(-\frac{1}{2}\right)\end{align}

Answer:

\begin{align} Let \;\; tan^{-1}(1)=x. \;\;Then\;\; tan x = 1 = tan\left(\frac{\pi}{4}\right).\end{align}

\begin{align} \therefore tan^{-1}(1)=tan\left(\frac{\pi}{4}\right)\end{align}

\begin{align} Let \;\;cos^{-1}\left(-\frac{1}{2}\right)=y. \;\;Then,\;\; cos y = -\frac{1}{2} = -cos\left(\frac{\pi}{3}\right)= cos\left(\pi - \frac{\pi}{3}\right) = cos\left(\frac{2\pi}{3}\right)\end{align}

\begin{align} \therefore cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}\end{align}

\begin{align} Let \;\; sin^{-1}\left(-\frac{1}{2}\right)=z. \;\;Then,\;\; sin z = -\frac{1}{2} = -sin\left(\frac{\pi}{6}\right)= sin\left(-\frac{\pi}{6}\right)\end{align}

\begin{align} \therefore sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\end{align}

\begin{align} \therefore tan^{-1} (1) + cos^{-1}\left(-\frac{1}{2}\right) + sin^{-1}\left(-\frac{1}{2}\right)\end{align}

\begin{align} =\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}\end{align}

\begin{align} =\frac{3\pi + 8\pi -2\pi}{12}=\frac{9\pi}{12}=\frac{3\pi}{4}\end{align}

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