Question

Find the p values of the following questions:
$ta{n}^{-1}\left(1\right)+co{s}^{-1}(-\frac{1}{2})+si{n}^{-1}(-\frac{1}{2})$
Given expression is not a standard identity, so we separately find the value of $ta{n}^{-1}\left(1\right),co{s}^{-1}(-\frac{1}{2}),si{n}^{-1}(-\frac{1}{2})$ and simplify it.

Answer 1

Let $ta{n}^{-1}\left(1\right)=x\Rightarrow tanx=1=tan\frac{\pi }{4}\Rightarrow x=\frac{\pi }{4}$
where principal value $x\epsilon (-\frac{\pi }{2},\frac{\pi }{2})\therefore ta{n}^{-1}\left(1\right)=\frac{\pi }{4}$
Let $co{s}^{-1}(-\frac{1}{2})=y\Rightarrow cosy=-\frac{1}{2}=-cos\left(\frac{\pi }{3}\right)=cos(\pi -\frac{\pi }{3})=cos\left(\frac{2\pi }{3}\right)(\because cos(\pi -\theta )=-cos\theta )$
$\Rightarrow y=\frac{2\pi }{3}$, where principal value of $y\epsilon [0,\pi ]$
$\therefore co{s}^{-1}(-\frac{1}{2})=\frac{2\pi }{3}$
Let $si{n}^{-1}(-\frac{1}{2})=z\Rightarrow sinz=-\frac{1}{2}=-sin\left(\frac{\pi }{6}\right)=sin(-\frac{\pi }{6})$
$\Rightarrow z=-\frac{\pi }{6}$, where principal value $zϵ[-\frac{\pi }{2},\frac{\pi }{2}]$
$\therefore si{n}^{-1}(-\frac{1}{2})=-\frac{\pi }{6}\therefore ta{n}^{-1}\left(1\right)+co{s}^{-1}(-\frac{1}{2})+si{n}^{-1}(-\frac{1}{2})=x+y+z=\frac{\pi }{4}+\frac{2\pi }{3}-\frac{\pi }{6}=\frac{3\pi +8\pi -2\pi }{12}=\frac{9\pi }{12}=\frac{3\pi }{4}$