# Find the principal value of :

Question:

Find the principal value of :

(i) $\sin ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$

(ii) $\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$

(iii) $\tan ^{-1}(-\sqrt{3})$

(iv) $\sec ^{-1}(-2)$

(v) $\operatorname{cosec}^{-1}(-\sqrt{2})$

(vi) $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$

Solution:

(i) Let $\sin ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=x$

$\Rightarrow-\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=x\left[\right.$ Formula: $\left.\sin ^{-1}(-x)=-\sin ^{-1} x\right]$

$\Rightarrow \frac{1}{\sqrt{2}}=-\sin x$ [We know which value of $x$ when put in this expression will give us this result]

$\therefore \mathrm{X}=-\frac{\pi}{4}$

(ii) $\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\pi-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ [Formula: $\cos ^{-1}(-x)=\pi-\cos ^{-1} x$ ]

Let $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\mathrm{X}$

$\Rightarrow\left(\frac{\sqrt{3}}{2}\right)=\cos x$ [We know which value of $x$ when put in this expression will give us this result]

$\therefore \mathrm{x}=\frac{\pi}{6}$

Putting this value back in the equation

$\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$

(iii) Let $\tan ^{-1}(-\sqrt{3})=\mathrm{x}$

$\Rightarrow-\tan ^{-1}(\sqrt{3})=x\left[\right.$ Formula: $\left.\tan ^{-1}(-x)=-\tan ^{-1}(x)\right]$

$\Rightarrow \sqrt{3}=-\tan x$ [We know which value of $x$ when put in this expression will give us this result]

$\therefore \mathrm{x}=\frac{-\pi}{3}$

(iv) $\sec ^{-1}(-2)=\pi-\sec ^{-1}(2) \ldots$ (i) [ Formula:sec $^{-1}(-x)=\pi-\sec ^{-1}(x)$ ]

Let $\sec ^{-1}(2)=x$

$\Rightarrow 2=\sec x$ [We know which value of $x$ when put in this expression will give us this result]

$\therefore \mathrm{x}=\frac{\pi}{3}$

Putting the value in (i)

$\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$

(v) Let $\operatorname{cosec}^{-1}(-\sqrt{2})=x$

$\Rightarrow-\operatorname{cosec}^{-1}(\sqrt{2})=x\left[\right.$ Formula: $\left.\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}(x)\right]$

$\Rightarrow \sqrt{2}=-\operatorname{cosec} x$

$\therefore \mathrm{x}=-\frac{\pi}{4}$

(vi) $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\pi-\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right) \ldots$ (i)

Let $\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\mathrm{x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\cot ^{-1} x$ [We know which value of $x$ when put in this expression will give us this result]

$\Rightarrow \mathrm{X}=\frac{\pi}{3}$

Putting in (i)

$\pi-\frac{\pi}{3}$

$=\frac{2 \pi}{3}$