Rd sharma solution class 12 chapter Inverse trigonometric functions

Exercise 4.1

Question 1

Find the principal value of each of the following:

(i) $sin^{-1} \left(-\dfrac{\sqrt{3}}{~~2}\right)$

Sol :

$=sin^{-1}\left[sin~\left(-\dfrac{\pi}{3}\right)\right]$

$=-\dfrac{\pi}{3}$

(ii) $sin^{-1} \left(cos~\dfrac{2\pi}{3}\right)$

Sol :

$=sin^{-1} \left(-\dfrac{1}{2}\right)$

$=sin^{-1} \left[sin~\left(-\dfrac{\pi}{6}\right)\right]$

$=-\dfrac{\pi}{6}$

(iii) $sin^{-1} \left(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)$

Sol :

$=sin^{-1}\left(\dfrac{\sqrt{3}}{2\sqrt{2}}-\dfrac{1}{2\sqrt{2}}\right)$

$=sin^{-1}\left(\dfrac{\sqrt{3}}{2}\times\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\times\dfrac{1}{\sqrt{2}}\right)$

$=sin^{-1}\left[\dfrac{\sqrt{3}}{2}\times \sqrt{1-\left(\dfrac{1}{\sqrt{2}}\right)^2}-\dfrac{1}{2}\times\sqrt{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}\right]$ $\bigg[\because sin^{-1}~x-sin^{-1}~y=sin^{-1}\left(x~\sqrt{1-\left(y\right)^2}-y~\sqrt{1-\left(x\right)^2}\right)\bigg]$

$=sin^{-1}~\left(\dfrac{\sqrt{3}}{2}\right)-sin^{-1}~\left(\dfrac{1}{\sqrt{2}}\right)$

$=\dfrac{\pi}{3}-\dfrac{\pi}{4}$

$=\dfrac{\pi}{12}$

(iv) $sin^{-1} \left(\dfrac{\sqrt{3}+1}{2\sqrt{2}}\right)$

Sol :

$=sin^{-1}\left(\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}}\right)$

$=sin^{-1}\left(\dfrac{\sqrt{3}}{2}\times\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}\times\dfrac{1}{\sqrt{2}}\right)$

$=sin^{-1}\left[\dfrac{\sqrt{3}}{2}\times \sqrt{1-\left(\dfrac{1}{\sqrt{2}}\right)^2}+\dfrac{1}{2}\times\sqrt{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}\right]$ $\bigg[\because sin^{-1}~x+sin^{-1}~y=sin^{-1}\left(x~\sqrt{1-\left(y\right)^2}+y~\sqrt{1-\left(x\right)^2}\right)\bigg]$

$=sin^{-1}~\left(\dfrac{\sqrt{3}}{2}\right)+sin^{-1}~\left(\dfrac{1}{\sqrt{2}}\right)$

$=\dfrac{\pi}{3}+\dfrac{\pi}{4}$

$=\dfrac{7\pi}{12}$

(v) $sin^{-1} \left(cos~\dfrac{3\pi}{4}\right)$

Sol :

$=sin^{-1} \left(-\dfrac{1}{\sqrt{2}}\right)$

$=sin^{-1}\left[sin \left(-\dfrac{\pi}{4}\right)\right]$

$=-\dfrac{\pi}{4}$

(vi) $sin^{-1} \left(tan~\dfrac{5\pi}{4}\right)$

Sol :

Let $y=sin^{-1}\left(tan\dfrac{5\pi}{4}\right)$

Therefore $sin~y=\left(tan\dfrac{5\pi}{4}\right)$ $=tan\left(\pi+\dfrac{\pi}{4}\right)$ $=tan\left(\dfrac{\pi}{4}\right)$ $=1=sin\dfrac{\pi}{2}$

We know that principal value of $sin^{-1}\text{ is }\left[\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ and $sin\left(\dfrac{\pi}{2}\right)=tan\left(\dfrac{5\pi}{4}\right)$