Question 513 Marks
Write the value of $\sin^{-1}(\sin(-600^\circ))\sin(-600^\circ).$
Answer$\sin^{-1}\{\sin(-600^\circ)\}$
$=\sin^{-1}\{\sin(-600^\circ+360\times2)\}$ $\{\text{Since},\sin(2\text{n}\pi+\theta)=\sin\theta\}$
$=\sin^{-1}\{\sin120^\circ\}$
$=180^\circ-120^\circ$
$\begin{Bmatrix}\text{Since},\sin^{-1}(\sin\theta)=\begin{cases}-\pi-\theta,&\text{ if }\theta\in\Big[\frac{-3\pi}{2},\frac{-\pi}{2}\Big]\\\theta,&\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\\\pi}{2}\Big]\\\pi-\theta,&\text{ if }\theta\in\Big[\frac{\pi}{2},\frac{3\pi}{2}\Big]\\\pi-\theta,&\text{ if }\theta\in\Big[\frac{3\pi}{2},\frac{5\pi}{2}\Big]\end{cases}\end{Bmatrix}$
$=60^\circ$
Hence,
$\sin^{-1}\{\sin(-600^\circ)\}=60^\circ$
View full question & answer→Question 523 Marks
Write the value of $\cos\Big(2\sin^{-1}\frac{1}{3}\Big).$
AnswerLet $\text{y}=\sin^{-1}\frac{1}{3}$
Then, $\sin\text{y}=\frac{1}{3}$
Now, $\cos\text{y}=\sqrt{1-\sin^2\text{y}}$
$\Rightarrow\cos\text{y}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt2}{3}$
$\cos\Big(2\sin^{-1}\frac{1}{3}\Big)=\cos(2\text{y})$
$=\cos^2\text{y}-\sin^2\text{y}$ $\big[\because\ \cos2\text{x}=\cos^2\text{x}-\sin^2\text{x}\big]$
$=\Big(\frac{2\sqrt2}{3}\Big)^2-\Big(\frac{1}{3}\Big)^2$
$=\frac{8}{9}-\frac{1}{9}$
$=\frac{7}{9}$
$\because\ \cos\Big(2\sin^{-1}\frac{1}{3}\Big)=\frac{7}{9}$
View full question & answer→Question 533 Marks
Prove the following results:
$\frac{9\pi}{4}-\frac{9}{4}\sin^{-1}\frac{1}{3}=\frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}$
Answer$\text{L.H.S}=\frac{9\pi}{4}-\frac{9}{4}\sin^{-1}\frac{1}{3}$
$=\frac{9}{4}\Big(\frac{\pi}{2}-\sin^{-1}\frac{1}{3}\Big)...(1)$ $\Big[\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
Now, let $\cos^{-1}\frac{1}{3}=\text{x}.$ Then,
$\cos\text{x}=\frac{1}{3}\Rightarrow\sin\text{x}=\sqrt{1-\Big(\frac{1}{3}\Big)^2}=\frac{2\sqrt2}{3}.$
$\therefore\ \text{x}=\sin^{-1}\frac{2\sqrt2}{3}\Rightarrow\cos^{-1}\frac{1}{3}=\frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}$
$\therefore\ \text{L.H.S}=\frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}=\text{R.H.S}$
View full question & answer→Question 543 Marks
Evaluate the following:
$\cos^{-1}\Big\{\cos\Big(-\frac{\pi}{4}\Big)\Big\}$
AnswerWe know that,
$\cos^{-1}\big(\cos\theta\big)=\begin{cases}-\theta,&\text{if }\theta\in[-\pi,0]\\\theta,&\text{if }\theta\in[0,\pi]\\2\pi-\theta,&\text{if }\theta\in[\pi,2\pi]\\-2\pi+\theta,&\text{if }\theta\in[2\pi,3\pi]\end{cases}$
$\therefore\ \cos^{-1}\Big\{\cos\Big(-\frac{\pi}{4}\Big)\Big\}$
$=-\Big(-\frac{\pi}{4}\Big)$ $\Big\{\because-\frac{\pi}{4}\in[-\pi,0]\Big\}$
$=\frac{\pi}{4}$
Hence,
$\therefore\ \cos^{-1}\Big\{\cos\Big(-\frac{\pi}{4}\Big)\Big\}=\frac{\pi}{4}$
View full question & answer→Question 553 Marks
Find the principal value of the following:
$\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)$
AnswerLet $\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)=\text{y}$ Then, $\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)=\text{y}$We know that the range of the principal value branch is $[0,\pi].$
Thus, $\cos\text{y}=-\frac{\sqrt3}{2}=\cos\Big(\frac{5\pi}{6}\Big)$ $\Rightarrow\text{y}=\frac{5\pi}{6}\in[0,\pi]$ Hence, the pricipal value of $\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)$ is $\frac{5\pi}{6}.$
View full question & answer→Question 563 Marks
Evaluate the following:
$\tan^{-1}\Big(\tan\frac{5\pi}{6}\Big)+\cos^{-1}\Big\{\cos\Big(\frac{13\pi}{6}\Big)\Big\}$
Answer$\tan^{-1}\Big(\tan\frac{5\pi}{6}\Big)+\cos^{-1}\Big\{\cos\Big(\frac{13\pi}{6}\Big)\Big\}$
$=\tan^{-1}\Big\{\tan\Big(\pi-\frac{5\pi}{6}\Big)\Big\}+\cos^{-1}\Big\{\cos\Big(2\pi+\frac{\pi}{6}\Big)\Big\}$
$=\tan^{-1}\Big\{-\tan\Big(\frac{\pi}{6}\Big)\Big\}+\cos^{-1}\Big\{\cos\Big(\frac{\pi}{6}\Big)\Big\}$
$=-\tan^{-1}\Big\{\tan\Big(\frac{\pi}{6}\Big)\Big\}+\cos^{-1}\Big\{\cos\Big(\frac{\pi}{6}\Big)\Big\}$
$=-\frac{\pi}{6}+\frac{\pi}{6}$
$=0$
View full question & answer→Question 573 Marks
Evaluate: $\sin^{-1}\Big(\sin\frac{3\pi}{5}\Big).$
Answer$\sin^{-1}\Big(\sin\frac{3\pi}{5}\Big).$
$ =\pi-\frac{3\pi}{5}$
$\begin{Bmatrix}\text{Since},\sin^{-1}(\sin\theta)=\begin{cases}-\pi-\theta,&\text{if }\theta\in\Big[-\frac{3\pi}{2},-\frac{\pi}{2}\Big]\\\theta,&\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\\\pi-\theta,&\text{if }\theta\in\Big[\frac{\pi}{2},\frac{\\\pi}{2}\Big]\\-2\pi+\theta,&\text{if }\theta\in\Big[\frac{3\pi}{2},\frac{5\pi}{2}\Big]\end{cases}\end{Bmatrix}$
$=\frac{2\pi}{5}$
View full question & answer→Question 583 Marks
Find the domain of $\text{f(x)}=2\cos^{-1}2\text{x}=\sin^{-1}\text{x}.$
AnswerDomain of $\cos^{-1}\text{x}$ lies in the interval [-1, 1]
$\therefore$ Domain of $\cos^{-1}(2\text{x})$ lies in the interval [-1, 1]
$\Rightarrow-1\leq2\text{x}\leq1$
$\Rightarrow\frac{-1}{2}\leq\text{x}\leq\frac{1}{2}$
Domain of $\cos^{-1}(2\text{x})$ is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$
Domain of $\cos^{-1}\text{x}$ lies in the interval [-1, 1]
$\therefore$ Domain of $\cos^{-1}(2\text{x})+\sin^{-1}\text{x}$ lies in the interval $\Big[-\frac{1}{2},\frac{1}{2}\Big].$
View full question & answer→Question 593 Marks
Write the value of $\sin\Big\{\frac{\pi}{3}-\sin^{-1}\Big(-\frac{1}{2}\Big)\Big\}.$
Answer$\sin\Big\{\frac{\pi}{3}-\sin^{-1}\Big(-\frac{1}{2}\Big)\Big\}$
$=\sin\Big\{\frac{\pi}{3}+\sin^{-1}\Big(\frac{1}{2}\Big)\Big\}$ $\big\{\text{Since},\sin^{-1}(-\theta)=-\sin^{-1}\theta\big\}$
$=\sin\Big\{\frac{\pi}{3}+\frac{\pi}{6}\Big\}$ $\Big\{\text{Since},\sin^{-1}\text{x}=\text{An angle in}\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\text{whose sine is x}\Big\}$
$=\sin\Big(\frac{\pi}{2}\Big)$
$=1$
Hence,
$\sin\Big\{\frac{\pi}{3}-\sin^{-1}\Big(-\frac{1}{2}\Big)\Big\}=1$
View full question & answer→Question 603 Marks
If $\big(\sin^{-1}\text{x}\big)^2+\big(\sin^{-1}\text{y}\big)^2+\big(\sin^{-1}\text{z}\big)^2=\frac{3}{4}\pi^2,$ find the value of $x^2 + y^2 + z^2$
AnswerRange of $\sin^{-1}$ is $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big].$
Given that $\big(\sin^{-1}\text{x}\big)^2+\big(\sin^{-1}\text{y}\big)^2+\big(\sin^{-1}\text{z}\big)^2=\frac{3}{4}\pi^2$
$\Rightarrow$ Each of $\sin^{-1}\text{x},\sin^{-1}\text{y}$ and $\sin^{-1}\text{z}$ takes value of $\frac{\pi}{2}.$
$\Rightarrow x = 1, y = 1$ and $z = 1$
$x^2 + y^2 + z^2 $
$= 1 + 1 + 1 $
$= 3$
View full question & answer→Question 613 Marks
For the principal values, evaluate the following:
$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)+\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)$
Answer$\sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)+\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)$
$=\sin^{-1}\Big\{\sin\Big(-\frac{\pi}{3}\Big)\Big\}=\cos^{-1}\Big(\cos\frac{\pi}{6}\Big)$
$=-\frac{\pi}{3}+\frac{\pi}{6}$ $\begin{bmatrix}\because\text{Range of sine is }\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big];-\frac{\pi}{3}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\\\ \\\text{and range of cosine is }[0,\pi];\frac{\pi}{6}\in[0,\pi]\end{bmatrix}$
$=-\frac{\pi}{6}$
$\therefore\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)+\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)=-\frac{\pi}{6}$
View full question & answer→Question 623 Marks
Write the following in the simplest form:
$\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}$
Answer$\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}$
$=\sin\begin{Bmatrix}\sin^{-1}\begin{pmatrix}\frac{2\sqrt{\frac{1-\text{x}}{1+\text{x}}}}{1+\Big(\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big)^2}\end{pmatrix}\end{Bmatrix}$ $\Big\{\text{Since},2\tan^{-1}\text{x}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}\Big\}$
$=\sin\begin{Bmatrix}\sin^{-1}\begin{pmatrix}\frac{2\sqrt{\frac{1-\text{x}}{1+\text{x}}}}{\frac{1+\text{x}+1-\text{x}}{1+\text{x}}}\end{pmatrix}\end{Bmatrix}$
$=2\sqrt{\frac{1-\text{x}}{1+\text{x}}}\times\frac{1+\text{x}}{2}$
$=\sqrt{1-\text{x}}\sqrt{1+\text{x}}$
$=\sqrt{1-\text{x}^2}$
Hence, $\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}=\sqrt{1-\text{x}^2}$
View full question & answer→Question 633 Marks
If $\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}+\sin^{-1}\frac{2\text{b}}{1+\text{b}^2}=2\tan^{-1}\text{x},$ prove that $\text{x}=\frac{\text{a}+\text{b}}{1-\text{ab}}.$
AnswerLet: $\text{a}=\tan\text{z}$ $\text{b}=\tan\text{y}$ Then, $\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}+\sin^{-1}\frac{2\text{b}}{1+\text{b}^2}=2\tan^{-1}\text{x}$$\Rightarrow\sin^{-1}\frac{2\tan\text{z}}{1+\tan^2\text{z}}+\sin^{-1}\frac{2\tan\text{y}}{1+\tan^2\text{y}}=2\tan^{-1}\text{x}$
$\Rightarrow\sin^{-1}(\sin2\text{z})+\sin^{-1}(\sin2\text{y})=2\tan^{-1}\text{x}$ $\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big]$ $\Rightarrow2\text{z}+2\text{y}=2\tan^{-1}\text{x}$ $\Rightarrow\tan^{-1}\text{a}+\tan^{-1}\text{b}=\tan^{-1}\text{x}$ $[\because\ \text{a}=\tan\text{z}\text{ and }\text{b}=\tan\text{y}]$ $\Rightarrow\tan^{-1}\frac{\text{a}+\text{b}}{1-\text{ab}}=\tan^{-1}\text{x}$ $\Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big]$ $\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{1-\text{ab}}$
View full question & answer→Question 643 Marks
Solve the following equation for x:
$\cos^{-1}\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)+\frac{1}{2}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{2\pi}{3}$
Answer$\cos^{-1}\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)+\frac{1}{2}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{2\pi}{3}$
$\Rightarrow\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\frac{1}{2}\times2\tan^{-1}\text{x}=\frac{2\pi}{3}$ $\Big[\because\ \tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=2\tan^{-1}\text{x}\Big]$
$\Rightarrow2\tan^{-1}\text{x}+\tan^{-1}\text{x}=\frac{2\pi}{3}$ $\Big[\because\ \cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)=2\tan^{-1}\text{x}\Big]$
$\Rightarrow3\tan^{-1}\text{x}=\frac{2\pi}{3}$
$\Rightarrow\tan^{-1}\text{x}=\frac{2\pi}{9}$
$\Rightarrow\text{x}=\tan\Big(\frac{2\pi}{9}\Big)$
View full question & answer→Question 653 Marks
Evaluate $\sin\Big(\tan^{-1}\frac{3}{4}\Big).$
AnswerWe know that
$\tan^{-1}\text{x}=\sin^{-1}\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
$\therefore\ \sin\Big(\tan^{-1}\frac{3}{4}\Big)=\sin\Bigg\{\sin^{-1}\Bigg(\frac{\frac{3}{4}}{\sqrt{1+\frac{9}{10}}}\Bigg)\Bigg\}$
$=\sin\Bigg\{\sin^{-1}\Bigg(\frac{\frac{3}{4}}{\frac{5}{4}}\Bigg)\Bigg\}$
$=\sin\Big(\sin^{-1}\frac{3}{5}\Big)$
$=\frac{3}{5}$ $\big[\because\ \sin\big(\sin^{-1}\text{x}\big)=\text{x}\big]$
$\therefore\ \Big(\tan^{-1}\frac{3}{4}\Big)=\frac{3}{5}$
View full question & answer→Question 663 Marks
Write the value of $\cos^2\Big(\frac{1}{2}\cos^{-1}\frac{3}{5}\Big).$
AnswerLet $\text{y}=\cos^{-1}\Big(\frac{3}{5}\Big)$
$\Rightarrow\cos\text{y}=\frac{3}{5}$
Now,
$\cos^2\Big(\frac{1}{2}\cos^{-1}\frac{3}{5}\Big)=\cos^{2}\Big(\frac{1}{2}\text{y}\Big)$
$=\frac{\cos\text{y}+1}{2}$ $\big[\because\ \cos2\text{x}=\cos^2\text{x}-1\big]$
$=\frac{\frac{3}{5}+1}{2}$
$=\frac{\frac{8}{5}}{2}$
$=\frac{4}{5}$
$\therefore\ \cos^2\Big(\frac{1}{2}\cos^{-1}\frac{3}{5}\Big)=\frac{4}{5}$
View full question & answer→Question 673 Marks
Write the difference between maximum and minimum values of $\sin^{-1}\text{x}$ for $\text{x}\in[-1,1].$
AnswerWe have to find the difference between maximum and minimum values of $\sin^{-1}\text{x}$ for $\text{x}\in[-1,1]$ We know that, $\sin^{-1}\text{x}=$ An angle in $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$ whose sin is x. So, minimum value of $\sin^{-1}\text{x}=-\frac{\pi}{2}$ maximum value of $\sin^{-1}\text{x}=\frac{\pi}{2}$Difference between maximum and minimum values of
$\sin^{-1}\text{x}=\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)$ $=\frac{\pi}{2}+\frac{\pi}{2}$ $=\pi$ The required difference $=\pi.$
View full question & answer→Question 683 Marks
For the principal values, evaluate the following:
$\sin^{-1}[\cos\{2\text{cosec}^{-1}(-2)\}]$
Answer$\text{cosec}^{-1}\text{x}$ represents an angle in $\Big[-\frac{\pi}{2},0\Big)\cup\Big(0,\frac{\pi}{2}\Big]$ whose cosecant is x. Let $\text{x}=\text{cosec}^{-1}(-2)$ $\Rightarrow\text{cosec x}=-2=\text{cosec c}\Big(-\frac{\pi}{6}\Big)$ $\Rightarrow\text{x}=-\frac{\pi}{6}$ $\sin^{-1}[\cos\{2\text{cosec}^{-1}(-2)\}]=\sin^{-1}\Big[\cos\Big\{2\times\Big(-\frac{\pi}{6}\Big)\Big\}\Big]$ $=\sin^{-1}\Big[\cos\Big(-\frac{\pi}{3}\Big)\Big]=\sin^{-1}\Big[\frac{1}{2}\Big]$ $\sin^{-1}\text{x}$ represents an angle in $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$ whose sin is x.Let $\text{x}=\sin^{-1}\Big[\frac{1}{2}\Big]$
$\Rightarrow\sin\text{x}=\frac{1}{2}=\sin\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=\frac{\pi}{6}$
View full question & answer→Question 693 Marks
Evaluate:
$\cos\Big(\tan^{-1}\frac{3}{4}\Big)$
AnswerWe have
$\cos\Big(\tan^{-1}\frac{3}{4}\Big)$
$=\cos\Bigg[\frac{1}{2}\cos^{-1}\Bigg(\frac{1-\big(\frac{3}{4}\big)^2}{1+\big(\frac{3}{4}\big)^2}\Bigg)\Bigg]$ $\Big[\because\ 2\tan^{-1}\text{x}=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\Big]$
$=\cos\Big[\frac{1}{2}\cos^{-1}\Big(\frac{7}{25}\Big)\Big]$
Let
$\text{y}=\cos^{-1}\Big(\frac{7}{25}\Big)$
$\Rightarrow\cos\text{y}=\frac{7}{25}$
Now,
$=\cos\Big[\frac{1}{2}\cos^{-1}\Big(\frac{7}{25}\Big)\Big]=\cos\Big[\frac{1}{2}\text{y}\Big]$
$=\sqrt{ \frac{\cos\text{y}+1}{2}}$ $\big[\because\ \cos2\text{x}=2\cos^2\text{x}-1\big]$
$=\sqrt{\frac{\frac{7}{25}+1}{2}}$
$=\sqrt{\frac{32}{50}}$
$=\frac{4}{5}$
$\therefore\ \cos\Big[\tan^{-1}\Big(\frac{3}{4}\Big)\Big]=\frac{4}{5}$
View full question & answer→Question 703 Marks
Prove the following results:
$\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}=\sin^{-1}\frac{1}{\sqrt5}$
Answer$\text{L.H.S}=\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}$
$=\tan^{-1}\Bigg(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\times\frac{2}{9}}\Bigg)$ $\Big[\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x+y}}{1-\text{xy}}\Big)\Big]$
$=\tan^{-1}\Bigg(\frac{\frac{17}{36}}{\frac{34}{36}}\Bigg)$
$=\tan^{-1}\frac{1}{2}$
$=\sin^{-1}\frac{\frac{1}{2}}{\sqrt{1+\big(\frac{1}{2}\big)^2}}$
$=\sin^{-1}\frac{1}{\sqrt5}=\text{R.H.S}$
View full question & answer→Question 713 Marks
If $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{\pi}{4},$ then write the value of x + y + xy.
AnswerGiven,
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)=\frac{\pi}{4}$ $\Big\{\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big\}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)=\tan^{-1}(1)$
$\Rightarrow\frac{\text{x}+\text{y}}{1-\text{xy}}=1$
$\Rightarrow\text{x}+\text{y}=1-\text{xy}$
$\Rightarrow\text{x}+\text{y}+\text{xy}=1$
So,
$\Rightarrow\text{x}+\text{y}+\text{xy}=1$
View full question & answer→Question 723 Marks
Solve: $\cos\big(\sin^{-1}\text{x}\big)=\frac{1}{6}$
Answer$\cos\big(\sin^{-1}\text{x}\big)=\frac{1}{6}$
$\Rightarrow\cos\Big(\cos^{-1}\sqrt{1-\text{x}^2}\Big)=\frac{1}{6}$
$\Rightarrow\sqrt{1-\text{x}^2}=\frac{1}{6}$
$\Rightarrow1-\text{x}^2=\frac{1}{36}$
$\Rightarrow1-\frac{1}{36}=\text{x}^2$
$\Rightarrow\text{x}^2=\frac{35}{36}$
$\Rightarrow\text{x}=\pm\frac{\sqrt{35}}{6}$
View full question & answer→Question 733 Marks
Write the value of $\sin\big(\cot^{-1}\text{x}\big).$
AnswerWe know$\cot^{-1}\text{x}=\tan^{-1}\frac{1}{\text{x}}$
Now, we have $\sin\big(\cot^{-1}\text{x}\big)=\sin\Big(\tan^{-1}\frac{1}{\text{x}}\Big)$ $=\sin\Bigg[\sin^{-1}\Bigg(\frac{\frac{1}{\text{x}}}{\sqrt{1+\frac{1}{\text{x}^2}}}\Bigg)\Bigg]$ $\Big[\because\ \tan^{-1}\text{x}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}}}\Big)\Big]$ $=\sin\Bigg[\sin^{-1}\Bigg(\frac{\frac{1}{\text{x}}}{\frac{\sqrt{\text{x}^2+1}}{\text{x}}}\Bigg)\Bigg]$ $=\sin\bigg(\sin^{-1}\frac{1}{\sqrt{\text{x}^2+1}}\bigg)$ $=\frac{1}{\sqrt{\text{x}^2+1}}$ $\big[\because\ \sin\big(\sin^{-1}\text{x}=\text{x}\big)\big]$ Hence, $\sin\big(\cot^{-1}\text{x}\big)=\frac{1}{\sqrt{\text{x}^2-1}}.$
View full question & answer→Question 743 Marks
If $4\sin^{-1}\text{x}+\cos^{-1}\text{x}=\pi,$ then what is the value of x?
AnswerGiven,
$4\sin^{-1}\text{x}+\cos^{-1}\text{x}=\pi$
$\Rightarrow4\sin^{-1}\text{x}+\frac{\pi}{2}-\sin^{-1}\text{x}=\pi$ $\Big\{\text{Since},\cos^{-1}\text{x}=\frac{\pi}{2}-\sin^{-1}\text{x}\Big\}$
$\Rightarrow3\sin^{-1}\text{x}=\pi-\frac{\pi}{2}$
$\Rightarrow3\sin^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\sin^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\sin^{-1}\text{x}=\sin^{-1}\Big(\frac{1}{2}\Big)$
$\Rightarrow\text{x}=\frac{1}{2}$
Hence,
$\text{x}=\frac{1}{2}$
View full question & answer→Question 753 Marks
Find the principal values of the following:
$\text{cosec}^{-1}\Big(\frac{2}{\sqrt3}\Big)$
AnswerLet $\text{cosec}^{-1}\Big(\frac{2}{\sqrt3}\Big)=\text{y}$ Then, $\text{cosec y}=\frac{2}{\sqrt3}$ We know that the range of the principal value branch is $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]-\{0\}$Thus,
$\text{cosec y}=\frac{2}{\sqrt3}=\text{cosec}\Big(\frac{\pi}{3}\Big)$ $\Rightarrow\text{y}=\frac{\pi}{3}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big],\text{y}\neq0$ Hence, the principal value of $\text{cosec}^{-1}\Big(\frac{2}{\sqrt3}\Big)$ is $\frac{\pi}{3}.$
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