Question 512 Marks
Evaluate the following:
$\sin\Big(\frac{1}{2}\cos^{-1}\frac{4}{5}\Big)$
Answer$\sin\Big(\frac{1}{2}\cos^{-1}\frac{4}{5}\Big)$
$=\sin\Bigg\{\frac{1}{2}\times2\sin^{-1}\pm\sqrt{\frac{1-\frac{4}{5}}{2}}\Bigg\}$ $\bigg[\because\ \cos^{-1}\text{x}=2\sin^{-1}\pm\sqrt{\frac{1-\text{x}}{2}}\bigg]$
$ =\sin\Big(\sin^{-1}\pm\frac{1}{\sqrt{10}}\Big)$
$\pm\frac{1}{\sqrt{10}}$
View full question & answer→Question 522 Marks
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{25\pi}{6}\Big)$
AnswerWe have
$\sec^{-1}\Big(\sec\frac{25\pi}{6}\Big)=\sec^{-1}\Big[\sec\Big(4\pi+\frac{\pi}{6}\Big)\Big]$
$=\sec^{-1}\Big[\sec\Big(\frac{\pi}{6}\Big)\Big]$
$=\frac{\pi}{6}$
View full question & answer→Question 532 Marks
Find the values:
$\tan^{-1}\bigg(\tan\frac{3\pi}{4}\bigg)$
AnswerFor $\tan^{-1}\left(\tan x\right)$ type of problem we have to always check whether the angle is in the principle range or not. This angle must be in the principle range $\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].$
$=\tan^{-1}\bigg(\tan\frac{4\pi-\pi}{4}\bigg)$
$=\tan^{-1}\bigg[\tan\bigg(\pi-\frac{\pi}{4}\bigg)\bigg]=\tan^{-1}\bigg[-\tan\frac{\pi}{4}\bigg]$
$=\tan^{-1}\tan\bigg(-\frac{\pi}{4}\bigg)=-\frac{\pi}{4}$
View full question & answer→Question 542 Marks
For the principal values, evaluate the following:
$\cos^{-1}\Big(\frac{1}{2}\Big)-2\sin^{-1}\Big(-\frac{1}{2}\Big)$
AnswerLet $\cos^{-1}\frac{1}{2}=\text{x}.$ Then, $\cos\text{x}=\frac{1}{2}=\cos\Big(\frac{\pi}{3}\Big)$ $\therefore\cos^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{3}$ Let $\sin^{-1}\Big(-\frac{1}{2}\Big)=\text{y}$ Then, $\sin\text{y}=-\frac{1}{2}=-\sin\Big(\frac{\pi}{6}\Big)$ $=\sin\Big(-\frac{\pi}{6}\Big)$ $\therefore\sin^{-1}\Big(-\frac{1}{2}\Big)=-\frac{\pi}{6}$
$\therefore\cos^{-1}\Big(\frac{1}{2}\Big)-2\sin^{-1}\Big(-\frac{1}{2}\Big)$ $=\frac{\pi}{3}-\Big(-\frac{2\pi}{6}\Big)=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}$
View full question & answer→Question 552 Marks
Find the values of the following:
$\tan^{-1}\Big\{2\cos\Big(2\sin^{-1}\frac{1}{2}\Big)\Big\}$
Answer$\tan^{-1}\Big\{2\cos\Big(2\sin^{-1}\frac{1}{2}\Big)\Big\}$ $=\tan^{-1}\Big\{2\cos\Big(2\times\frac{\pi}{6}\Big)\Big\}$ $=\tan^{-1}\Big\{2\cos\frac{\pi}{3}\Big\}$ $=\tan^{-1}\Big\{2\times\frac{1}{2}\Big\}$ $=\tan^{-1}(1)$ $=\frac{\pi}{4}$ Hence, $\tan^{-1}\Big\{2\cos\Big(2\sin^{-1}\frac{1}{2}\Big)\Big\}=\frac{\pi}{4}$
View full question & answer→Question 562 Marks
Write the value of $\cos\Big(\frac{\tan^{-1}\text{x}+\cot^{-1}\text{x}}{3}\Big),$ when $\text{x}=-\frac{1}{\sqrt3}$
Answer$\cos\Big(\frac{\tan^{-1}\text{x}+\cot^{-1}\text{x}}{3}\Big)=\cos\Big(\frac{\pi}{6}\Big)$ $\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=\frac{\sqrt3}{2}$
View full question & answer→Question 572 Marks
For the principal values of the following:
$\cot^{-1}\Big(-\frac{1}{\sqrt3}\Big)$
AnswerLet $\cot^{-1}\Big(-\frac{1}{\sqrt3}\Big)=\text{y}$
Then,
$\cot\text{y}=-\frac{1}{\sqrt3}$
We know that the range of the principal value branch is $(0,\pi).$
Thus,
$\cot\text{y}=-\frac{1}{\sqrt3}=\cot\Big(\frac{2\pi}{3}\Big)$
$\Rightarrow\text{y}=\frac{2\pi}{3}\in(0,\pi)$
Hence, the principal value of $\cot^{-1}\Big(-\frac{1}{\sqrt3}\Big)$ is $\frac{2\pi}{6}.$
View full question & answer→Question 582 Marks
Evaluate:
$\cot\Big(\sin^{-1}\frac{3}{4}+\sec^{-1}\frac{4}{3}\Big)$
Answer$\cot\Big(\sin^{-1}\frac{3}{4}+\sec^{-1}\frac{4}{3}\Big)$
$=\cot\Big(\sin^{-1}\frac{3}{4}+\cos^{-1}\frac{3}{4}\Big)$ $\Big[\because\ \sec^{-1}\text{x}=\cos^{-1}\frac{1}{\text{x}}\Big]$
$=\cot\Big(\frac{\pi}{2}\Big)$ $\Big[\because\ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=0$
View full question & answer→Question 592 Marks
Write the value of $\sin^{-1}\Big(\cos\frac{\pi}{6}\Big).$
Answer$\sin^{-1}\Big(\cos\frac{\pi}{6}\Big)=\sin^{-1}\Big\{\sin\Big(\frac{\pi}{2}-\frac{\pi}{9}\Big)\Big\}$ $\Big[\because\ \cos\text{x}=\sin\Big(\frac{\pi}{2}-\text{x}\Big)\Big]$
$=\sin^{-1}\Big\{\sin\Big(\frac{7\pi}{18}\Big)\Big\}$
$=\frac{7\pi}{18}$ $[\because\ \sin^{-1}(\sin\text{x})=\text{x}]$
$\therefore\ \sin^{-1}\Big(\cos\frac{\pi}{9}\Big)=\frac{7\pi}{18}$
View full question & answer→Question 602 Marks
Solve the following equation:
$\tan^{-1}\frac{1-x}{1+x}=\frac{1}{2}\tan^{-1}x,\left(x>0\right)$
Answer$\tan^{-1}\frac{1-x}{1+x}=\frac{1}{2}\tan^{-1}x$
$\Rightarrow\tan^{-1}1-\tan^{-1}x=\frac{1}{2}\tan^{-1}x$ $\left[\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}\right]$
$\Rightarrow\frac{\pi}{4}=\frac{3}{2}\tan^{-1}x$
$\Rightarrow\tan^{-1}x=\frac{\pi}{6}$
$\Rightarrow x=\tan\frac{\pi}{6}$
$\therefore x=\frac{1}{\sqrt{3}}$
View full question & answer→Question 612 Marks
Evaluate the following:
$\sin\Big(\tan^{-1}\frac{24}{7}\Big)$
Answer$\sin\Big(\tan^{-1}\frac{24}{7}\Big)$
$=\sin\begin{pmatrix}\sin^{-1}\frac{\frac{24}{7}}{\sqrt{1+\big(\frac{24}{7}\big)^2}}\end{pmatrix}$ $\Big[{\therefore\ \tan^{-1}}\text{x}=\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big]$
$=\sin\begin{pmatrix}\sin^{-1}\frac{\frac{24}{7}}{\sqrt{1+\frac{576}{49}}}\end{pmatrix}$
$=\sin\begin{pmatrix}\sin^{-1}\frac{\frac{24}{7}}{\sqrt{\frac{625}{49}}}\end{pmatrix}$
$=\sin\begin{pmatrix}\sin^{-1}\frac{\frac{24}{7}}{\frac{25}{7}}\end{pmatrix}$
$=\frac{24}{25}$
View full question & answer→Question 622 Marks
Write the value of $\cos^{-1}(\cos6).$
AnswerWe know that $\cos^{-1}(\cos\text{x})=\text{x}$
Now,
$\cos^{-1}(\cos6)=\cos^{-1}\{\cos(2\pi-6)\}$
$=2\pi-6$
View full question & answer→Question 632 Marks
Evaluate the following:
$\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)+\cot^{-1}\Big(\frac{1}{\sqrt3}\Big)+\tan^{-1}\Big(\sin\Big(-\frac{\pi}{2}\Big)\Big)$
Answer$\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)+\cot^{-1}\Big(\frac{1}{\sqrt3}\Big)+\tan^{-1}\Big(\sin\Big(-\frac{\pi}{2}\Big)\Big)$
$=\tan^{-1}\Big[\tan\Big(-\frac{\pi}{6}\Big)\Big]+\cot^{-1}\Big(\cot\frac{\pi}{3}\Big)+\tan^{-1}(-1)$
$=\tan^{-1}\Big[\tan\Big(-\frac{\pi}{6}\Big)\Big]+\cot^{-1}\Big(\cot\frac{\pi}{3}\Big)+\tan^{-1}\Big[\tan\Big(-\frac{\pi}{4}\Big)\Big]$
$=-\frac{\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4}$
$=-\frac{\pi}{12}$
View full question & answer→Question 642 Marks
Evaluate the following:
$\cos^{-1}\Big\{\cos\frac{5\pi}{4}\Big\}$
AnswerWe know that,
$\cos^{-1}\big(\cos\theta\big)=\begin{cases}-\theta,&\text{if }\theta\in[-\pi,0]\\\theta,&\text{if }\theta\in[0,\pi]\\2\pi-\theta,&\text{if }\theta\in[\pi,2\pi]\\-2\pi+\theta,&\text{if }\theta\in[2\pi,3\pi]\end{cases}$
We have
$\cos^{-1}\Big\{\cos\Big(\frac{5\pi}{4}\Big)\Big\}=\cos^{-1}\Big\{\cos\Big(2\pi-\frac{3\pi}{4}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\Big(\frac{3\pi}{4}\Big)\Big\}$
$=\frac{3\pi}{4}$
View full question & answer→Question 652 Marks
Evaluate the following:
$\cos\Big(\tan^{-1}\frac{24}{7}\Big)$
Answer$\cos\Big(\tan^{-1}\frac{24}{7}\Big)$
$=\cos\begin{bmatrix}\cos^{-1}\frac{1}{\sqrt{1+\big(\frac{24}{7}\big)^2}}\end{bmatrix}$ $\bigg[\because\ \tan^{-1}\text{x}=\cos^{-1}\frac{1}{\sqrt{1+\text{x}^2}}\bigg]$
$=\cos\begin{bmatrix}\cos^{-1}\frac{1}{\sqrt{1+\frac{576}{49}}}\end{bmatrix}$
$=\cos\bigg[\cos^{-1}\frac{1}{\frac{25}{7}}\bigg]$
$=\cos\Big[\cos^{-1}\frac{7}{25}\Big]$
$=\frac{7}{25}$
View full question & answer→Question 662 Marks
Solve the following equation for x:
$\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)-\frac{1}{2}\tan^{-1}\text{x}=0,$ where x>0
Answer$\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)=\frac{1}{2}\tan^{-1}\text{x}$
$\Rightarrow\tan^{-1}1-\tan^{-1}\text{x}=\frac{1}{2}\tan^{-1}\text{x}$
$\Big[\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$
$\Rightarrow\frac{\pi}{4}=\frac{3}{2}\tan^{-1}\text{x}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\tan\frac{\pi}{6}$
$\therefore\ \text{x}=\frac{1}{\sqrt3}$
View full question & answer→Question 672 Marks
Find the principal values:
$\cos^{-1}\left(-\frac{1}{2}\right)$
Answer$\text{Let}=\cos^{-1}\bigg(-\frac{1}{2}\bigg)=\text{Y where}\ 0\leq\text{Y}\leq{\pi}$
$ \therefore\ \ \ \ \cos\text{Y}=-\frac{1}{2} \text{where}\ 0\leq\text{Y} \leq {\pi}$
$\Rightarrow\ \ \ \text{Y}=\frac{2\pi}{3}$, $\left[\because\ \cos\frac{2\pi}{3}=\cos\left({\pi}-\frac{{\pi}}{3}\right)=-\cos\frac{{\pi}}{3}=-\frac{1}{2}\right]$
$\therefore$ required principal value $=\frac{2\pi}{3}$
View full question & answer→Question 682 Marks
Find the principal values:
$\cot^{-1}\left(\sqrt{3}\right)$
Answer$\text{Let y}=\cot^{-1}\left(\sqrt{3}\right)$, $\text{where}\ 0<\text{y}<{\pi}$
$ \therefore\cot\text{y}=\sqrt{3}$, $ \text{where}\ 0<\text{y}<{\pi}$
$ \therefore\text{Y}=\frac{{\pi}}{6}$
$\therefore$ required principal value $ =\frac{\pi}{6}$
View full question & answer→Question 692 Marks
Find the values of the following:
$\cos\big(\sec^{-1}\text{x}+\text{cosec}^{-1}\text{x}\big),|\text{x}|\geq1$
AnswerWe have
$\cos\big(\sec^{-1}\text{x}+\text{cosec}^{-1}\text{x}\big)$
$=\cos\frac{\pi}{2}$ $\Big[\because\ \sec^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=0$
$\therefore\ \cos\big(\sec^{-1}\text{x}+\text{cosec}^{-1}\text{x}\big),|\text{x}|\geq1$
View full question & answer→Question 702 Marks
For the principal values, evaluate the following:
$\cos^{-1}\frac{1}{2}+2\sin^{-1}\Big(\frac{1}{2}\Big)$
AnswerLet $\cos^{-1}\Big(\frac{1}{2}\Big)=\text{x}.$ Then, $\cos\text{x}=\frac{1}{2}=\cos\Big(\frac{\pi}{3}\Big)$ $\therefore\cos^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{3}$ Let $\sin^{-1}\Big(\frac{1}{2}\Big)=\text{y}$ Then, $\sin\text{y}=\frac{1}{2}=\sin\Big(\frac{\pi}{6}\Big)$ $\therefore\sin^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{6}$
$\therefore\cos^{-1}\Big(\frac{1}{2}\Big)+2\sin^{-1}\Big(\frac{1}{2}\Big)$ $=\frac{\pi}{3}+\frac{2\pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}$
View full question & answer→Question 712 Marks
Solve:
$4\sin^{-1}\text{x}={\pi}-\cos^{-1}\text{x}$
Answer$4\sin^{-1}\text{x}={\pi}-\cos^{-1}\text{x}$
$\Rightarrow4\sin^{-1}\text{x}={\pi}-\Big(\frac{\pi}{2}-\sin^{-1}\text{x}\Big)$
$\Big[\because\ \cos^{-1}\text{x}=\frac{\pi}{2}-\sin^{-1}\text{x}\Big]$
$\Rightarrow4\sin^{-1}\text{x}=\frac{\pi}{2}+\sin^{-1}\text{x}$
$\Rightarrow3\sin^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\sin^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\sin\frac{\pi}{6}=\frac{1}{2}$
View full question & answer→Question 722 Marks
Show that $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)=2\sin^{-1}\text{x}.$
AnswerWe have $\text{L.H.S}=\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ Putting $\text{x}=\sin\text{a},$ we get
$=\sin^{-1}\Big(2\sin\text{a}\sqrt{1-\sin^2\text{a}}\Big)$
$=\sin^{-1}(2\sin\text{a}\cos\text{a})$
$=\sin^{-1}(\sin2\text{a})$
$=2\text{a}$
$=2\sin^{-1}\text{a}$ $(\because\ \text{x}=\sin\text{a})$ View full question & answer→Question 732 Marks
Find the domain of $\text{f(x)}=\cot\text{x}+\cot^{-1}\text{x}$
AnswerDomain of $\cot\text{x}$ is $(0,\pi)$
Domain of $\cot^{-1}\text{x}$ is R.
So domain of $\cot\text{x}+\cot^{-1}\text{x}$ is R.
View full question & answer→Question 742 Marks
For the principal values, evaluate the following:
$\tan^{-1}\Big\{2\sin\Big(4\cos^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
Answer$\tan^{-1}\Big\{2\sin\Big(4\cos^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
$=\tan^{-1}\Big\{2\sin\Big[4\cos^{-1}\Big(\cos\frac{\pi}{6}\Big)\Big]$
$=\tan^{-1}\Big\{2\sin\Big[4\times\frac{\pi}{6}\Big]\Big\}$
$=\tan^{-1}\Big(2\sin\frac{2\pi}{3}\Big)$
$=\tan^{-1}\Big[2\times\Big(\frac{\sqrt3}{2}\Big)\Big]$
$=\tan^{-1}\big(\sqrt3\big)$
$=\tan^{-1}\Big[\tan\Big(\frac{\pi}{3}\Big)\Big]$
$=\frac{\pi}{3}$
View full question & answer→Question 752 Marks
Prove the following results:
$\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{13}=\tan^{-1}\frac{2}{9}$
Answer$\text{L.H.S=}\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{13}$
$=\tan^{-1}\Bigg(\frac{\frac{1}{7}+\frac{1}{13}}{1-\frac{1}{7}\times\frac{1}{13}}\Bigg)$
$\Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x+y}}{1-\text{xy}}\Big)\Big]$
$=\tan^{-1}\Bigg(\frac{\frac{20}{91}}{\frac{90}{91}}\Bigg)$
$=\tan^{-1}\frac{2}{9}=\text{R.H.S}$
View full question & answer→Question 762 Marks
Evaluate:
$\tan\Big\{\cos^{-1}\Big(-\frac{7}{25}\Big)\Big\}$
Answer$\tan\Big\{\cos^{-1}\Big(-\frac{7}{25}\Big)\Big\}$
$=\tan\Big\{\cos^{-1}\Big(\pi-\frac{7}{25}\Big)\Big\}$
$=-\tan\Big\{\cos^{-1}\Big(\frac{7}{25}\Big)\Big\}$
$=-\tan\begin{Bmatrix}\tan^{-1}\begin{bmatrix}\frac{\sqrt{1-\Big(\frac{7}{25}\Big)^3}}{\frac{7}{25}}\end{bmatrix}\end{Bmatrix}$
$=-\tan\Big\{\tan\frac{24}{7}\Big\}$
$=-\frac{24}{7}$
View full question & answer→Question 772 Marks
Evaluate:
$\text{cosec}\Big\{\cot^{-1}\Big(-\frac{12}{5}\Big)\Big\}$
Answer$\text{cosec}\Big\{\cot^{-1}\Big(-\frac{12}{5}\Big)\Big\}$
$=\text{cosec}\Big\{-\cot^{-1}\Big(\frac{12}{5}\Big)\Big\}$ $\big[\because\ \cot^{-1}(\text{x})=-\cot^{-1}(\text{x})\big]$
$=\text{cosec}\Big\{-\text{cosec}^{-1}\Big(\frac{13}{12}\Big)\Big\}$
$\Big[\therefore\ \cot^{-1}\Big(\frac{\text{b}}{\text{p}}\Big)=\text{cosec}^{-1}\Big(\frac{\text{h}}{\text{p}}\Big)\Big]$
$=\text{cosec}\Big\{\text{cosec}^{-1}\Big(\frac{13}{12}\Big)\Big\}$ $\big[\because\ \text{cosec}(-\text{x})=-\text{cosec}(\text{x})\big]$
$=-\frac{13}{12}$
View full question & answer→Question 782 Marks
Write the principal value of $\sin^{-1}\Big\{\cos\Big(\sin^{-1}\frac{1}{2}\Big)\Big\}$
Answer$\sin^{-1}\Big\{\cos\Big(\sin^{-1}\frac{1}{2}\Big)\Big\}$
$=\sin^{-1}\Big\{\cos\Big[\sin^{-1}\Big(\sin\frac{\pi}{3}\Big)\Big]\Big\}$
$=\sin^{-1}\Big[\cos\Big(\frac{\pi}{3}\Big)\Big]$
$=\sin^{-1}\Big[\frac{1}{2}\Big]$
$=\sin^{-1}\Big[\sin\Big(\frac{\pi}{3}\Big)\Big]$
$=\frac{\pi}{3}$
View full question & answer→Question 792 Marks
Evaluate:
$\sec\Big\{\cot^{-1}\Big(-\frac{5}{12}\Big)\Big\}$
Answer$\sec\Big\{\cot^{-1}\Big(-\frac{5}{12}\Big)\Big\}$
$\sec\Big\{-\cot^{-1}\Big(\frac{5}{12}\Big)\Big\}$
$\big[\because\ \cot^{-1}(-\text{x})=-\cot^{-1}(\text{x})\text{ for all }\text{x}\in(-1,1)\big]$
$=\sec\Big\{-\sec^{-1}\Big(\frac{13}{5}\Big)\Big\}$ $\Big[\because\ \cot^{-1}\Big(\frac{\text{b}}{\text{p}}\Big)=\sec^{-1}\Big(\frac{\text{h}}{\text{b}}\Big)\Big]$
$=\sec\Big\{\sec^{-1}\Big(\frac{13}{5}\Big)\Big\}$ $[\because\ \sec(\text{-x})=\sec\text{x}]$
$=\frac{13}{5}$
View full question & answer→Question 802 Marks
Find the values:
If $\sin\left(\sin^{-1}\frac{1}{5}+\cos^{-1} x\right)=1,$ then find the value of x
AnswerGiven: $\sin\bigg(\sin^{-1}\frac{1}{5}+\cos^{-1}x\bigg)=1=\sin\frac{\pi}{2}$
$ \Rightarrow\sin^{-1}\frac{1}{5}+\cos^{-1}x=\frac{\pi}{2}$
$ \Rightarrow\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}\frac{1}{5}$
$\Rightarrow\cos^{-1}x=\cos^{-1}\frac{1}{5}$ $\bigg[\because\sin^{-1}t+\cos^{-1}t=\frac{\pi}{2}\bigg]$
$ \Rightarrow\ x=\frac{1}{5}$
View full question & answer→Question 812 Marks
Write the value of $\tan^{-1}\Big(\frac{1}{\text{x}}\Big)$ for x < 0 in terms of $\cot^{-1}\text{x}.$
Answer$\tan^{-1}\Big(\frac{1}{\text{x}}\Big)=\tan^{-1}\Big(-\frac{1}{\text{x}}\Big)$ for x < 0
$=-\tan^{-1}\Big(\frac{1}{\text{x}}\Big)$
$=-\cot^{-1}\text{x}$
$=-\big(\pi-\cot^{-1}\text{x}\big)$
$=-\pi-\cot^{-1}\text{x}$
View full question & answer→Question 822 Marks
Evaluate the following:
$\cot\Big(\cot^{-1}\frac{3}{5}\Big)$
AnswerLet $\cot\Big(\cot^{-1}\frac{3}{5}\Big)=\text{y}$ where $\text{y}\in \Big[0,\frac{\pi}{2}\Big]$
$\Rightarrow \cos \text{y}=\frac{3}{5}$
$\cot \Big(\cos^{-1}\frac{3}{5}\Big)=\cot\text{y}$
To fing:
$\Rightarrow \text{As}\ 1+\tan^2\theta=\sec^2\theta$
$\Rightarrow \tan \text{y}=\sqrt{\sec^2\text{y}-1}$ where $\text{y}\in \Big[0, \frac{\pi}{2}\Big]$
$\Rightarrow \frac{1}{\cot\text{y}}=\sqrt{\Big(\frac{1}{\cos^2\text{y}}\Big)-1}$
$\Rightarrow \frac{1}{\cot\text{y}}=\sqrt{\Big(\frac{5}{3}\Big)^2-1}$
$\Rightarrow \frac{1}{\cot\text{y}}=\sqrt{\frac{16}{9}}$
$\Rightarrow \cot\text{y}=\frac{3}{4}$
$\Rightarrow \cot\Big(\cos^{-1}\frac{3}{5}\Big)=\frac{3}{4}$
View full question & answer→Question 832 Marks
Evaluate:
$\cos\Big\{\sin^{-1}\Big(-\frac{7}{25}\Big)\Big\}$
Answer$\cos\Big\{\sin^{-1}\Big(-\frac{7}{25}\Big)\Big\}$
$=\cos\Big\{-\sin^{-1}\Big(\frac{7}{25}\Big)\Big\}$
$=\cos\Big\{\sin^{-1}\Big(\frac{7}{25}\Big)\Big\}$
$=\cos\Big\{\cos^{-1}\sqrt{1-\big(\frac{7}{25}\big)^2}\Big\}$
$=\cos\Big\{\cos^{-1}\frac{24}{25}\Big\}$
$=\frac{24}{25}$
View full question & answer→Question 842 Marks
Evaluate the following:
$\sin^{-1}\Big(\sin\frac{17\pi}{8}\Big)$
AnswerWe know
$\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
$\sin^{-1}\Big(\sin\frac{17\pi}{8}\Big)$
$=\sin^{-1}\Big(\sin\Big(2\pi+\frac{\pi}{8}\Big)\Big)$
$=\sin^{-1}\Big(\sin\Big(\frac{\pi}{8}\Big)\Big)$
$=\frac{\pi}{8}$
View full question & answer→Question 852 Marks
Evaluate the following:
$\sin^{-1}(\sin2)$
AnswerWe know $\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$ We have
$\sin^{-1}(\sin2)=\sin^{-1}\{\sin(\pi-2)\}=\pi-2$ View full question & answer→Question 862 Marks
Evaluate the following:
$\cot^{-1}\Big(\cot\frac{19\pi}{6}\Big)$
Answer$\cot^{-1}\Big(\cot\frac{19\pi}{6}\Big)$
$=\cot^{-1}\Big[\cot\Big(3\pi+\frac{\pi}{6}\Big)\Big]$
$=\cot^{-1}\Big(\cot\frac{\pi}{6}\Big)$
$=\frac{\pi}{6}$
View full question & answer→Question 872 Marks
If $\tan^{-1}\big(\sqrt{3}\big)+\cot^{-1}\text{x}=\frac{\pi}{2},$ find x.
AnswerWe know that $\tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2} $
We have
$\Rightarrow\tan^{-1}\big(\sqrt3\big)+\cot^{-1}\text{x}=\frac{\pi}{2}$
$ \Rightarrow\tan^{-1}\big(\sqrt3\big)=\frac{\pi}{2}-\cot^{-1}\text{x}$
$ \Rightarrow\tan^{-1}\big(\sqrt3\big)=\tan^{-1}\text{x}$
$\Rightarrow\text{x}=\sqrt3$
View full question & answer→Question 882 Marks
Write the value of $\cot^{-1}(-\text{x})$ for all $\text{x}\in\text{R}$ in terms of $\cot^{-1}\text{x}$
AnswerWe know that $\cot^{-1}(-\text{x})=\pi-\cot^{-1}(\text{x})$
Therefore, the value of $\cot^{-1}(-\text{x})$ for all $\text{x}\in\text{R}$ in term of $\cot^{-1}\text{x}$ is $\pi-\cot^{-1}(\text{x}).$
View full question & answer→Question 892 Marks
Evaluate the following:
$\sin^{-1}\Big(\sin\frac{7\pi}{6}\Big)$
AnswerWe know
$\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
$\sin^{-1}\Big(\sin\frac{7\pi}{6}\Big)$
$=\sin^{-1}\Big(\sin\Big(\pi+\frac{\pi}{6}\Big)\Big)$
$=\sin^{-1}\Big(\sin\Big(-\frac{\pi}{6}\Big)\Big)$
$=-\frac{\pi}{6}$
View full question & answer→Question 902 Marks
Write the following function in the simplest form:
$\tan^{-1}\bigg(\frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}}\bigg), x<{\pi}$
Answer$\tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}=\tan^{-1}\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}=\tan^{-1}\tan\frac{x}{2}=\frac{x}{2}$
View full question & answer→Question 912 Marks
Find the values:
$\sin^{-1}\bigg(\sin\frac{2\pi}{3}\bigg)$
AnswerFor $\sin^{-1}\left(\sin x\right)$ type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range $\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].$
$=\sin^{-1}\bigg(\sin\frac{3\pi-\pi}{3}\bigg)$
$=\sin^{-1}\bigg[\sin\bigg(\pi-\frac{\pi}{3}\bigg)\bigg]$
$=\sin^{-1}\sin\frac{\pi}{3}=\frac{\pi}{3}$
View full question & answer→Question 922 Marks
Find the principal values:
$\tan^{-1}\left(1\right)+\cos^{-1}\bigg(-\frac{1}{2}\bigg)+\sin^{-1}\bigg(-\frac{1}{2}\bigg)$
Answer$ \tan^{-1}\left(1\right)+\cos^{-1}\bigg(\frac{-1}{2}\bigg)+\sin^{-1}\bigg(\frac{-1}{2}\bigg)$ $=\tan^{-1}\tan\frac{\pi}{4}+\cos^{-1}\bigg(-\cos\frac{\pi}{3}\bigg)+\sin^{-1}\bigg(-\sin\frac{\pi}{6}\bigg)$
$=\frac{\pi}{4}+\cos\bigg({\pi}-\frac{\pi}{3}\bigg)+\sin^{-1}\sin\bigg(-\frac{\pi}{6}\bigg)=\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}$
$=\frac{3\pi+8\pi-2\pi}{12}=\frac{9\pi}{12}=\frac{3\pi}{4}$
View full question & answer→Question 932 Marks
Find the principal value of the following:
$\cos^{-1}\Big(\tan\frac{3\pi}{4}\Big)$
AnswerLet $\cos^{-1}\Big(\tan\frac{3\pi}{4}\Big)=\text{y}$
Then,
$\cos\text{y}=\tan\frac{3\pi}{4}$
We know that the principal value branch is $[0,\pi].$
Thus,
$\cos\text{y}=\tan\frac{3\pi}{4}=-1=\cos(\pi)$
$\Rightarrow\text{y}=\pi\in[0,\pi]$
Hence the principal value of $\cos^{-1}\Big(\tan\frac{3\pi}{4}\Big)$ is $\pi.$
View full question & answer→Question 942 Marks
Evaluate the following:
$\sin\Big(2\tan^{-1}\frac{2}{3}\Big)+\cos\Big(\tan^{-1}\sqrt3\Big)$
Answer$\sin\Big(2\tan^{-1}\frac{2}{3}\Big)+\cos\Big(\tan^{-1}\sqrt3\Big)$
$=\sin\Bigg(\sin^{-1}\frac{2\times\frac{2}{3}}{1+\frac{4}{9}}\Bigg)+\cos\Bigg(\cos^{-1}\frac{1}{\sqrt{1+\big(\sqrt3\big)^2}}\Bigg)$
$=\sin\Big(\sin^{-1}\frac{12}{13}\Big)+\cos\Big(\cos^{-1}\frac{1}{2}\Big)$
$=\sin\Big(\sin^{-1}\frac{12}{13}\Big)+\cos\Big(\cos^{-1}\frac{1}{2}\Big)$
$=\frac{12}{13}+\frac{1}{2}$
$=\frac{37}{26}$
View full question & answer→Question 952 Marks
Evaluate the following:
$\cos^{-1}\Big\{\cos\Big(\frac{13\pi}{6}\Big)\Big\}$
AnswerWe know that,
$\cos^{-1}\big(\cos\theta\big)=\begin{cases}-\theta,&\text{if }\theta\in[-\pi,0]\\\theta,&\text{if }\theta\in[0,\pi]\\2\pi-\theta,&\text{if }\theta\in[\pi,2\pi]\\-2\pi+\theta,&\text{if }\theta\in[2\pi,3\pi]\end{cases}$
We have
$\cos^{-1}\Big\{\cos\Big(\frac{13\pi}{6}\Big)\Big\}=\cos^{-1}\Big\{\cos\Big(2\pi+\frac{\pi}{6}\Big)\Big\}$
$\cos^{-1}\Big\{\cos\Big(\frac{\pi}{6}\Big)\Big\}$
$=\frac{\pi}{6}$
View full question & answer→Question 962 Marks
Write the value of $\cos^{-1}(\cos1540^\circ).$
AnswerWe know that
$\cos^{-1}(\cos\text{x})=\text{x}$
Now,
$\cos^{-1}(\cos1540^\circ)=\cos^{-1}\{\cos(1440+100^\circ)\}$
$=\cos^{-1}\{\cos(100^\circ)\}$ $[\because\ \cos(4\pi+100^\circ)=\cos100^\circ]$
$=100^\circ$
View full question & answer→Question 972 Marks
Evaluate the following:
$\cot^{-1}\Big\{\cot\Big(\frac{21\pi}{4}\Big)\Big\}$
AnswerWe have
$\cot^{-1}\Big\{\cot\Big(\frac{21\pi}{4}\Big)\Big\}$
$=\cot^{-1}\Big[\cot\Big(5\pi+\frac{\pi}{4}\Big)\Big]$
$=\cot^{-1}\Big(\cot\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
View full question & answer→Question 982 Marks
Find the values:
$\cot\left(\tan^{-1}a+\cot^{-1}a\right)$
Answer$\cot\left(\tan^{-1}a+\cot^{-1}a\right)=\cot\frac{\pi}{2}=0$ $\bigg[\because\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\bigg]$
View full question & answer→Question 992 Marks
Solve:
$\sin^{-1}\text{x}=\frac{\pi}{6}+\cos^{-1}\text{x}$
Answer$\frac{\pi}{2}-\cos^{-1}\text{x}=\frac{\pi}{6}+\cos^{-1}\text{x}$
$\frac{\pi}{3}=2\cos^{-1}\text{x}$
$\cos^{-1}\text{x}=\frac{\pi}{6}$
$\text{x}=\frac{\sqrt3}{2}$
View full question & answer→Question 1002 Marks
Solve:
$5\tan^{-1}\text{x}+3\cot^{-1}\text{x}={2\pi}$
Answer$5\tan^{-1}\text{x}+3\cot^{-1}\text{x}={2\pi}$
$\Rightarrow5\tan^{-1}\text{x}+3\Big(\frac{\pi}{2}-\tan^{-1}\text{x}\Big)={2\pi}$
$\Big[\because\ \cot^{-1}\text{x}=\frac{\pi}{2}-\tan^{-1}\text{x}\Big]$
$\Rightarrow5\tan^{-1}\text{x}+\frac{3\pi}{2}-3\tan^{-1}\text{x}={2\pi}$
$\Rightarrow2\tan^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{4}$
$\Rightarrow\text{x}=\tan\frac{\pi}{4}=1$
View full question & answer→