Question 512 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 522 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 532 Marks
Find the value of $2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$
Answer$2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$
$=2\sec^{-1}\Big(\sec\frac{\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{\pi}{6}\Big)$
$=2\times\frac{\pi}{3}+\frac{\pi}{6}$
$=\frac{5\pi}{6}$
View full question & answer→Question 542 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 552 Marks
If $\cos\Big(\sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}\Big)=0$ find the value of X.
Answer$\cos\Big(\sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}\Big)=0$
$\Rightarrow\cos\Big(\sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}\Big)=\cos\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$\therefore\ \text{x}=\frac{2}{5}$ $\Big[\because\ \sin^{-1}\text{y}+\cos^{-1}\text{y}=\frac{\pi}{2}\Big]$
View full question & answer→Question 562 Marks
Find the principal value of the following:
$\sec^{-1}(2)$
AnswerWe know that, for any $\text{x}\in\text{R},\sec^{-1}\text{x}$ represents an angle in $[0,\pi]-\Big\{\frac{\pi}{2}\Big\}.$$\sec^{-1}(2)=$ An angle is $[0,\pi]-\Big\{\frac{\pi}{2}\Big\}$ whose secant is 2
$=\frac{\pi}3{}$
$\therefore\sec^{-1}(2)=\frac{\pi}{3}$
View full question & answer→Question 572 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 582 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 592 Marks
Find the domain of$\sec^{-1}(3\text{x}-1)$
AnswerDomain of $\sec^{-1}\text{x}$ lies in the interval $(-\infty,-1]\cup[1,\infty)$
$\therefore$ Domain of $\sec^{-1}(3\text{x}-1)$ lies in the interval $(-\infty,-1]\cup[1,\infty)$
$\Rightarrow-\infty\leq3\text{x}-1\leq-1$ and $1\leq3\text{x}-1\leq\infty$
$\Rightarrow-\infty\leq3\text{x}\leq0$ and $2\leq3\text{x}\leq\infty$
$\Rightarrow-\infty\leq\text{x}\leq0$ and $\frac{2}{3}\leq\text{x}\leq\infty$
Domain of $\sec^{-1}\text{x}$ lies in the interval $(-\infty,0]\cup\Big[\frac{2}{3},\infty\Big).$
View full question & answer→Question 602 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 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\big(\sin^{-1}\text{x}+\cos^{-1}\text{x}\big),|\text{x}|\leq1$
AnswerWe have$|\text{x}|\leq1$
$\Rightarrow\pm\text{x}\leq1$
$\Rightarrow\text{x}\leq1$ or $\Rightarrow-\text{x}\leq1$
$\Rightarrow\text{x}\leq1$ or $\Rightarrow\text{x}\geq1$
$\Rightarrow\in[-1, 1]$
Now,
$\cos\big(\sin^{-1}\text{x}+\cos^{-1}\text{x}\big)=\cos\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 632 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 642 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 652 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 662 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 672 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 682 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 692 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 702 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 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
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 732 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 742 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 752 Marks
If $\cos\big(\tan^{-1}\text{x}+\cot^{-1}\sqrt{3}\big)=0,$ find the value of x.
Answer$\cos\big(\tan^{-1}\text{x}+\cot^{-1}\sqrt{3}\big)=0$
$\Rightarrow\cos\big(\tan^{-1}\text{x}+\cot^{-1}\sqrt{3}\big)=\cos\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\tan^{-1}\text{x}+\cot^{-1}\sqrt3=\frac{\pi}{2}$
$\Rightarrow\text{x}=\sqrt3$ $\Big[\tan^{-1}\text{y}+\cot^{-1}\text{y}=\frac{\pi}{2}\Big]$
View full question & answer→Question 762 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 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
Find the domain of definition of $\text{f(x)}=\cos^{-1}\big(\text{x}^2-4\big).$
AnswerFor $\cos^{-1}\big(\text{x}^2-4\big)$ to be defined$-1\leq\text{x}^2-4\leq1$
$\Rightarrow3\leq\text{x}^2\leq5$
$\Rightarrow\text{x}\in\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
Hence, the domain of f(x) is $\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
View full question & answer→Question 792 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 802 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 812 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 822 Marks
Write the value of $\sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big).$
AnswerWe know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$ and $\cos^{-1}(-\text{x})=\pi-\cos^{-1}\text{x}.$
$\therefore\ \sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big)$
$=\sin^{-1}\Big(\frac{1}{3}\Big)-\Big[\pi-\cos^{-1}\Big(\frac{1}{3}\Big)\Big]$
$=\sin\Big(\frac{1}{3}\Big)-\pi+\cos^{-1}\Big(\frac{1}{3}\Big)$
$=\Big[\sin^{-1}\Big(\frac{1}{3}\Big)+\cos^{-1}\Big(\frac{1}{3}\Big)\Big]-\pi$
$=\frac{-\pi}{2}$ $\Big[\because\ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=-\frac{\pi}{2}$
$\therefore\ \sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big)=-\frac{\pi}{2}$
View full question & answer→Question 832 Marks
For the principal values of the following:
$\cot^{-1}\Big(\sqrt3\Big)$
Answer$\cot^{-1}\Big(\sqrt3\Big)$ represents an angle in $(0,\pi)$ whose cotagent is x.
Let $\text{x}=\cot^{-1}\big(\sqrt3\big)$
$\Rightarrow\cot\text{x}=\sqrt3=\cot\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=\frac{\pi}{6}$
$\therefore$ Principal value of $\cot^{-1}\Big(\sqrt3\Big)$ is $\frac{\pi}{6}.$
View full question & answer→Question 842 Marks
For the principal values of the following:
$\cot^{-1}\Big(\tan\frac{3\pi}{4}\Big)$
AnswerLet $\cot^{-1}\Big(\tan\frac{3\pi}{4}\Big)=\text{y}$
Then,
$\cot\text{y}=\tan\frac{3\pi}{4}$
We know that the range of the principal value branch is $(0,\pi).$
Thus,
$\cot\text{y}=\tan\frac{3\pi}{4}=-1=\cot\Big(\frac{3\pi}{4}\Big)$
$\Rightarrow\text{y}=\frac{3\pi}{4}\in(0,\pi)$
Hence, the principal value of $\cot^{-1}\Big(\tan\frac{3\pi}{4}\Big)$ is $\frac{3\pi}{4}.$
View full question & answer→Question 852 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 862 Marks
Write the value of $\tan^{-1}\Big\{2\sin\Big(2\cos^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
Answer$\tan^{-1}\Big\{2\sin\Big(2\cos^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
$=\tan^{-1}\Big\{2\sin\Big[\cos^{-1}2\Big(\frac{\sqrt3}{2}\Big)^2-1\Big]\Big\}$
$=\tan^{-1}\Big[2\sin\Big(\cos^{-1}\frac{1}{2}\Big)\Big]$
View full question & answer→Question 872 Marks
What is the principal value of $\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)$
AnswerWe have, $\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)$$=\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big\{\sin\Big(\pi-\frac{\pi}{3}\Big)\Big\}$ $\Big[\because\ \Big(\pi-\frac{2\pi}{3}\Big)\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$=\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big\{\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{2\pi}{3}+\frac{\pi}{3}$
$=\pi$
$\therefore\ \cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)=\pi$
View full question & answer→Question 882 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 892 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 902 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 912 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
$\sin^{-1}(\sin2)=\sin^{-1}\{\sin(\pi-2)\}=\pi-2$
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Find the principal values of the following:
$\text{cosec}^{-1}(-2)$
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 $\text{x}=\text{cosec}^{-1}(-2)$
$\Rightarrow\text{cosec x}=-2=\text{cosec}\Big(-\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=\Big(-\frac{\pi}{6}\Big)$
$\therefore$ Principal value of $\text{cosec}^{-1}(-2)$ is $-\frac{\pi}{6}.$
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If -1 < x < 0, then write the value of $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{cx}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big).$
AnswerLet $\text{x}=-\tan\text{y}$
Where $0<\text{y}<\frac{\pi}{2}$
Then,
$\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$=\sin^{-1}\Big(\frac{-2\tan\text{y}}{1+\tan^2\text{y}}\Big)+\cos^{-1}\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)$
$=\sin^{-1}\{-\sin(2\text{y})\}+\cos^{-1}\{\cos(2\text{y})\}$
$=-\sin^{-1}\{\sin(2\text{y})\}+\cos^{-1}\{\cos(2\text{y})\}$
$=-2\text{y}+2\text{y}$
$=0$
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Evaluate the following:
$\cot^{-1}\Big(\cot\frac{19\pi}{6}\Big)$
Answer$\cot^{-1}\Big(\cot\frac{19\pi}{6}\Big)$
$=\cot^{-1}\Big[\cot\Big(3\pi+\frac{\pi}{6}\Big)\Big]$
$=\cot^{-1}\Big(\cot\frac{\pi}{6}\Big)$
$=\frac{\pi}{6}$
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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}$
$\Rightarrow\text{x}=\sqrt3$
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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 $\pi-\cot^{-1}(\text{x}).$
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What is the principal value of $\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)$
Answer$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)$ $-\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)$ $\Big\{\text{Since},\sin^{-1}(-\theta)=-\sin^{-1}(\theta)\Big\} $$=-\frac{\pi}{3} $ $\Big\{\text{Since},\sin^{-1}\text{x}=\text{An angle in }\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\text{ whose sine is x}\Big\}$
Hence, $\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)=-\frac{\pi}{3}$
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Evaluate the following:
$\sin^{-1}\Big(\sin\frac{7\pi}{6}\Big)$
AnswerWe know
$\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
$\sin^{-1}\Big(\sin\frac{7\pi}{6}\Big)$
$=\sin^{-1}\Big(\sin\Big(\pi+\frac{\pi}{6}\Big)\Big)$
$=\sin^{-1}\Big(\sin\Big(-\frac{\pi}{6}\Big)\Big)$
$=-\frac{\pi}{6}$
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Write the following function in the simplest form:
$\tan^{-1}\bigg(\frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}}\bigg), x<{\pi}$
Answer$\tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}=\tan^{-1}\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}=\tan^{-1}\tan\frac{x}{2}=\frac{x}{2}$
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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 $\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].$
$=\sin^{-1}\bigg(\sin\frac{3\pi-\pi}{3}\bigg)$
$=\sin^{-1}\bigg[\sin\bigg(\pi-\frac{\pi}{3}\bigg)\bigg]$
$=\sin^{-1}\sin\frac{\pi}{3}=\frac{\pi}{3}$
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