Question 512 Marks
If $\cot\Big(\cos^{-1}\frac{3}{5}+\sin^{-1}\text{x}\Big)=0,$ find the values of x.
Answer$\cot(\text{z})=0$ means $\text{z}=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}.....$
$\cos^{-1}\Big(\frac{3}{5}\Big)+\sin^{-1}\text{x}=\text{n}\pi+\frac{\pi}{2}$
$\sin^{-1}\text{x}=\text{n}\pi+\frac{\pi}{2}-\cos^{-1}\Big(\frac{3}{5}\Big)$
$\sin^{-1}\text{x}=\text{n}\pi+\sin^{-1}\Big(\frac{3}{5}\Big)$
$\text{x}=\sin\Big(\text{n}\pi+\sin^{-1}\Big(\frac{3}{5}\Big)\Big)$
$=(-1)^{\text{n}}\sin\Big(\sin^{-1}\Big(\frac{3}{5}\Big)\Big)$
$\text{x}=(-1)^{\text{n}}\frac{3}{5}$
View full question & answer→Question 522 Marks
If x > 1, then write the value of $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ in terms of $\tan^{-1}\text{x.}$
Answer$\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=\pi-2\tan^{-1}\text{x}$ $\Big[\because\ 2\tan^{-1}\text{x}=\pi-\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{ for }\text{x}>1\Big]$
View full question & answer→Question 532 Marks
Find the set values of $\text{cosec}^{-1}\Big(\frac{\sqrt3}{2}\Big)$
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. Domain of $\text{cosec}^{-1}\text{x}$
is $(-\infty,-1]\cup[1,\infty)$ $\frac{\sqrt3}{2}\notin(-\infty,-1]\cup[1,\infty)$
Hence,
$\text{cosec}^{-1}\Big(\frac{\sqrt3}{2}\Big)$ does not exist or its $\phi.$ View full question & answer→Question 542 Marks
Evaluate the following:
$\cot^{-1}\Big\{\cot\Big(-\frac{8\pi}{3}\Big)\Big\}$
AnswerWe have
$\cot^{-1}\Big[\cot\Big(-\frac{8\pi}{3}\Big)\Big]$
$=\cot^{-1}\Big[-\cot\Big(\frac{8\pi}{3}\Big)\Big]$
$=\cot^{-1}\Big[-\cot\Big(3\pi-\frac{\pi}{3}\Big)\Big]$
$=\cot^{-1}\Big(\cot\frac{\pi}{3}\Big)$
$=\frac{\pi}{3}$
View full question & answer→Question 552 Marks
Write the value of $\sin^{-1}(\sin1550^\circ).$
AnswerWe know that $\sin^{-1}(\sin\text{x})=\text{x}.$
Now,
$\sin^{-1}(\sin1550^\circ)=\sin^{-1}\{\sin(1620^\circ-1550^\circ)\}$
$[\because\ \sin\text{x}=\sin(1620^\circ-\text{x})]$
$=\sin^{-1}\{\sin(70^\circ)\}$
$=70^\circ$
$\because\ \sin^{-1}(\sin1550^\circ)=70^\circ$
View full question & answer→Question 562 Marks
Find the domain of $\text{f(x)}=\cos^{-1}\text{x}=\cos\text{x}$
AnswerFor $\cos^{-1}\text{x} $ to be defined. $-1\leq\text{x}\leq1$ Now, $\cos\text{x}$ is defined for all real values. So, domain of $\cos\text{x}$ is R. Domain of f(x) is
$\text{R}\cap[-1,1]=[-1,1].$ View full question & answer→Question 572 Marks
If $\cos^{-1}\text{x}+\cos^{-1}\text{y}=\frac{\pi}{4},$ find the value of $\sin^{-1}\text{x}+\sin^{-1}\text{y}$
Answer$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\frac{\pi}{4},$
$\Rightarrow\frac{\pi}{2}-\sin^{-1}\text{x}+\frac{\pi}{2}-\sin^{-1}\text{y}=\frac{\pi}{4}$
$\Big[\because\ \cos^{-1}\text{x}=\frac{\pi}{2}-\sin^{-1}\text{x}\Big]$
$\Rightarrow{\pi}-\big(\sin^{-1}\text{x}+\sin^{-1}\text{y}\big)=\frac{\pi}{4}$
$\Rightarrow\sin^{-1}\text{x}+\sin^{-1}\text{y}=\frac{3\pi}{44}$
View full question & answer→Question 582 Marks
Find the domain of the following functions:
$\text{f(x)}=\sin^{-1}\text{x}+\sin\text{x}$
AnswerLet f(x) = g(x) + h(x), where
Therefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)
The domain of g(x) is [-1, 1]
The domain of h(x) is $(-\infty,\infty)$
Therefore, the intersection of g(x) and h(x) is [-1, 1]
Hence, the domain is [-1, 1]
View full question & answer→Question 592 Marks
Write the value of $\tan^{-1}\Big\{\tan\Big(\frac{15\pi}{4}\Big)\Big\}.$
AnswerWe have
$\tan^{-1}\Big\{\tan\Big(\frac{15\pi}{4}\Big)\Big\}$
$=\tan^{-1}\Big\{\tan\Big(4\pi-\frac{\pi}{4}\Big)\Big\}$
$=\tan^{-1}\Big\{-\tan\Big(\frac{\pi}{4}\Big)\Big\}$ $[\because\ \tan(4\pi-\text{x})=-\tan\text{x}]$
$=\tan^{-1}\Big\{\tan\Big(-\frac{\pi}{4}\Big)\Big\}$
$=-\frac{\pi}{4}$ $\big[\because\ \tan^{-1}(\tan\text{x})=\text{x}\big]$
$\therefore\ \tan^{-1}\Big\{\tan\Big(\frac{15\pi}{4}\Big)\Big\}=-\frac{\pi}{4}$
View full question & answer→Question 602 Marks
Evaluate the following:
$\sec\Big(\sin^{-1}\frac{12}{13}\Big)$
Answer$\sec\Big(\sin^{-1}\frac{12}{13}\Big)$
$=\sec\Bigg[\cos^{-1}\sqrt{1-\Big(\frac{12}{13}\Big)^2}\Bigg]$ $\Big[\because\ \sin^{-1}\text{x}=\cos^{-1}\sqrt{1-\text{x}^2}\Big]$
$=\sec\Big[\cos^{-1}\Big(\sqrt{1-\frac{144}{169}}\Big)\Big]$
$=\sec\Big[\cos^{-1}\Big(\sqrt{\frac{25}{169}}\Big)\Big]$
$=\sec\Big[\cos^{-1}\frac{5}{13}\Big]$
$=\sec\Big[\sec^{-1}\frac{13}{5}\Big]$
$=\frac{13}{5}$
View full question & answer→Question 612 Marks
Evaluate the following:
$\sin^{-1}(\sin3)$
AnswerWe know $\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$ We have
$\sin^{-1}(\sin3)=\sin^{-1}\{\sin(\pi-3)\}=\pi-3$ View full question & answer→Question 622 Marks
Write the value of $\cos^{-1}\Big(\cos\frac{5\pi}{4}\Big).$
Answer$\cos^{-1}\Big(\cos\frac{5\pi}{4}\Big)\neq\frac{5\pi}{4}$ as $\frac{5\pi}{4}$ does not lie between 0 and $\pi.$
We have
$\cos^{-1}\Big(\cos\frac{5\pi}{4}\Big)$
$=\cos^{-1}\Big\{\cos\Big(2\pi-\frac{3\pi}{4}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\Big(\frac{3\pi}{4}\Big)\Big\}$
$=\frac{3\pi}{4}$
View full question & answer→Question 632 Marks
Write the value of $\cos^{-1}\Big(\cos\frac{14\pi}{3}\Big)$
Answer$\cos^{-1}\Big(\cos\frac{14\pi}{3}\Big)=\cos\Big[\cos\Big(4\pi+\frac{2\pi}{3}\Big)\Big]$
$=\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)$
$=\frac{2\pi}{3}$
View full question & answer→Question 642 Marks
Evaluate:
$\cot\Big\{\sec^{-1}\Big(-\frac{13}{5}\Big)\Big\}$
Answer$\cot\Big\{\sec^{-1}\Big(-\frac{13}{5}\Big)\Big\}$
$\cot\Big\{\sec^{-1}\Big(\pi-\frac{13}{5}\Big)\Big\}$
$=-\cot\Big\{\sec^{-1}\Big(\frac{13}{5}\Big)\Big\}$
$=-\cot\begin{Bmatrix}\tan^{-1}\begin{pmatrix}\frac{\sqrt{1-\Big(\frac{5}{13}\Big)^3}}{\frac{5}{13}}\end{pmatrix}\end{Bmatrix}$
$=-\cot\Big\{\tan^{-1}\Big(\frac{12}{5}\Big)\Big\}$
$=-\cot\Big\{\cot^{-1}\Big(\frac{5}{12}\Big)\Big\}$
$=-\frac{5}{12}$
View full question & answer→Question 652 Marks
Evaluate the following:
$\sec^{-1}\Big\{\sec\Big(-\frac{7\pi}{3}\Big)\Big\}$
Answer$\sec^{-1}\Big\{\sec\Big(-\frac{7\pi}{3}\Big)\Big\}$
$=\sec^{-1}\Big\{\sec\Big(-2\pi-\frac{\pi}{3}\Big)\Big\}$
$=\sec^{-1}\Big\{\sec\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{\pi}{3}$
View full question & answer→Question 662 Marks
Evaluate the following:
$\sin^{-1}\Big(\sin\frac{13\pi}{7}\Big)$
AnswerWe know
$\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
We have
$\sin^{-1}\Big(\sin\frac{13\pi}{7}\Big)=\sin^{-1}\Big\{\sin\Big(2\pi-\frac{\pi}{7}\Big)\Big\}$
$=\sin^{-1}\Big(\sin-\frac{\pi}{7}\Big)$
$=-\frac{\pi}{7}$
View full question & answer→Question 672 Marks
Find the principal value of the following:
$\cos^{-1}\Big(\sin\frac{4\pi}{3}\Big)$
AnswerLet $\cos^{-1}\Big(\sin\frac{4\pi}{3}\Big)=\text{y}$ Then, $\cos\text{y}=\sin\frac{4\pi}{3}$ We know that the range of the principal value branch is $[0,\pi].$ Thus,
$\cos\text{y}=\sin\frac{4\pi}{3}$ $=-\frac{\sqrt3}{2}=\cos\Big(\frac{5\pi}{6}\Big)$ $\Rightarrow\text{y}=\frac{5\pi}{6}\in[0,\pi]$ Hence, the principal value of $\cos^{-1}\Big(\sin\frac{4\pi}{3}\Big)$ is $\frac{5\pi}{6}.$ View full question & answer→Question 682 Marks
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{5\pi}{4}\Big)$
AnswerWe have
$\sec^{-1}\Big(\sec\frac{5\pi}{4}\Big)$
$=\sec^{-1}\Big(\sec\Big(\pi+\frac{\pi}{4}\Big)\Big)$
$=\sec^{-1}\Big(-\sec\Big(\frac{\pi}{4}\Big)\Big)$
$=\sec^{-1}\big(-\sqrt2\big)$
$=\frac{3\pi}{4}$
View full question & answer→Question 692 Marks
Evaluate the following:
$\sin^{-1}\Big\{\Big(\sin-\frac{17\pi}{8}\Big)\Big\}$
AnswerWe know
$\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
$\sin^{-1}\Big(\sin-\frac{17\pi}{8}\Big)=\sin^{-1}\Big(-\sin\frac{17\pi}{8}\Big)$
$=\sin^{-1}\Big\{-\sin\Big(2\pi+\frac{\pi}{8}\Big)\Big\}$
$=\sin^{-1}\Big(-\sin\frac{\pi}{8}\Big)$
$=-\frac{\pi}{8}$
View full question & answer→Question 702 Marks
For the principal values, evaluate the following:
$\sec^{-1}\big(\sqrt2\big)+2\text{cosec}^{-1}\big(-\sqrt2\big)$
Answer$\sec^{-1}\big(\sqrt2\big)+2\text{cosec}^{-1}\big(-\sqrt2\big)$
$\sec^{-1}\Big(\sec\frac{\pi}{4}\Big)+2\text{cosec}^{-1}\Big[\text{cosec}\Big(-\frac{\pi}{4}\Big)\Big]$
$=\frac{\pi}{4}-2\times\frac{\pi}{4}$
$=\frac{\pi}{4}-\frac{\pi}{2}$
$=-\frac{\pi}{4}$
View full question & answer→Question 712 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 722 Marks
Evaluate the following:
$\cot^{-1}\Big\{2\cos\Big(\sin^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
Answer$\cot^{-1}\Big\{2\cos\Big(\sin^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
$=\cot^{-1}\Big\{2\cos\Big[\sin^{-1}\Big(\sin\frac{\sqrt3}{2}\Big)\Big]\Big\}$
$=\cot^{-1}\Big(2\cos\frac{\pi}{3}\Big)$
$=\cot^{-1}\Big(2\times\frac{1}{2}\Big)$
$=\cot^{-1}(1)$
$=\cot^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
View full question & answer→Question 732 Marks
For the principal values of the following:
$\cot^{-1}\Big(-\sqrt3\Big)$
AnswerLet $\cot^{-1}\Big(-\sqrt3\Big)=\text{y}$
Then,
$\cot\text{y}=-\sqrt3$
We know that the range of the principal value branch is $(0,\pi).$
Thus,
$\cot\text{y}=-\sqrt3=\cot\Big(\frac{5\pi}{6}\Big)$
$\Rightarrow\text{y}=\frac{5\pi}{6}\in(0,\pi)$
Hence, the principal value of $\cot^{-1}\Big(-\sqrt3\Big)$ is $\frac{5\pi}{6}.$
View full question & answer→Question 742 Marks
Find the domain of the following functions:
$\text{f(x)}=\sin^{-1}\text{x}^2$
AnswerTo the domain of sin-1y which is [-1, 1]
$\therefore$ x2 $\in$ [0, 1] as x2 can, not be negative
$\because$ x $\in$ [-1, 1]
Hence, the domaine is [-1, 1]
View full question & answer→Question 752 Marks
For the principal values, evaluate the following:
$\text{cosec}^{-1}\Big(2\tan\frac{11\pi}{6}\Big)$
Answer$\text{cosec}^{-1}\Big(2\tan\frac{11\pi}{6}\Big)$
$=\text{cosec}^{-1}\Big[2\times\Big(-\frac{1}{\sqrt3}\Big)\Big]$
$=\text{cosec}^{-1}\Big[-\frac{2}{\sqrt3}\Big]$
$=\text{cosec}^{-1}\Big[\text{cosec}\Big(-\frac{\pi}{3}\Big)\Big]$
$=-\frac{\pi}{3}$
View full question & answer→Question 762 Marks
Evaluate the following:
$\text{cosec}^{-1}\Big(-\frac{2}{\sqrt3}\Big)+2\cot^{-1}(-1)$
Answer$\text{cosec}^{-1}\Big(-\frac{2}{\sqrt3}\Big)+2\cot^{-1}(-1)$
$=-\frac{\pi}{3}+2\times\Big(\frac{3\pi}{4}\Big)$
$-\frac{\pi}{3}+\frac{3\pi}{4}$
$=\frac{7\pi}{6}$
View full question & answer→Question 772 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 782 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 792 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 802 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 812 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 822 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 832 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 842 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 852 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 862 Marks
Evaluate:
$\sin\Big(\tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}\Big)\text{ for }\text{x}<0$
Answer$\sin\Big(\tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}\Big)$
$=\sin\Bigg(-\pi+\tan^{-1}\Bigg(\frac{\text{x}+\frac{1}{\text{x}}}{\text{x}-\frac{1}{\text{x}}}\Bigg)\Bigg)$
$=\sin\big(-\pi+\tan^{-1}(\infty)\big)$
$=\sin\Big(-\pi+\frac{\pi}{2}\Big)$
$=\sin\Big(-\frac{\pi}{2}\Big)$
$=-1$
View full question & answer→Question 872 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 882 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 892 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 902 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 912 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 922 Marks
Evaluate the following:
$\text{cosec}\Big(\cos^{-1}\frac{3}{5}\Big)$
Answer$\text{cosec}\Big(\cos^{-1}\frac{3}{5}\Big)$
$=\text{cosec}\Big(\text{cosec}^{-1}\frac{5}{4}\Big)$ $\Big[\therefore\ \cos^{-1}\Big(\frac{\text{p}}{\text{h}}\Big)=\text{cosec}^{-1}\Big(\frac{\text{h}}{\text{p}}\Big)\Big]$
$=\frac{5}{4}$
View full question & answer→Question 932 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 942 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 952 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 962 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 972 Marks
Evaluate the following:
$\cos^{-1}\Big\{\cos\Big(\frac{4\pi}{3}\Big)\Big\}$
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
$\cos^{-1}\Big\{\cos\Big(\frac{4\pi}{3}\Big)\Big\}=\cos^{-1}\Big\{\cos\Big(2\pi-\frac{2\pi}{3}\Big)\Big\}$
$\cos^{-1}\Big\{\cos\Big(\frac{2\pi}{3}\Big)\Big\}$
$=\frac{2\pi}{4}$
View full question & answer→Question 982 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 992 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 1002 Marks
Evaluate:
$\text{cosec}\Big\{\cot^{-1}\Big(-\frac{12}{5}\Big)\Big\}$
Answer$\text{cosec}\Big\{\cot^{-1}\Big(-\frac{12}{5}\Big)\Big\}$
$=\text{cosec}\Big\{-\cot^{-1}\Big(\frac{12}{5}\Big)\Big\}$ $\big[\because\ \cot^{-1}(\text{x})=-\cot^{-1}(\text{x})\big]$
$=\text{cosec}\Big\{-\text{cosec}^{-1}\Big(\frac{13}{12}\Big)\Big\}$
$\Big[\therefore\ \cot^{-1}\Big(\frac{\text{b}}{\text{p}}\Big)=\text{cosec}^{-1}\Big(\frac{\text{h}}{\text{p}}\Big)\Big]$
$=\text{cosec}\Big\{\text{cosec}^{-1}\Big(\frac{13}{12}\Big)\Big\}$ $\big[\because\ \text{cosec}(-\text{x})=-\text{cosec}(\text{x})\big]$
$=-\frac{13}{12}$
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