Question 12 Marks
Prove that:
$3\sin^{-1}\text{x}=\sin^{-1}(3\text{x}-4\text{x}^3),\text{x}\in\Big[-\frac{1}{2},\frac{1}{2}\Big]$
Answer$3 \sin^{-1}\text{x}=\sin^{-1}(3\text{x}-4\text{x}^3)$
$\text{Let}\ \sin^{-1}\text{x}=\theta$
$\text{x}=\sin\theta$
$\therefore\sin3\theta=3\sin\theta-4\sin^3\theta$
$\sin3\theta=3\text{x}-4\text{x}^3$
Case I$\text{ When}\frac{-1}{2}\leq\text{x}\leq\frac{1}{2}$
$\frac{-1}{2}\leq\sin\theta\leq\frac{1}{2}$
$\frac{-\pi}{6}\leq\theta\leq\frac{\pi}{6}$
$\frac{-\pi}{2}\leq3\theta\leq\frac{\pi}{2}$
$\text{Also}\ \frac{-1}{2}\leq\text{x}\leq\frac{1}{2}\Rightarrow-1\leq3\text{x}-4\text{x}^3\leq1$
$\sin3\theta=3\text{x}-4\text{x}^3$
$3\theta=\sin^{-1}(3\text{x}-4\text{x}^3)$
$3\sin^{-1}\text{x}=\sin^{-1}(3\text{x}-4\text{x}^3)$
View full question & answer→Question 22 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 32 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 42 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 52 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 62 Marks
Write the principal value of $\tan^{-1}1+\cos^{-1}\Big(-\frac{1}{2}\Big)$
Answer$\tan^{-1}1+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)+\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)$
$=\frac{\pi}{4}+\frac{2\pi}{3}$
$=\frac{11\pi}{3}$
View full question & answer→Question 72 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 82 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 92 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 102 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 112 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 122 Marks
Evaluate $\cos\Big[\cos^{-1}\Big(\frac{-\sqrt{3}}{2}\Big)+\frac{\pi}{6}\Big]$
AnswerWe have, $\cos\Big[\cos^{-1}\Big(\frac{-\sqrt{3}}{2}\Big)+\frac{\pi}{6}\Big]$
$=\cos\Big[\cos^{-1}\Big(\cos\frac{5\pi}{6}\Big)+\frac{\pi}{6}\Big]$
$\Big[\because\cos\frac{5\pi}{6}=\frac{-\sqrt{3}}{2}\Big]$
$=\cos\Big(\frac{5\pi}{6}+\frac{\pi}{6}\Big)$
$\{\because\ \cos^{-1}\cos\text{x}=\text{x};\ \text{x}\in[0,\pi]\}$
$=\cos\Big(\frac{6\pi}{6}\Big)$
$=\cos(\pi)=-1$
View full question & answer→Question 132 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 142 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 152 Marks
What is the principal value of $\sin^{-1}\Big(-\frac{1}{2}\Big)?$
AnswerLet $ \text{y}=\sin^{-1}\Big(-\frac{1}{2}\Big)$
Then,
$ \sin\text{y}=-\frac{1}{2}=\sin\Big(-\frac{\pi}{6}\Big) $
$\text{y}=-\frac{\pi}{6}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$
Here, $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$ is the range of the pricipal value branch of the inverse sine function.
$\therefore\ \sin^{-1}\Big(-\frac{1}{2}\Big)=-\frac{\pi}{6}$
View full question & answer→Question 162 Marks
Find the principal values:
$\cos^{-1}\bigg(\frac{1}{2}\bigg)+2\sin^{-1}\bigg(\frac{1}{2}\bigg)$
Answer$\cos^{-1}\bigg(\frac{1}{2}\bigg)+2\sin^{-1}\bigg(\frac{1}{2}\bigg)$ $=\cos^{-1}\cos\frac{\pi}{3}+2\sin^{-1}\sin\frac{\pi}{6}$
$=\frac{\pi}{3}+2\bigg(\frac{\pi}{6}\bigg)=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}$
View full question & answer→Question 172 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 182 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 192 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 202 Marks
$\sin^{-1}\Big\{\cos\Big(\sin^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
Answer$\sin^{-1}\Big\{\cos\Big(\sin^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
$=\sin^{-1}\Big\{\cos\Big(\frac{\pi}{3}\Big)\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt3}{2}\Big\}=\frac{\pi}{6}$
View full question & answer→Question 212 Marks
Prove the following:
$3\cos^{-1}x=\cos^{-1}\left(4x^{3}-3x\right), x\in\bigg[\frac{1}{2},1\bigg]$
AnswerWe know that: $\cos3\theta=4\cos^{3}\theta-3\cos\theta$
Putting $\cos\theta=x\ \ \Rightarrow\ \ \theta=\cos^{-1}x$
$\therefore\ \ \cos3\theta=4x^{3}-3x\ \ \Rightarrow\ \ \ 3\theta=\cos^{-1}\left(3x-4x^{3}\right)$
Putting $\theta=\cos^{-1}x,$
$3\cos^{-1}x=\cos^{-1}\left(4x^{3}-3x\right)$ Proved.
View full question & answer→Question 222 Marks
Prove the following:
$3\sin^{-1}x=\sin^{-1}\left(3x-4x^{3}\right), x\in\bigg[-\frac{1}{2},\frac{1}{2}\bigg]$
AnswerWe know that: $\sin3\theta=3\sin\theta-4\sin^{3}\theta$
Putting $\sin\theta=x\ \ \Rightarrow\ \ \theta=\sin^{-1}x$
$\therefore\ \ \sin3\theta=3x-4x^{3}\ \ \Rightarrow\ \ 3\theta=\sin^{-1}\left(3x-4x^{3}\right)$
Putting $\theta=\sin^{-1}x,$
$ 3\sin^{-1}x=\sin^{-1}\left(3x-4x^{3}\right)$ Proved.
View full question & answer→Question 232 Marks
Prove that:
$\tan^{-1}\sqrt{x}=\frac{1}{2}\cos^{-1}\bigg(\frac{1-x}{1+x}\bigg),x\in\left[0,1\right]$
Answer$\text{Let}x=\tan^2\theta.\text{Then},\sqrt{x}=\tan\theta\Rightarrow\theta=\tan^{-1}\sqrt{x}.$
$\therefore\frac{1-x}{1+x}=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta$
Now, we have:
$\text{R.H.S.}=\frac{1}{2}\cos^{-1}\bigg(\frac{1-x}{1+x}\bigg)=\frac{1}{2}\cos^{-1}\left(\cos2\theta\right)$
$=\frac{1}{2}\times2\theta=\theta=\tan^{-1}\sqrt{x}=\text{L.H.S.}$
View full question & answer→Question 242 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 252 Marks
Find the principal values:
$\tan^{-1}\left({-\sqrt{3}}\right)$
Answer$\text{Let Y}=\tan^{-1}\left(-\sqrt{3}\right)$, $\text{where} -\frac{{\pi}}{2}<\text{Y}<\frac{{\pi}}{2}$
$\therefore \ \tan\text{Y}=-\sqrt{3}$, $\text{where} -\frac{{\pi}}{2}<\text{Y}<\frac{{\pi}}{2}$
$\therefore \ \tan\text{Y}=-\tan \frac{{\pi}}{3}=\tan \left(-\frac{{\pi}}{3}\right) \text{where}-\frac{{\pi}}{2}<\text{Y}<\frac{{\pi}}{2}$
$\therefore \ \text{Y}=-\frac{{\pi}}{3}$
$\therefore$ required principal value $=-\frac{{\pi}}{3}.$
View full question & answer→Question 262 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 272 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 282 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 292 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 302 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 312 Marks
Find the domain of the following functions:
$\text{f(x)}=\sin^{-1}\text{x}^2$
AnswerTo the domain of $sin^{-1}y$ which is $[-1, 1]$
$\therefore x^2 \in [0, 1]$ as $x^2$ can, not be negative
$\because$ x $\in$ [-1, 1]
Hence, the domaine is [-1, 1]
View full question & answer→Question 322 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 332 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 342 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 352 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 362 Marks
Find the principal values:
$\text{cosec}^{-1}(2)$
Answer$\text{Let y}==\text{cosec}^{-1}(2)$, $\text{where}\ \text{y}\in\bigg[\frac{-\pi}{2},0\bigg]\cup\bigg(0,\frac{{\pi}}{2}\bigg)$
$\therefore\ \ \text{cosec y}=2$, $\text{where y}\in\bigg[\frac{-\pi}{2},0\bigg]\cup\bigg(0,\frac{\pi}{2}\bigg)$
$\therefore\text{Y}=\frac{\pi}{6}$
$\therefore$ required principal value $ =\frac{\pi}{6}$
View full question & answer→Question 372 Marks
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{13\pi}{4}\Big)$
AnswerWe have
$\sec^{-1}\Big(\sec\frac{13\pi}{4}\Big)=\sec^{-1}\Big[\sec\Big(4\pi-\frac{3\pi}{4}\Big)\Big]$
$=\sec^{-1}\Big[\sec\Big(\frac{3\pi}{4}\Big)\Big]$
$=\frac{3\pi}{4}$
View full question & answer→Question 382 Marks
Find the principal value of the following:
$\cos^{-1}\Big(-\frac{1}{\sqrt2}\Big)$
AnswerLet $\cos^{-1}\Big(-\frac{1}{\sqrt2}\Big)=\text{y}.$ Then, $\cos\text{y}=-\frac{1}{\sqrt2}=-\cos\Big(\frac{\pi}{4}\Big)$ $=\cos\Big(\pi-\frac{\pi}{4}\Big)=\cos\Big(\frac{3\pi}{4}\Big).$
We know that the range of the principal value branch of $\cos^{-1}$ is $[0,\pi]$ and $\cos\Big(\frac{3\pi}{4}\Big)=-\frac{1}{\sqrt2}$
Therefore, the principal value of $\cos^{-1}\Big(-\frac{1}{\sqrt2}\Big)$ is $\frac{3\pi}{4}.$
View full question & answer→Question 392 Marks
Find the principal values:
$\sin^{-1}\bigg(-\frac{1}{2}\bigg)$
Answer$\text{Let y}= \sin^{-1}\left(\frac{-1}{2}\right)\text {where}-\frac{\pi}{2}\leq\text{Y}\leq\frac{\pi}{2}$
$\therefore\ \text{sin Y} =-\frac{1}{2}\ \ \ \text{where}\ -\frac{\pi}{2}\leq \text{Y}\ \leq\ \frac{\pi}{2}$
$\therefore\text{Y}=\ -\frac{\pi}{6}\ \bigg[\because\sin\bigg(-\frac{\pi}{6}\bigg)=-\sin\frac{\pi}{6}=-\frac{1}{2}\bigg]$
$\therefore$ required principal value $ =-\frac{\pi}{6}$
View full question & answer→Question 402 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 412 Marks
Evaluate:
$\cot\Big(\tan^{-1}\text{a}+\cot^{-1}\text{a}\Big)$
Answer$\cot\Big(\tan^{-1}\text{a}+\cot^{-1}\text{a}\Big)$
$=\cot\Big(\frac{\pi}{2}\Big)$ $\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=0$
View full question & answer→Question 422 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 432 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 442 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 452 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 462 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 472 Marks
Write the value of $\tan^{-1}\sqrt3+\cot^{-1}\sqrt3$
Answer$\tan^{-1}\sqrt3+\cot^{-1}\sqrt3$
$=\frac{\pi}{3}+\frac{\pi}{6}$
$=\frac{2\pi+\pi}{6}$
$=\frac{3\pi}{6}$
$=\frac{\pi}{2}$
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Find the principal values:
$\text{cosec}^{-1}(-\sqrt{2})$
Answer$\text{Let cosec}^{-1}(-\sqrt{2})=\text{y}$
$-\sqrt{2}=\text{cosec y}-\text{cosec}\frac{\pi}{4}=\text{cosec y}$
$\text{Hence y}=-\frac{\pi}{4}$
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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 502 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 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 $\frac{2\pi}{6}.$
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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}}$
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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\}$ $\Big[\because\ \Big(\pi-\frac{2\pi}{3}\Big)\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$=\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big\{\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{2\pi}{3}+\frac{\pi}{3}$
$=\pi$
$\therefore\ \cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)=\pi$
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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
$\sin^{-1}(\sin2)=\sin^{-1}\{\sin(\pi-2)\}=\pi-2$
View full question & answer→Question 922 Marks
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 $-\frac{\pi}{6}.$
View full question & answer→Question 932 Marks
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,
$\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$=\sin^{-1}\Big(\frac{-2\tan\text{y}}{1+\tan^2\text{y}}\Big)+\cos^{-1}\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)$
$=\sin^{-1}\{-\sin(2\text{y})\}+\cos^{-1}\{\cos(2\text{y})\}$
$=-\sin^{-1}\{\sin(2\text{y})\}+\cos^{-1}\{\cos(2\text{y})\}$
$=-2\text{y}+2\text{y}$
$=0$
View full question & answer→Question 942 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 952 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 962 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 972 Marks
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}$
View full question & answer→Question 982 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 992 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 1002 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 1012 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 1022 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 1032 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 1042 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 1052 Marks
Write the value of $\sin^{-1}\Big(\sin\frac{3\pi}{5}\Big)$
Answer$\sin^{-1}\Big(\sin\frac{3\pi}{5}\Big)=\sin^{-1}\Big[\sin\Big(\pi-\frac{2\pi}{5}\Big)\Big]$
$=\sin^{-1}\Big(\sin\frac{2\pi}{5}\Big)$
$=\frac{2\pi}{5}$
View full question & answer→Question 1062 Marks
Find the domain of$\sec^{-1}\text{x}-\tan^{-1}\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 $\Big[0,\frac{\pi}{2}\Big)\cup\Big[\pi,\frac{3\pi}{2}\Big)$
The domain of h(x) is $\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$
Therefore, the intersection of g(x) and h(x) is $\text{R}-(\text{n}\pi,\text{n}\in\text{Z})$
View full question & answer→Question 1072 Marks
Write the orincipal value of $\cos^{-1}(\cos680^\circ)$
Answer$\cos^{-1}(\cos680^\circ)=\cos^{-1}[\cos(720^\circ-680^\circ)]$
$=\cos^{-1}(\cos40^\circ)$
$=40^\circ$
View full question & answer→Question 1082 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 1092 Marks
Find the principal values:
$\sec^{-1}\bigg(\frac{2}{\sqrt{3}}\bigg)$
Answer$\text{Let Y} =\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$, $\sec\text{Y}= \frac{2}{\sqrt{3}} $
$\sec\text{Y}=\sec\frac{\pi}{6}$
$\text{Y}=\frac{\pi}{6}$
View full question & answer→Question 1102 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 1112 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→Question 1122 Marks
If $\sin^{-1}\Big(\frac{1}{3}\Big)+\cos^{-1}\text{x}=\frac{\pi}{2},$ find x.
AnswerWe know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$We have
$\sin^{-1}\Big(\frac{1}{3}\Big)+\cos^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\sin^{-1}\Big(\frac{1}{3}\Big)=\frac{\pi}{2}-\cos^{-1}\text{x}$
$\Rightarrow\sin^{-1}\Big(\frac{1}{3}\Big)=\sin^{-1}\text{x}$
$\Rightarrow\text{x}=\frac{1}{3}$
View full question & answer→Question 1132 Marks
Evaluate the following:
$\tan\Big(\cos^{-1}\frac{8}{17}\Big)$
Answer$\tan\Big(\cos^{-1}\frac{8}{17}\Big)$
$=\begin{Bmatrix}\tan^{-1}\frac{\sqrt{1-\Big(\frac{8}{17}\Big)^2}}{\frac{8}{17}}\end{Bmatrix}$ $\bigg[\because\ \cos^{-1}\text{x}=\tan^{-1}\bigg(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\bigg)\bigg]$
$=\tan\Bigg(\tan^{-1}\frac{\frac{15}{17}}{\frac{8}{17}}\Bigg)$
$=\frac{15}{8}$
View full question & answer→Question 1142 Marks
Find the domain of the following function:$\text{f(x)}=\sin^{-1}\text{x}+\sin^{-1}2\text{x}$
AnswerLet f(x) = g(x) + h(x), where
Therefore the domain of f(x) is given by intersection of the domain of g(x) and h(x)
The domain of g(x) is [-1, 1]
The domain of h(x) is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$
Therefore, the intersection of g(x) and h(x) is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$
Hence, the domain is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$
View full question & answer→Question 1152 Marks
Write the value of $\cos\Big(2\sin^{-1}\frac{1}{2}\Big).$
Answer$\cos\Big(2\sin^{-1}\frac{1}{2}\Big)$
$=\cos\Big(2\times\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\}$
$=\cos\Big(\frac{\pi}{3}\Big)$
$=\frac{1}{2}$
Hence,
$\cos\Big(2\sin^{-1}\frac{1}{2}\Big)=\frac{1}{2}$
View full question & answer→Question 1162 Marks
Write the value of $2\sin^{-1}\frac{1}{2}+\cos^{-1}\Big(-\frac{1}{2}\Big).$
Answer$2\sin^{-1}\frac{1}{2}+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$=\sin^{-1}2\times\frac{1}{2}\sqrt{1-\Big(\frac{1}{2}\Big)^2}+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$=\sin^{-1}\frac{\sqrt3}{2}+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$=\sin^{-1}\Big(\sin\frac{\pi}{3}\Big)+\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)$
$=\frac{\pi}{3}+\frac{2\pi}{3}$
$=\pi$
View full question & answer→Question 1172 Marks
Find the principal values:
$\cos^{-1}\bigg(-\frac{1}{\sqrt{2}}\bigg)$
Answer$\text{Let Y}=\cos^{-1}\bigg(-\frac{1}{\sqrt2}\bigg) \text{where}\ 0\leq\text{Y}\leq {\pi}$,
$ \therefore \cos\text{Y}=-\frac{1}{\sqrt{2}}\ \ \text{where}\ 0\leq\text{Y}\leq {\pi}$
$\therefore\cos\text{Y}=-\cos\frac{{\pi}}{4}=\cos\bigg({\pi}-\frac{{\pi}}{4}\bigg)=\cos\frac{3\pi}{4}$
$\therefore\text{Y}=\frac{3\pi}{4}$
$\therefore$ required principal value $=\frac{3\pi}{4} $
View full question & answer→Question 1182 Marks
Evaluate:
$\cos\Big(\sec^{-1}\text{x}+\text{cosec}^{-1}\text{x}\Big)$
Answer$\cos\Big(\sec^{-1}\text{x}+\text{cosec}^{-1}\text{x}\Big)$
$=\cos\Big(\frac{\pi}{2}\Big)$
$=0$
View full question & answer→Question 1192 Marks
Evaluate the following:
$\cot^{-1}\frac{1}{\sqrt3}-\text{cosec}^{-1}(-2)+\sec^{-1}\Big(\frac{2}{\sqrt3}\Big)$
Answer$\cot^{-1}\frac{1}{\sqrt3}-\text{cosec}^{-1}(-2)+\sec^{-1}\Big(\frac{2}{\sqrt3}\Big)$$=\frac{\pi}{6}-\Big(-\frac{\pi}{6}\Big)+\frac{\pi}{3}$
$=\frac{\pi}{6}+\frac{\pi}{6}+\frac{\pi}{3}$
$=\frac{4\pi}{6}$
$=\frac{2\pi}{3}$
View full question & answer→Question 1202 Marks
$\sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt2}$
Answer$\sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt2}$
$=\sin^{-1}\frac{1}{2}-\sin^{-1}\Bigg(2\times\frac{1}{\sqrt2}\sqrt{1-\Big(\frac{1}{\sqrt2}\Big)^2}\Bigg)$
$=\sin^{-1}\frac{1}{2}-\sin^{-1}(1)$
$=\frac{\pi}{6}-\frac{\pi}{2}$
$=-\frac{\pi}{3}$
View full question & answer→Question 1212 Marks
For the principal values, evaluate the following:
$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)+\text{cosec}^{-1}\Big(-\frac{2}{\sqrt3}\Big)$
Answer$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)+\text{cosec}^{-1}\Big(-\frac{2}{\sqrt3}\Big)$
$=-\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)+\text{cosec}^{-1}\Big(-\frac{2}{\sqrt3}\Big)$
$=-\sin^{-1}\Big(\sin\frac{\pi}{3}\Big)+\text{cosec}^{-1}\Big[\text{cosec}\Big(-\frac{\pi}{3}\Big)\Big]$
$=-\frac{\pi}{3}-\frac{\pi}{3}$
$=-\frac{2\pi}{3}$
View full question & answer→Question 1222 Marks
For the principal values, evaluate the following:
$\tan^{-1}(-1)+\cos^{-1}\Big(-\frac{1}{2}\Big)$
Answer$\tan^{-1}(-1)+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$\tan^{-1}\Big\{\tan\Big(-\frac{\pi}{4}\Big)\Big\}+\cos^{-1}\Big(\cos\frac{3\pi}{4}\Big)$$\begin{bmatrix}\because\text{Range of tan is }\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big);-\frac{\pi}{4}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\\\ \\\text{and range of cosine is }[0,\pi];\frac{3\pi}{4}\in[0,\pi]\end{bmatrix}$
$=-\frac{\pi}{4}+\frac{3\pi}{4}$
$=\frac{\pi}{2}$
$\therefore\tan^{-1}(-1)+\cos^{-1}\Big(-\frac{1}{2}\Big)=\frac{\pi}{2}$
View full question & answer→Question 1232 Marks
Evaluate the following:
$\text{cosec}^{-1}\Big(\text{cosec}\frac{11\pi}{6}\Big)$
Answer$\text{cosec}^{-1}\Big(\text{cosec}\frac{11\pi}{6}\Big)$
$=\text{cosec}^{-1}\Big(\text{cosec}\Big(2\pi-\frac{\pi}{6}\Big)\Big)$
$=\text{cosec}^{-1}\Big(\text{cosec}\Big(-\frac{\pi}{6}\Big)\Big)$
$=\frac{\pi}{6}$
View full question & answer→Question 1242 Marks
Find the domain of the following function:$\text{f(x)}=\sin^{-1}\sqrt{\text{x}^2-1}$
AnswerTo the domain of $\sin^{-1}y$ which is $[-1, 1]$
$\therefore x^2 -1 \in [0, 1]$ as square root can not be negative
$\Rightarrow x^2\in [1, 2]$
$\Rightarrow\text{x}\in\big[-\sqrt2,-1\big]\cup\big[1,\sqrt2\big]$
Hence, the domain is $\big[-\sqrt2,-1\big]\cup\big[1,\sqrt2\big]$
View full question & answer→Question 1252 Marks
Find the principal values:
$\tan^{-1}(-1)$
Answer$\text{Let}\ \text{y}=\tan^{-1}\left(-1\right)$, $ \text{where}-\frac{\pi}{2}<\text{Y}<\frac{\pi}{2}$
$\therefore\ \tan\text{y}=-1$, $\text{where}-\frac{\pi}{2}<\text{Y}<\frac{\pi}{2}$
$\therefore\ \ \text{Y}=-\frac{\pi}{4}$, $ \bigg[\because\tan\bigg(-\frac{{\pi}}{4}\bigg)=-\tan\frac{\pi}{4}=-1\bigg]$
$\therefore$ required principal value $=-\frac{\pi}{4}$
View full question & answer→Question 1262 Marks
Write the value of $\tan\Big(2\tan^{-1}\frac{1}{5}\Big)$
AnswerLet $\tan\theta=\frac{1}{5}$$\tan\Big(2\tan^{-1}\frac{1}{5}\Big)$
$=\tan2\theta$
$=\frac{2\tan\theta}{1-\tan^2\theta}$
$=\frac{2\times\frac{1}{5}}{1-\frac{1}{25}}$
$=\frac{\frac{2}{5}}{\frac{24}{25}}$
$=\frac{5}{12}$
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