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}$
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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}$
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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}$
$\therefore\cos^{-1}\Big(\frac{1}{2}\Big)-2\sin^{-1}\Big(-\frac{1}{2}\Big)$ $=\frac{\pi}{3}-\Big(-\frac{2\pi}{6}\Big)=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}$
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