Question 12 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 22 Marks
Find the principal values of the following:
$\tan^{-1}\Big(2\cos\frac{2\pi}{3}\Big)$
Answer$\tan^{-1}\Big(2\cos\frac{2\pi}{3}\Big)=\tan^{-1}\Big(2\times\frac{-1}{2}\Big)=\tan^{-1}(-1)$ We know that for any $\text{x}\in\text{R},$ $\tan^{-1}\text{x}$ represents an angle in $\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$ whose tangent is x.
$\therefore\tan^{-1}(-1)=-\frac{\pi}{4}$
$\therefore$ Principal value of $\tan^{-1}\Big(2\cos\frac{2\pi}{3}\Big)$ is $-\frac{\pi}{4}.$
View full question & answer→Question 32 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 42 Marks
Find the principal values of the following:
$\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)$
AnswerWe know that, for any $\text{x}\in\text{R},\tan^{-1}\text{x}$ represents an angle in $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$ whose tangent is x. So,
$\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)=$ An angle in $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$ whose tangent is $\frac{1}{\sqrt3}$
$=\frac{\pi}{6}$
$\therefore\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)=\frac{\pi}{6}$
View full question & answer→Question 52 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 62 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 72 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 82 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 92 Marks
What is the principal value of $\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)$
AnswerWe have, $\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)$ $=\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big\{\sin\Big(\pi-\frac{\pi}{3}\Big)\Big\}$ $\Big[\because\ \Big(\pi-\frac{2\pi}{3}\Big)\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$=\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big\{\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{2\pi}{3}+\frac{\pi}{3}$
$=\pi$
$\therefore\ \cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)=\pi$
View full question & answer→Question 102 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 112 Marks
Find the domain of $\sec^{-1}(3\text{x}-1)$
AnswerDomain of $\sec^{-1}\text{x}$ lies in the interval $(-\infty,-1]\cup[1,\infty)$
$\therefore$ Domain of $\sec^{-1}(3\text{x}-1)$ lies in the interval $(-\infty,-1]\cup[1,\infty)$
$\Rightarrow-\infty\leq3\text{x}-1\leq-1$ and $1\leq3\text{x}-1\leq\infty$
$\Rightarrow-\infty\leq3\text{x}\leq0$ and $2\leq3\text{x}\leq\infty$
$\Rightarrow-\infty\leq\text{x}\leq0$ and $\frac{2}{3}\leq\text{x}\leq\infty$
Domain of $\sec^{-1}\text{x}$ lies in the interval $(-\infty,0]\cup\Big[\frac{2}{3},\infty\Big).$
View full question & answer→Question 122 Marks
Write the value of $\tan^{-1}\sqrt3+\cot^{-1}\sqrt3$
Answer$\tan^{-1}\sqrt3+\cot^{-1}\sqrt3$
$=\frac{\pi}{3}+\frac{\pi}{6}$
$=\frac{2\pi+\pi}{6}$
$=\frac{3\pi}{6}$
$=\frac{\pi}{2}$
View full question & answer→Question 132 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 142 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 152 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 162 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 172 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 182 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 192 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 202 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 212 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 222 Marks
Find the value of $2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$
Answer$2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$
$=2\sec^{-1}\Big(\sec\frac{\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{\pi}{6}\Big)$
$=2\times\frac{\pi}{3}+\frac{\pi}{6}$
$=\frac{5\pi}{6}$
View full question & answer→Question 232 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 242 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 252 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 262 Marks
Write the value of $\tan\Big(2\tan^{-1}\frac{1}{5}\Big)$
AnswerLet $\tan\theta=\frac{1}{5}$ $\tan\Big(2\tan^{-1}\frac{1}{5}\Big)$
$=\tan2\theta$
$=\frac{2\tan\theta}{1-\tan^2\theta}$
$=\frac{2\times\frac{1}{5}}{1-\frac{1}{25}}$
$=\frac{\frac{2}{5}}{\frac{24}{25}}$
$=\frac{5}{12}$
View full question & answer→Question 272 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 282 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 292 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 302 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 312 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 322 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 332 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 342 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 352 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 362 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 372 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 382 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 392 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 402 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 412 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 422 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 432 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 442 Marks
Find the domain of the following function: $\text{f(x)}=\sin^{-1}\sqrt{\text{x}^2-1}$
AnswerTo the domain of sin-1y which is [-1, 1] $\therefore$ x2 -1 $\in$ [0, 1] as square root can not be negative ⇒ 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 452 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 462 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 472 Marks
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{13\pi}{4}\Big)$
AnswerWe have
$\sec^{-1}\Big(\sec\frac{13\pi}{4}\Big)=\sec^{-1}\Big[\sec\Big(4\pi-\frac{3\pi}{4}\Big)\Big]$
$=\sec^{-1}\Big[\sec\Big(\frac{3\pi}{4}\Big)\Big]$
$=\frac{3\pi}{4}$
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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 $\cos^{-1}(-\text{x})=\pi-\cos^{-1}\text{x}.$
$\therefore\ \sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big)$
$=\sin^{-1}\Big(\frac{1}{3}\Big)-\Big[\pi-\cos^{-1}\Big(\frac{1}{3}\Big)\Big]$
$=\sin\Big(\frac{1}{3}\Big)-\pi+\cos^{-1}\Big(\frac{1}{3}\Big)$
$=\Big[\sin^{-1}\Big(\frac{1}{3}\Big)+\cos^{-1}\Big(\frac{1}{3}\Big)\Big]-\pi$
$=\frac{-\pi}{2}$ $\Big[\because\ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=-\frac{\pi}{2}$
$\therefore\ \sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big)=-\frac{\pi}{2}$
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Write the principal value of $\tan^{-1}1+\cos^{-1}\Big(-\frac{1}{2}\Big)$
Answer$\tan^{-1}1+\cos^{-1}\Big(-\frac{1}{2}\Big)$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)+\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)$
$=\frac{\pi}{4}+\frac{2\pi}{3}$
$=\frac{11\pi}{3}$
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Evaluate:
$\sin\Big(\tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}\Big)\text{ for }\text{x}>0$
Answer$\sin\Big(\tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}\Big)$
$\sin\Big(\tan^{-1}\text{x}+\cot^{-1}{\text{x}}\Big)$ $\Big[\because\ \tan^{-1}\text{x}=\cot^{-1}\frac{1}{\text{x}}\Big]$
$=\sin\Big(\frac{\pi}{2}\Big)$
$=1$
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