Question 11 Mark
Evaluate the following:
$\tan^{-1}(\tan4)$
AnswerWe know that
$\tan^{-1}(\tan\theta)=\theta,-\frac{\pi}{2}<\theta<\frac{\pi}{2}$
We have
$\tan^{-1}(\tan4)$
$=4-\pi$ $\Big(\because\ \tan^{-1}(\tan\theta)=\theta-\pi,\text{if }\theta\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)\Big)$
View full question & answer→Question 21 Mark
Evaluate the following:
$\tan^{-1}\Big(\tan\frac{6\pi}{7}\Big)$
AnswerWe know that $\tan^{-1}(\tan\theta)=\theta,-\frac{\pi}{2}<\theta<\frac{\pi}{2}$We have
$\tan^{-1}\Big(\tan\frac{6\pi}{7}\Big)$
$=\frac{6\pi}{7}-\pi$ $\Big(\because\ \tan^{-1}(\tan\theta)=\theta-\pi,\ \text{if }\theta\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)\Big)$ $=-\frac{\pi}{7}$
View full question & answer→Question 31 Mark
The set of values of $\text{cosec}^{-1}\Big(\frac{\sqrt3}{2}\Big)$
AnswerThe values of $\text{cosec}^{-1}\Big(\frac{\sqrt3}{2}\Big)$ is undefined as it is outside the range i.e., R - (-1, 1).
View full question & answer→Question 41 Mark
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{2\pi}{3}\Big)$
AnswerWe have
$\sec^{-1}\Big(\sec\frac{2\pi}{3}\Big)=\frac{2\pi}{3}$
View full question & answer→Question 51 Mark
Evaluate the following:
$\tan^{-1}(\tan1)$
AnswerWe know that $\tan^{-1}(\tan\theta)=\theta,-\frac{\pi}{2}<\theta<\frac{\pi}{2}$$\tan^{-1}(\tan1)$
$=1$ $\Big(\because\ \tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big)$
View full question & answer→Question 61 Mark
Evaluate the following:
$\cos^{-1}(\cos3)$
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
$\cos^{-1}(\cos3)=3$
View full question & answer→Question 71 Mark
Evaluate the following:
$\tan^{-1}\Big(\tan\frac{7\pi}{6}\Big)$
AnswerWe know that $\tan^{-1}(\tan\theta)=\theta,-\frac{\pi}{2}<\theta<\frac{\pi}{2}$ We have$\tan^{-1}\Big(\tan\frac{7\pi}{6}\Big)=\tan^{-1}\Big[\tan\Big(\pi+\frac{\pi}{6}\Big)\Big]$
$=\tan^{-1}\Big[\tan\Big(\frac{\pi}{6}\Big)\Big]$ $=\frac{\pi}{6}$
View full question & answer→Question 81 Mark
Evaluate the following:
$\cot^{-1}\Big(\cot\frac{9\pi}{4}\Big)$
AnswerWe have
$\cot^{-1}\Big(\cot\frac{9\pi}{4}\Big)$
$=\cot^{-1}\Big[\cot\Big(2\pi+\frac{\pi}{4}\Big)\Big]$
$=\cot^{-1}\Big(\cot\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
View full question & answer→Question 91 Mark
Write the value of the expression $\tan\Big(\frac{\sin^{-1}\text{x}+\cos^{-1}\text{x}}{2}\Big),$ when $\text{x}=\frac{\sqrt3}{2}$
Answer$\tan\Big(\frac{\sin^{-1}\text{x}+\cos^{-1}\text{x}}{2}\Big)=\tan\Big(\frac{\pi}{4}\Big)$ $\big[\because\ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\big]$
$=1$
View full question & answer→Question 101 Mark
Evaluate the following:
$\text{cosec}^{-1}\Big\{\text{cosec}\Big(-\frac{9\pi}{4}\Big)\Big\}$
Answer$\text{cosec}^{-1}\Big\{\text{cosec}\Big(-\frac{9\pi}{4}\Big)\Big\}$
$=\text{cosec}^{-1}\Big[-\text{cosec}\Big(2\pi+\frac{\pi}{4}\Big)\Big]$
$=\text{cosec}^{-1}\Big(-\text{cosec}\frac{\pi}{4}\Big)$
$=\text{cosec}^{-1}\Big(\text{cosec}-\frac{\pi}{4}\Big)$
$=-\frac{\pi}{4}$
View full question & answer→Question 111 Mark
Find the principal value of the following:
$\sin^{-1}\Big(\cos\frac{3\pi}{4}\Big)$
Answer$\sin^{-1}\Big(\cos\frac{3\pi}{4}\Big)=\sin^{-1}\Big(-\frac{\sqrt2}{2}\Big)$
$=\sin^{-1}\Big[\sin\Big(-\frac{\pi}{4}\Big)\Big]=-\frac{\pi}{4}$
View full question & answer→Question 121 Mark
Evaluate the following:
$\tan^{-1}\Big(\tan\frac{\pi}{3}\Big)$
AnswerWe know that $\tan^{-1}(\tan\theta)=\theta,-\frac{\pi}{2}<\theta<\frac{\pi}{2}$We have
$\tan^{-1}\Big(\tan\frac{\pi}{3}\Big)=\frac{\pi}{3}$
View full question & answer→Question 131 Mark
Evaluate the following:
$\sin^{-1}(\sin12)$
AnswerWe know $\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$We have
$\sin^{-1}(\sin12)=\sin^{-1}\{\sin(-\pi+12)\}$ $=12-\pi$
View full question & answer→Question 141 Mark
Write the value of $\sec^{-1}\Big(\frac{1}{2}\Big).$
AnswerThe value of $\sec^{-1}\Big(\frac{1}{2}\Big)$ is undefined outside the range i.e., R - (-1, 1).
View full question & answer→Question 151 Mark
Evaluate the following:
$\tan^{-1}(\tan2)$
AnswerWe know that
$\tan^{-1}(\tan\theta)=\theta,-\frac{\pi}{2}<\theta<\frac{\pi}{2}$
We have
$\tan^{-1}(\tan2)=\tan^{-1}[\tan(-\pi+2)]$
$=2-\pi$
View full question & answer→Question 161 Mark
Evaluate the following:
$\sin^{-1}\Big(\sin\frac{\pi}{6}\Big)$
AnswerWe know
$\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
We have
$\sin^{-1}\Big(\sin\frac{\pi}{6}\Big)=\frac{\pi}{6}$
View full question & answer→Question 171 Mark
Evaluate the following:
$\text{cosec}^{-1}\Big(\text{cosec}\frac{3\pi}{4}\Big)$
Answer$\text{cosec}^{-1}\Big(\text{cosec}\frac{3\pi}{4}\Big)=\text{cosec}^{-1}\Big[\text{cosec}\Big(\pi-\frac{\pi}{4}\Big)\Big]$
$=\text{cosec}^{-1}\Big(\text{cosec}\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
View full question & answer→Question 181 Mark
Find the principal value of the following:
$\sin^{-1}\Big(\frac{\sqrt3+1}{2\sqrt2}\Big)$
Answer$\sin^{-1}\Big(\frac{\sqrt3+1}{2\sqrt2}\Big)=\sin^{-1}\Big(\sin\frac{5\pi}{12}\Big)=\frac{5\pi}{12}$
View full question & answer→Question 191 Mark
Evaluate the following:
$\cos^{-1}(\cos5)$
Answer$\cos^{-1}(\cos\text{x})=\text{x}$
provided $\text{x}\in [0,\pi]\approx[0,3.14]$
And in our equation X is 5 which does not lie in the above range.
We know $\cos [2\pi -\text{x}]=\cos[\text{x}]$
$\therefore \cos (2\pi -5)=\cos (5)$
Also $2\pi -5$ belongs in $[0,\pi]$
$\therefore \cos^{-1}(\cos 5)=2\pi -5$
View full question & answer→Question 201 Mark
Evaluate the following:
$\cot^{-1}\Big(\cot\frac{4\pi}{3}\Big)$
AnswerWe have
$\cot^{-1}\Big(\cot\frac{4\pi}{3}\Big)$
$=\cot^{-1}\Big[\cot\Big(\pi+\frac{\pi}{3}\Big)\Big]$
$=\cot^{-1}\Big(\cot\frac{\pi}{3}\Big)$
$=\frac{\pi}{3}$
View full question & answer→Question 211 Mark
Evaluate the following:
$\tan^{-1}(\tan12)$
AnswerWe know that
$\tan^{-1}(\tan\theta)=\theta,-\frac{\pi}{2}<\theta<\frac{\pi}{2}$
We have
$\tan^{-1}(\tan12)$
$=12-4\pi$ $\Big(\because\ \tan^{-1}(\tan\theta)=\theta-4\pi,\text{if }\theta\in\Big(\frac{7\pi}{2},\frac{9\pi}{2}\Big)\Big)$
View full question & answer→Question 221 Mark
Evaluate the following:
$\sin^{-1}\Big(\sin\frac{5\pi}{6}\Big)$
AnswerWe know
$\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
We have
$\sin^{-1}\Big(\sin\frac{5\pi}{6}\Big)=\sin^{-1}\Big\{\sin\Big(\pi-\frac{\pi}{6}\Big)\Big\}$
$=\sin^{-1}\Big(\sin\frac{\pi}{6}\Big)$
$=\frac{\pi}{6}$
View full question & answer→Question 231 Mark
Write the range of $\tan^{-1}\text{x}.$
AnswerThe range of $\tan^{-1}\text{x}$ is $\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big).$
View full question & answer→Question 241 Mark
Evaluate the following:
$\text{cosec}^{-1}\Big(\text{cosec}\frac{6\pi}{5}\Big)$
Answer$\text{cosec}^{-1}\Big(\text{cosec}\frac{6\pi}{5}\Big)=\text{cosec}^{-1}\Big[\text{cosec}\Big(\pi+\frac{\pi}{5}\Big)\Big]$
$=\text{cosec}^{-1}\Big(\text{cosec}-\frac{\pi}{5}\Big)$
$=-\frac{\pi}{5}$
View full question & answer→Question 251 Mark
Evaluate the following:
$\sin^{-1}(\sin4)$
AnswerWe know $\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$We have
$=\sin^{-1}(\sin4)$ $=\pi-4$ $\Big(\because\ \sin^{-1}(\sin\theta)=\pi-\theta,\text{if}\ \theta\in\Big[\frac{\pi}{2},\frac{3\pi}{2}\Big]\Big)$
View full question & answer→Question 261 Mark
Evaluate the following:
$\sin\Big(\sin^{-1}\frac{7}{25}\Big)$
Answer$\sin\Big(\sin^{-1}\frac{7}{25}\Big)=\frac{7}{25}$
View full question & answer→Question 271 Mark
Evaluate the following:
$\text{cosec}^{-1}\Big(\text{cosec}\frac{13\pi}{6}\Big)$
Answer$\text{cosec}^{-1}\Big(\text{cosec}\frac{13\pi}{6}\Big)$
$=\text{cosec}^{-1}\Big[\text{cosec}\Big(2\pi+\frac{\pi}{6}\Big)\Big]$
$=\text{cosec}^{-1}\Big(\text{cosec}\frac{\pi}{6}\Big)$
$=\frac{\pi}{6}$
View full question & answer→Question 281 Mark
Evaluate the following:
$\text{cosec}^{-1}\Big(\text{cosec}\frac{\pi}{4}\Big)$
Answer$\text{cosec}^{-1}\Big(\text{cosec}\frac{\pi}{4}\Big)$
$=\text{cosec}^{-1}\big(\sqrt2\big)$
$=\frac{\pi}{4}$
View full question & answer→Question 291 Mark
Evaluate the following:
$\cos^{-1}(\cos12)$
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
$\cos^{-1}(\cos12)=\cos^{-1}\{\cos(4\pi-12)\}$
$=4\pi-12$
View full question & answer→Question 301 Mark
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{\pi}{3}\Big)$
AnswerWe have
$\sec^{-1}\Big(\sec\frac{\pi}{3}\Big)=\frac{\pi}{3}$
View full question & answer→Question 311 Mark
Find the principal value of the following:
$\sin^{-1}\Big(\tan\frac{5\pi}{4}\Big)$
Answer$\sin^{-1}\Big(\tan\frac{5\pi}{4}\Big)=\sin^{-1}(1)$
$\sin^{-1}\Big[\sin\Big(\frac{\pi}{2}\Big)\Big]=\frac{\pi}{2}$
View full question & answer→Question 321 Mark
Evaluate the following:
$\cot^{-1}\Big(\cot\frac{\pi}{3}\Big)$
Answer$\cot^{-1}\Big(\cot\frac{\pi}{3}\Big)$
$=\cot^{-1}\big(\sqrt3\big)$
$=\frac{\pi}{3}$
View full question & answer→Question 331 Mark
Find the principal value of the following:
$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)$
Answer$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)=\sin^{-1}\Big[\sin\big(-\frac{\pi}{3}\big)\Big]$
$=-\frac{\pi}{3}$
View full question & answer→Question 341 Mark
Find the principal value of the following:
$\sin^{-1}\Big(\frac{\sqrt3-1}{2\sqrt2}\Big)$
Answer$\sin^{-1}\Big(\frac{\sqrt3-1}{2\sqrt2}\Big)=\sin^{-1}\Big(\sin\frac{\pi}{12}\Big)=\frac{\pi}{12}$
View full question & answer→Question 351 Mark
Evaluate the following:
$\cos^{-1}(\cos4)$
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
$\cos^{-1}(\cos4)=\cos^{-1}\{\cos(2\pi-4)\}=2\pi-4$
View full question & answer→Question 361 Mark
Find the value of $\cos^{-1}\Big(\cos\frac{13\pi}{6}\Big)$
Answer$\cos^{-1}\Big(\cos\frac{13\pi}{6}\Big)=\cos^{-1}\Big[\cos\Big(2\pi+\frac{\pi}{6}\Big)\Big]$
$=\cos^{-1}\Big[\cos\Big(\frac{\pi}{6}\Big)\Big]$
$=\frac{\pi}{6}$
View full question & answer→Question 371 Mark
Evaluate the following:
$\tan^{-1}\Big(\tan\frac{9\pi}{4}\Big)$
AnswerWe know that $\tan^{-1}(\tan\theta)=\theta,-\frac{\pi}{2}<\theta<\frac{\pi}{2}$We have
$\tan^{-1}\Big(\tan\frac{9\pi}{4}\Big)=\tan^{-1}\Big[\tan\Big(2\pi+\frac{\pi}{4}\Big)\Big]$
$\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}\Big)\Big]$ $=\frac{\pi}{4}$
View full question & answer→Question 381 Mark
Find the value of $\tan^{-1}\Big(\tan\frac{9\pi}{8}\Big)$
Answer$\tan^{-1}\Big(\tan\frac{9\pi}{8}\Big)=\tan^{-1}\Big[\tan\Big(\pi+\frac{\pi}{8}\Big)\Big]$
$=\tan^{-1}\Big[\tan\Big(\frac{\pi}{8}\Big)\Big]$
$=\frac{\pi}{8}$
View full question & answer→Question 391 Mark
Find the principal value of the following:
$\sin^{-1}\Big(\cos\frac{2\pi}{3}\Big)$
Answer$\sin^{-1}\Big(\frac{\cos2\pi}{3}\Big)=\sin^{-1}\Big(-\frac{1}{2}\Big)$
$=\sin^{-1}\Big[\sin\Big(-\frac{\pi}{6}\Big)\Big]=-\frac{\pi}{6}$
View full question & answer→Question 401 Mark
Evaluate the following:
$\sin\Big(\cos^{-1}\frac{5}{13}\Big)$
Answer$\sin\Big(\cos^{-1}\frac{5}{13}\Big)$
$=\sin\Big(\sin^{-1}\frac{12}{13}\Big)$ $\Big[{\therefore\ \cos^{-1}}\text{x}=\sin^{-1}\sqrt{1-\text{x}^2}\Big]$
$ =\frac{12}{13}$
View full question & answer→