Question 11 Mark
If $\tan^{-1}\text{x} +\tan^{-1}\text{y} = \frac{\pi}{4},\text{xy} < 1,$then write the value of x + y + xy.
View full question & answer→Question 21 Mark
Write the value of $\tan^{-1}\bigg[2\sin\bigg(2\cos^{-1}\frac{\sqrt{3}}{2}\bigg)\bigg].$
Answer$\tan^{-1}\bigg[2\sin\bigg(2\cos^{-1}\frac{\sqrt{3}}{2}\bigg)\bigg] = \tan^{-1}\bigg(2\sin\bigg(2\times\frac{\pi}{6}\bigg)\bigg)\bigg[\because\cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}\bigg]$
$ = \tan^{-1}\bigg(2\sin\frac{\pi}{3}\bigg) = \tan^{-1}\bigg(2\times\frac{\sqrt{3}}{2}\bigg) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}.$
View full question & answer→Question 31 Mark
Write the principal value of $\tan^{-1}(\sqrt{3}) - \cot^{-1}( - \sqrt{3}).$
Answer$\tan^{-1}\sqrt{3} - \cot^{-1}(- \sqrt{3}) = \tan^{-1}\bigg(\tan\frac{\pi}{3}\bigg) -\cot^{-1}\bigg(- \cot\frac{\pi}{6}\bigg)$
$ =\tan^{-1}\bigg(\tan\frac{\pi}{3}\bigg) - \cot^{-1}\bigg(\cot\bigg(\pi - \frac{\pi}{6}\bigg)\bigg)$
$ = \tan^{-1}\bigg(\tan\frac{\pi}{3}\bigg) - \cot^{-1}\bigg(\cot\frac{5\pi}{6}\bigg)$
$ = \frac{\pi}{3} - \frac{5\pi}{6}\bigg[\because\frac{\pi}{3}\in\bigg( -\frac{\pi}{2} , \frac{\pi}{2}\bigg)\text{ and }\frac{5\pi}{6}\in(0,\pi)\bigg]$
$ = \frac{2\pi - 5 \pi}{6} = - \frac{\pi}{2}.$
View full question & answer→Question 41 Mark
Find the principal value of $\tan^{–1} \sqrt{3} – \sec^{–1}(– 2)$.
View full question & answer→Question 51 Mark
What is the principal value of $\cot^{-1} \Bigg(\cos\frac{2\pi}{3}\Bigg)+\sin^{-1}\Bigg(\sin\frac{2\pi}{3}\Bigg)?$
View full question & answer→Question 61 Mark
Write the principal value of $\sec^{–1} (–2)$.
View full question & answer→Question 71 Mark
Write the principal value of $\cos^{–1} \Bigg(\cos\frac{7\pi}{6}\Bigg).$
View full question & answer→Question 81 Mark
Using principal value, evaluate the following:
$\cos^{-1}\Bigg(\cos\frac{2\pi}{3}\Bigg)+\sin^{-1}\Bigg(\sin\frac{2\pi}{3}\Bigg)$.
Answer$ \cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)$
$=\frac{2\pi}{3}+\frac{2\pi}{3}=\frac{4\pi}{3}$.
View full question & answer→Question 91 Mark
If a line makes angles $90^{0}, 60^{0} $ and $\theta$ with x, y and z-axis respectively, where $\theta$ is acute, then find $\theta.$
Answer$\cos^{2} \frac{\pi}{2} + \cos^{2} \frac{\pi}{3} + \cos^{2} \theta = 1 \Rightarrow \theta = \frac{\pi}{6}$
View full question & answer→Question 101 Mark
If $\sin\bigg(\sin^{-1}\frac{1}{5}+\cos^{-1}\text{x}\bigg) = 1,$then find the vatue of x.
Answer$\text{x}=\frac{1}{5}.$
View full question & answer→Question 111 Mark
Write the value of $\tan \bigg(2 \tan^{-1}\frac{1}{5}\bigg).$
AnswerLet 2 $\tan^{-1}\frac{1}{5} =\theta$
$\Rightarrow\tan^{-1}\frac{1}{5} =\frac{\theta}{2}\Rightarrow\tan\frac{\theta}{2} = \frac{1}{5}.$
Now, $\tan\bigg(2\tan^{-1}\frac{1}{5}\bigg) = \tan\theta = \frac{2\tan\frac{\theta}{2}}{1-\tan^{2}\frac{\theta}{2}}$
$ = \frac{2\times\frac{1}{5}}{1-\big(\frac{1}{5}\big)^{2}} =\frac{2}{5}\times\frac{25}{24} = \frac{5}{12}.$
View full question & answer→Question 121 Mark
Write the principal value of $\tan^{-1} (1)+ \cos^{-1}\bigg(- \frac{1}{2}\bigg).$
Answer$\tan^{-1}(1) + \cos^{-1}\bigg(-\frac{1}{2}\bigg) =\tan^{-1}\bigg(\tan\frac{\pi}{4}\bigg) + \cos^{-1}\bigg(\cos\bigg(\pi -\frac{\pi}{3}\bigg)\bigg)$
$ = \tan^{-1}\bigg(\tan\frac{\pi}{4}\bigg) + \cos^{-1}\bigg(\cos\frac{2\pi}{3}\bigg)$
$ =\frac{\pi}{4} +\frac{2\pi}{3}\bigg[\because\frac{\pi}{4}\in\bigg( -\frac{\pi}{2},\frac{\pi}{2}\bigg)\text{ and }\frac{2\pi}{3}\in[0,\pi]\bigg]$
$ = \frac{3\pi +8 \pi}{12} = \frac{11\pi}{12}.$
View full question & answer→Question 131 Mark
Write the principal value of $\text{cos}^{-1}\Bigg(\frac{1}{2}\Bigg)-2\text{sin}^{-1}\Bigg(-\frac{1}{2}\Bigg).$
View full question & answer→Question 141 Mark
Write the value of sin $\Bigg[\frac{\pi}{3}-\sin^{-1}\Big(-\frac{1}{2}\Big)\Bigg]$
View full question & answer→Question 151 Mark
What is the principal value of $\sin^{–1} \Bigg(-\frac{\sqrt{3}}{2}\Bigg)?$
View full question & answer→Question 161 Mark
Using principal value, evaluate the following:$\sin^{-1} \bigg( \sin \frac{3{\pi}}{5}\bigg)$
Answer$\frac{3{\pi}}{5} = \pi - \frac{2{\pi}}{5}$$\therefore \sin^{-1} \bigg( \sin \frac{3{\pi}}{5}\bigg)$
$= \sin^{-1} \Bigg[ \sin\bigg( \pi - \frac{2{\pi}}{5}\bigg)\Bigg]$
$= \sin^{-1} \bigg[ \sin\frac{2\pi}{5}\bigg] = \frac{2\pi}{5} \in \bigg[ - \frac{\pi}{2}, \frac{\pi}{2}\bigg]$
View full question & answer→Question 171 Mark
$\text{Evaluate} : \sin \Bigg[\frac{\pi}{3} - \sin^{-1} \bigg( - \frac{1}{2}\bigg)\Bigg]$
Answer$\sin \Bigg[\frac{\pi}{3} - \sin^{-1} \bigg( - \frac{1}{2}\bigg)\Bigg] \sin \Bigg[\frac{\pi}{3} -\bigg( - \frac{\pi}{6}\bigg)\Bigg] = \sin\frac{\pi}{2} = 1$
View full question & answer→Question 181 Mark
Find the value of $\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3}).$
Answer$\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3}).$
$\frac{\pi}{3}-\frac{5\pi}{6}=\frac{2\pi-5\pi}{6}=\frac{-3\pi}{6}=\frac{-\pi}{2}$
View full question & answer→MCQ 191 Mark
Calculate the value of $ \sin−1\cos(\sin−1\text{x})+\cos−1\sin(\cos−1\text{x})$. where
- A
$ +\frac{\pi}{4}$
- B
$ -\frac{\pi}{4}$
- ✓
$ +\frac{\pi}{2}$
- D
$ -\frac{\pi}{2}$
AnswerCorrect option: C. $ +\frac{\pi}{2}$
Given, $ \displaystyle \sin^{-1} \cos \left ( \sin^{-1} \text{x}\right ) + \cos^{-1} \sin \left ( \cos^{-1} \text{x} \right )$
$ =\displaystyle \sin^{-1} \cos \left ( \cos^{-1} \sqrt{1-\text{x}^2}\right ) + \cos^{-1} \sin \left ( \sin^{-1} \sqrt{1-\text{x}^2} \right ) $
$ =\displaystyle \sin^{-1} \left(\sqrt{1-\text{x}^2}\right ) + \cos^{-1}\left ( \sqrt{1-\text{x}^2} \right ) =\frac{\pi}{2}$
Since $ \sin^{-1}\text{x}+\cos^{-1}\text{x} = \frac{\pi}{2} \forall |\text{x}|\leq $
View full question & answer→MCQ 201 Mark
If $\alpha $ and $\beta$ are two real values of $x$ which satisfy the equation $\sin^{-1} \text{x} + \sin^{-1} (1−\text{x})=\cos^{−1}\text{x}, $ then :
AnswerCorrect option: A. $ \alpha + \beta = \frac{1}{2}$
$ \sin^{-1}(\text{x})+\sin^{-1}(1-\text{x})=\cos^{-1}(\text{x})$
Now, $x\ \epsilon\ [-1, 1]$
and $-1 \leq (1 - x) \leq 1$
$x\ \epsilon \ [0, 2]$
Hence, $x\ \epsilon\ [0, 1]$
Now we get two solutions of the above equation as $0$ and $ \frac{1}{2}$
Thus, the two roots are
$x = 0$ and $ \text{x}=\frac{1}{2}$
$ \alpha +\beta=\frac{2}{1}$
and $ \alpha\beta = 0$
View full question & answer→Question 211 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 221 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 231 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→MCQ 241 Mark
$\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})$ is equal to
- A
$\pi$
- ✓
$-\frac{\pi}{2}$
- C
$0$
- D
$2\sqrt{3}$
AnswerCorrect option: B. $-\frac{\pi}{2}$
$\tan^{-1}\sqrt{3}-\cot^{-1}\bigg(-\sqrt{3}\bigg)=\tan^{-1}\tan\frac{\pi}{3}-\cot^{-1}\bigg(-\cot\frac{\pi}{6}\bigg)$
$=\frac{\pi}{3}-\cot^{-1}\bigg[\cot\bigg(\pi-\frac{\pi}{6}\bigg)\bigg]=\frac{\pi}{3}-\bigg(\pi-\frac{\pi}{6}\bigg)$
$=\frac{\pi}{3}-\frac{5\pi}{6}=\frac{2\pi-5\pi}{6}=-\frac{3\pi}{6}=-\frac{\pi}{2}$
Therefore, option $(b)$ is correct.
View full question & answer→Question 251 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 261 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→MCQ 271 Mark
$\sin\bigg(\frac{\pi}{3}-\sin^{-1}(-\frac{1}{2})\bigg)$ is equal to
- A
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{4}$
- ✓
$1$
Answer$\sin^{-1}\bigg(-\frac{1}{2}\bigg)=\sin^{-1}\bigg(-\sin\frac{\pi}{6}\bigg)$
$=\sin^{-1}\bigg[\sin\bigg(-\frac{\pi}{6}\bigg)\bigg]=-\frac{\pi}{6}$
$\therefore\sin\left[\frac{\pi}{3}-\sin^{-1}\left(-\frac{1}{2}\right)\right]=\sin\left[\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right]$
$=\sin\left[\frac{\pi}{3}+\frac{\pi}{6}\right]=\sin\frac{3\pi}{6}=\sin\frac{\pi}{2}=1 $
Therefore, option $(d)$ is correct.
View full question & answer→Question 281 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 291 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 301 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 311 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 321 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→MCQ 331 Mark
$\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$ is equal to :
- A
${\pi}$
- ✓
$-\frac{\pi}{3}$
- C
$\frac{\pi}{3}$
- D
$\frac{2\pi}{3}$
AnswerCorrect option: B. $-\frac{\pi}{3}$
$\tan^{-1}(\sqrt{3})=\text{x}$
$\sqrt{3}=\tan\text{x}$
$\tan\frac{\pi}{3}=\tan\text{x}$
$\frac{\pi}{3}=\text{x}$
or ${x}=\frac{\pi}{3}$
Similarly $ \sec^{-1}(-2)=\text{y}$
$-2=\sec\text{y}$
$\sec\frac{2\pi}{3}=\sec\text{y}$
$\text{y}=\frac{2\pi}{3}$
Now $ \tan^{-1}\sqrt{3}-\sec^{-1}(-2)=\text{x}-\text{y}$
$=\frac{\pi}{3}-\frac{2\pi}{3}$
$-\frac{\pi}{3}$
View full question & answer→Question 341 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 351 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→MCQ 361 Mark
$\sin\left(\tan^{-1}x\right),|x| < 1$ is equal to :
- A
$\frac{x}{\sqrt{1-x^{2}}}$
- B
$\frac{1}{\sqrt{1-x^{2}}}$
- C
$\frac{1}{\sqrt{1+x^{2}}}$
- ✓
$\frac{x}{\sqrt{1+x^{2}}}$
AnswerCorrect option: D. $\frac{x}{\sqrt{1+x^{2}}}$
Let $\tan^{-1}x=y.$ Then
$\tan y=x$
$\Rightarrow\sin y=\frac{x}{\sqrt{1+x^2}}$
$\therefore y=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$
$\Rightarrow\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$
$\therefore\sin\left(\tan^{-1}x\right)=\sin\left(\sin^{-1}\frac{x}{\sqrt{1+x^2}}\right)=\frac{x}{\sqrt{1+x^2}}$
The correct answer is $(d).$
View full question & answer→Question 371 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 381 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 391 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 401 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 411 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 421 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→MCQ 431 Mark
$\cos^{-1}\bigg(\cos\frac{7\pi}{6}\bigg)$ is equal to :
- A
$\frac{7\pi}{6}$
- ✓
$\frac{5\pi}{6}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: B. $\frac{5\pi}{6}$
$\cos^{-1}\bigg(\cos\frac{7\pi}{6}\bigg)=\cos^{-1}\bigg[\cos\bigg(2\pi-\frac{7\pi}{6}\bigg)\bigg]$
$\left[\because\cos\left(2\pi-\theta\right)=\cos\theta\right]$
$=2\pi-\frac{7\pi}{6}=\frac{12\pi-7\pi}{6}=\frac{5\pi}{6}$
View full question & answer→MCQ 441 Mark
$\tan^{-1}\bigg(\frac{x}{y}\bigg)-\tan^{-1}\frac{x-y}{x+y}$ is equal to :
- A
$\frac{\pi}{2}$
- B
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{4}$
- D
$\frac{-3\pi}{4}$
AnswerCorrect option: C. $\frac{\pi}{4}$
$\tan^{-1}\bigg(\frac{x}{y}\bigg)-\tan^{-1}\frac{x-y}{x+y}$
$=\tan^{-1}\Bigg[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\left(\frac{x}{y}\right)\left(\frac{x-y}{x+y}\right)}\Bigg]$
$\bigg[\tan^{-1}y-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}\bigg]$
$=\tan^{-1}\Bigg[\frac{\frac{x\left(x+y\right)-y\left(x-y\right)}{y\left(x+y\right)}}{\frac{y\left(x+y\right)+x\left(x-y\right)}{y\left(x+y\right)}}\Bigg]$
$=\tan^{-1}\bigg(\frac{x^2+xy-xy+y^2}{xy+y^2+x^2-xy}\bigg)$
$=\tan^{-1}\bigg(\frac{x^2+y^2}{x^2+y^2}\bigg)=\tan^{-1}1=\frac{\pi}{4}$
Hence, the correct answer is $(c).$
View full question & answer→Question 451 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 461 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 471 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 481 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 491 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 501 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→