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,
$\cos^{-1}(\cos1540^\circ)=\cos^{-1}\{\cos(1440+100^\circ)\}$
$=\cos^{-1}\{\cos(100^\circ)\}$ $[\because\ \cos(4\pi+100^\circ)=\cos100^\circ]$
$=100^\circ$
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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 $\text{R}-(\text{n}\pi,\text{n}\in\text{Z})$
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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} $
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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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