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
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 92 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 102 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)$$=\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big\{\sin\Big(\pi-\frac{\pi}{3}\Big)\Big\}$ $\Big[\because\ \Big(\pi-\frac{2\pi}{3}\Big)\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$=\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big\{\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{2\pi}{3}+\frac{\pi}{3}$
$=\pi$
$\therefore\ \cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)=\pi$
View full question & answer→Question 112 Marks
Find the value of $2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$
Answer$2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$
$=2\sec^{-1}\Big(\sec\frac{\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{\pi}{6}\Big)$
$=2\times\frac{\pi}{3}+\frac{\pi}{6}$
$=\frac{5\pi}{6}$
View full question & answer→Question 122 Marks
Write the value of $\tan^{-1}\Big\{2\sin\Big(2\cos^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
Answer$\tan^{-1}\Big\{2\sin\Big(2\cos^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
$=\tan^{-1}\Big\{2\sin\Big[\cos^{-1}2\Big(\frac{\sqrt3}{2}\Big)^2-1\Big]\Big\}$
$=\tan^{-1}\Big[2\sin\Big(\cos^{-1}\frac{1}{2}\Big)\Big]$
View full question & answer→Question 132 Marks
What is the principal value of $\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)$
Answer$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)$ $-\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)$ $\Big\{\text{Since},\sin^{-1}(-\theta)=-\sin^{-1}(\theta)\Big\} $$=-\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 142 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 152 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 162 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 172 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 182 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 192 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 202 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 212 Marks
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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$\sin^{-1}\Big\{\cos\Big(\sin^{-1}\frac{\sqrt3}{2}\Big)\Big\}$
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 232 Marks
Evaluate:
$\sin\Big(\tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}\Big)\text{ for }\text{x}>0$
Answer$\sin\Big(\tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}\Big)$
$\sin\Big(\tan^{-1}\text{x}+\cot^{-1}{\text{x}}\Big)$ $\Big[\because\ \tan^{-1}\text{x}=\cot^{-1}\frac{1}{\text{x}}\Big]$
$=\sin\Big(\frac{\pi}{2}\Big)$
$=1$
View full question & answer→Question 242 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 252 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 262 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 272 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 282 Marks
Write the value of $\sin^{-1}(\sin1550^\circ).$
AnswerWe know that $\sin^{-1}(\sin\text{x})=\text{x}.$
Now,
$\sin^{-1}(\sin1550^\circ)=\sin^{-1}\{\sin(1620^\circ-1550^\circ)\}$
$[\because\ \sin\text{x}=\sin(1620^\circ-\text{x})]$
$=\sin^{-1}\{\sin(70^\circ)\}$
$=70^\circ$
$\because\ \sin^{-1}(\sin1550^\circ)=70^\circ$
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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 302 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 312 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]
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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 332 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 342 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 352 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 362 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 372 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 382 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 392 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 402 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 412 Marks
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{5\pi}{4}\Big)$
AnswerWe have
$\sec^{-1}\Big(\sec\frac{5\pi}{4}\Big)$
$=\sec^{-1}\Big(\sec\Big(\pi+\frac{\pi}{4}\Big)\Big)$
$=\sec^{-1}\Big(-\sec\Big(\frac{\pi}{4}\Big)\Big)$
$=\sec^{-1}\big(-\sqrt2\big)$
$=\frac{3\pi}{4}$
View full question & answer→Question 422 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 432 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 442 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 452 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 462 Marks
For the principal values of the following:
$\cot^{-1}\Big(-\sqrt3\Big)$
AnswerLet $\cot^{-1}\Big(-\sqrt3\Big)=\text{y}$
Then,
$\cot\text{y}=-\sqrt3$
We know that the range of the principal value branch is $(0,\pi).$
Thus,
$\cot\text{y}=-\sqrt3=\cot\Big(\frac{5\pi}{6}\Big)$
$\Rightarrow\text{y}=\frac{5\pi}{6}\in(0,\pi)$
Hence, the principal value of $\cot^{-1}\Big(-\sqrt3\Big)$ is $\frac{5\pi}{6}.$
View full question & answer→Question 472 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 482 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 492 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 502 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)$
$\pm\frac{1}{\sqrt{10}}$
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