Question 1012 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 1022 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 1032 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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$\sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt2}$
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 1052 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 1062 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 1072 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 1082 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 1092 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 1102 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 1112 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 1122 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 1132 Marks
Find the principal values:
$\text{cosec}^{-1}(-\sqrt{2})$
Answer$\text{Let cosec}^{-1}(-\sqrt{2})=\text{y}$
$-\sqrt{2}=\text{cosec y}-\text{cosec}\frac{\pi}{4}=\text{cosec y}$
$\text{Hence y}=-\frac{\pi}{4}$
View full question & answer→Question 1142 Marks
For the principal values of the following:
$\cot^{-1}\Big(\sqrt3\Big)$
Answer$\cot^{-1}\Big(\sqrt3\Big)$ represents an angle in $(0,\pi)$ whose cotagent is x.
Let $\text{x}=\cot^{-1}\big(\sqrt3\big)$
$\Rightarrow\cot\text{x}=\sqrt3=\cot\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=\frac{\pi}{6}$
$\therefore$ Principal value of $\cot^{-1}\Big(\sqrt3\Big)$ is $\frac{\pi}{6}.$
View full question & answer→Question 1152 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 $\cos^{-1}(-\text{x})=\pi-\cos^{-1}\text{x}.$
$\therefore\ \sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big)$
$=\sin^{-1}\Big(\frac{1}{3}\Big)-\Big[\pi-\cos^{-1}\Big(\frac{1}{3}\Big)\Big]$
$=\sin\Big(\frac{1}{3}\Big)-\pi+\cos^{-1}\Big(\frac{1}{3}\Big)$
$=\Big[\sin^{-1}\Big(\frac{1}{3}\Big)+\cos^{-1}\Big(\frac{1}{3}\Big)\Big]-\pi$
$=\frac{-\pi}{2}$ $\Big[\because\ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=-\frac{\pi}{2}$
$\therefore\ \sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big)=-\frac{\pi}{2}$
View full question & answer→Question 1162 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 1172 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 1182 Marks
For the principal values of the following:
$\cot^{-1}\Big(\tan\frac{3\pi}{4}\Big)$
AnswerLet $\cot^{-1}\Big(\tan\frac{3\pi}{4}\Big)=\text{y}$
Then,
$\cot\text{y}=\tan\frac{3\pi}{4}$
We know that the range of the principal value branch is $(0,\pi).$
Thus,
$\cot\text{y}=\tan\frac{3\pi}{4}=-1=\cot\Big(\frac{3\pi}{4}\Big)$
$\Rightarrow\text{y}=\frac{3\pi}{4}\in(0,\pi)$
Hence, the principal value of $\cot^{-1}\Big(\tan\frac{3\pi}{4}\Big)$ is $\frac{3\pi}{4}.$
View full question & answer→Question 1192 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 $-\frac{\pi}{6}.$
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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 1212 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 1222 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 1232 Marks
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{13\pi}{4}\Big)$
AnswerWe have
$\sec^{-1}\Big(\sec\frac{13\pi}{4}\Big)=\sec^{-1}\Big[\sec\Big(4\pi-\frac{3\pi}{4}\Big)\Big]$
$=\sec^{-1}\Big[\sec\Big(\frac{3\pi}{4}\Big)\Big]$
$=\frac{3\pi}{4}$
View full question & answer→Question 1242 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 1252 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 1262 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}} $
$\sec\text{Y}=\sec\frac{\pi}{6}$
$\text{Y}=\frac{\pi}{6}$
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