Question 13 Marks
Solve the following equation for x:
$2\tan^{-1}(\sin\text{x})=\tan^{-1}(2\sin\text{x}),\text{x}\neq\frac{\pi}{2}.$
Answer$2\tan^{-1}(\sin\text{x})=\tan^{-1}(2\sec\text{x})$
$\tan^{-1}\Big(\frac{2\sin\text{x}}{1-\sin^{2}\text{x}}\Big)=\tan^{-1}(2\sec\text{x})$ $\Big[\text{Since }2\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\frac{2\sin\text{x}}{\cos^2\text{x}}=2\sec\text{x}$
$\frac{\sin\text{x}}{\cos\text{x}.\cos\text{x}}=\sec\text{x}$
$\tan\text{x}\sec\text{x}=\sec\text{x}$
$\tan\text{x}=1$
$\text{x}=\frac{\pi}{4}$
Hence the value of x is $\frac{\pi}{4}$
Thus, the solution is $\text{x}=\text{n}\pi+\frac{\pi}{4}$
View full question & answer→Question 23 Marks
Prove the following results
$\tan\Big(\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{2}{3}\Big)=\frac{17}{6}$
Answer$\text{L.H.S=}\tan\Big(\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{2}{3}\Big)$
$=\tan\begin{pmatrix}\tan^{-1}\frac{\sqrt{1-\big(\frac{4}{5}\big)^2}}{\frac{4}{5}}+\tan^{-1}\frac{2}{3}\end{pmatrix}$
$\bigg[\because\ \cos^{-1}\text{x}=\tan^{-1}\bigg(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\bigg)\bigg]$
$=\tan\Big(\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{2}{3}\Big)$
$=\tan\Bigg[\tan^{-1}\bigg(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times\frac{2}{3}}\bigg)\Bigg]$
$\Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x+y}}{1-\text{xy}}\Big)\Big]$
$=\tan\Bigg[\tan^{-1}\bigg(\frac{\frac{17}{12}}{\frac{6}{12}}\bigg)\Bigg]$
$=\tan\Big[\tan^{-1}\frac{17}{6}\Big]$
$=\frac{17}{6}=\text{R.H.S}$
View full question & answer→Question 33 Marks
Prove that $\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$
Answer$\text{L.H.S}=\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}$
$ =\cos^{-1}\Bigg[\frac{4}{5}\times\frac{12}{13}-\sqrt{1-\Big(\frac{4}{5}\Big)^2}\sqrt{1-\Big(\frac{12}{13}\Big)^2}\Bigg)$
$\Big[\because\ \cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)\Big]$
$ =\cos^{-1}\Big[\frac{48}{65}-\frac{3}{5}\times\frac{5}{12}\Big]$
$=\cos^{-1}\Big(\frac{48-15}{65}\Big)$
$=\cos^{-1}\frac{33}{65}=\text{R.H.S}$
View full question & answer→Question 43 Marks
Write the following in the simplest form:
$\tan^{-1}\Big\{\sqrt{1+\text{x}^2}-\text{x}\Big\},\text{x}\in\text{R}$
AnswerLet $\text{x}=\cot\theta$
Now,
$\tan^{-1}\Big\{\sqrt{1+\text{x}^2}-\text{x}\Big\}$
$\tan^{-1}\Big\{\sqrt{1+\cot^2\theta}-\cot\theta\Big\}$
$=\tan^{-1}\{\text{cosec }\theta-\cot\theta\}$
$=\tan^{-1}\Big\{\frac{1-\cos\theta}{\sin\theta}\Big\}$
$=\tan^{-1}\Bigg\{\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\Bigg\}$
$=\tan^{-1}\Big\{\tan\Big(\frac{\theta}{2}\Big)\Big\}$
$=\frac{\theta}{2}$
$=\frac{\cot^{-1}\text{x}}{2}$
View full question & answer→Question 53 Marks
Write the following in the simplest form:
$\tan^{-1}\Big\{\text{x}+\sqrt{1+\text{x}^2}\Big\},\text{x}\in\text{R}$
AnswerLet $\text{x}=\cot\theta$
Now,
$\tan^{-1}\Big\{\text{x}+\sqrt{1+\text{x}^2}\Big\}$
$=\tan^{-1}\Big\{\cot\theta+\sqrt{1+\cot^2\theta}\Big\}$
$=\tan^{-1}\{\cot\theta+\text{cosec}\theta\}$
$=\tan^{-1}\Big\{\frac{\cos\theta+1}{\sin\theta}\Big\}$
$=\tan^{-1}\Bigg\{\frac{2\cos^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\Bigg\}$
$=\tan^{-1}\Big\{\cot\frac{\theta}{2}\Big\}$
$=\tan^{-1}\Big\{\tan\Big(\frac{\pi}{2}-\frac{\theta}{2}\Big)\Big\}$
$=\Big(\frac{\pi}{2}-\frac{\theta}{2}\Big)$
$=\frac{\pi}{2}-\frac{\cot^{-1}\text{x}}{2}$
View full question & answer→Question 63 Marks
Find the principal value of the following:
$\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)$
AnswerLet $\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)=\text{y}$ Then, $\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)=\text{y}$We know that the range of the principal value branch is $[0,\pi].$
Thus, $\cos\text{y}=-\frac{\sqrt3}{2}=\cos\Big(\frac{5\pi}{6}\Big)$ $\Rightarrow\text{y}=\frac{5\pi}{6}\in[0,\pi]$ Hence, the pricipal value of $\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)$ is $\frac{5\pi}{6}.$
View full question & answer→Question 73 Marks
For the principal values, evaluate the following:
$\sin^{-1}\Big(-\frac{1}{2}\Big)+2\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)$
Answer$\sin^{-1}\Big(-\frac{1}{2}\Big)+2\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)$
$=\sin^{-1}\Big\{\sin\Big(-\frac{\pi}{6}\Big)\Big\}+2\cos^{-1}\Big(\cos\frac{5\pi}{6}\Big)$ $\begin{bmatrix}\because\text{Range of shine is}\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big];-\frac{\pi}{6}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big] \\\ \\\text{and range of cosine is}[0,\pi];\frac{5\pi}{6}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big] \end{bmatrix}$
$=-\frac{\pi}{6}+2\Big(\frac{5\pi}{6}\Big)$
$=-\frac{\pi}{6}+\frac{5\pi}{3}$
$=\frac{9\pi}{6}$
$=\frac{3\pi}{2}$
$\therefore\sin^{-1}\Big(-\frac{1}{2}\Big)+2\cos^{-1}\Big(-\frac{\sqrt3}{2}\Big)=\frac{3\pi}{2}$$$
View full question & answer→Question 83 Marks
For the principal values, evaluate the following:
$\tan^{-1}\big(\sqrt3\big)-\sec^{-1}(-2)$
AnswerLet $\tan^{-1}\big(\sqrt3\big)=\text{x}.$ Then, $\tan\text{x}=\sqrt3=\tan\Big(\frac{\pi}{3}\Big)$
$\therefore\tan^{-1}\big(\sqrt3\big)=\frac{\pi}{3}$
Let $\sec^{-1}(-2)=\text{y}.$ Then, $\sec\text{y}=-2=\sec\Big(\pi-\frac{\pi}{3}\Big)$
$\therefore\sec^{-1}(-2)=\frac{2\pi}{3}$
$\therefore\tan^{-1}\big(\sqrt3\big)-\sec^{-1}(-2)=\frac{\pi}{3}-\frac{2\pi}{3}$
$=\frac{\pi-2\pi}{3}=-\frac{\pi}{3}$
View full question & answer→Question 93 Marks
What is the value of $\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)$
Answer$\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)$
$=\frac{2\pi}{3}+\Big(\pi-\frac{2\pi}{3}\Big)=\pi$
$\begin{Bmatrix}\text{Since},\sin^{-1}(\sin\theta)=\begin{cases}-\pi-\theta,&\text{if }\theta\in\Big[\frac{-3\pi}{2},\frac{-\pi}{2}\Big]\\\theta,&\text{if }\theta\in\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]\\\pi-\theta,&\text{if }\theta\in\Big[\frac{\pi}{2},\frac{3\pi}{2}\Big]\\-2\theta+\theta&\text{if }\theta\in\Big[\frac{3\pi}{2},\frac{5\pi}{2}\Big]\end{cases}\\\text{And }\cos^{-1}(\cos\theta)\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\theta+\theta&\text{if }\theta\in[2\pi,3\pi]\end{cases}\end{Bmatrix}$
Hence,
$\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{2\pi}{3}\Big)=\pi$
View full question & answer→Question 103 Marks
Evaluate the following:
$\tan^{-1}1+\cos^{-1}\Big(-\frac{1}{2}\Big)+\sin^{-1}\Big(-\frac{1}{2}\Big)$
AnswerLet $\tan^{-1}=\text{x}.$ Then, $\tan\text{x}=1=\tan\frac{\pi}{4}$ $\therefore\tan^{-1}(1)=\frac{\pi}{4}$Let $\cos^{-1}\Big(-\frac{1}{2}\Big)=\text{y.}$ Then,
$\cos\text{y}=-\frac{1}{2}=-\cos\Big(\frac{\pi}{3}\Big)=\cos\Big(\pi-\frac{\pi}{3}\Big)=\cos\Big(\frac{2\pi}{3}\Big)$
$\therefore\cos^{-1}\Big(-\frac{1}{2}\Big)=\frac{2\pi}{3}$Let $\sin^{-1}\Big(-\frac{1}{2}\Big)=\text{z.}$ Then, $\sin\text{z}=-\frac{1}{2}=-\sin\Big(\frac{\pi}{6}\Big)=\sin\Big(-\frac{\pi}{6}\Big)$
$\therefore\sin^{-1}\Big(-\frac{1}{2}\Big)=-\frac{\pi}{6}$
$\therefore\tan^{-1}(1)+\cos^{-1}\Big(-\frac{1}{2}\Big)+\sin^{-1}\Big(-\frac{1}{2}\Big)$
$=\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 113 Marks
Write the value of $\cos^{-1}(\cos350^\circ)-\sin^{-1}(\sin350^\circ)$
Answer$\cos^{-1}(\cos350^\circ)-\sin^{-1}(\sin350^\circ)$
$=\cos^{-1}\{\cos(360^\circ-10^\circ)\}-\sin^{-1}\{\sin(360^\circ-10^\circ)\}$
$=\cos^{-1}\{\cos10^\circ\}-\sin^{-1}\{\sin10^\circ\}$
$\{\text{Since},\cos(2\pi-\theta)=\cos\theta,\sin(2\pi-\theta)=-\sin\theta\}$
$=10^\circ-\sin^{-1}\{\sin(-10^\circ)\}$
$\{\text{Since},\cos^{-1}(\cos\theta),\text{ if }\theta\in[0,\pi]\text{ and }\sin(-\theta)=-\sin\theta\}$
$=10^\circ-(-10^\circ)$ $\Big\{\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big\}$
$=10^\circ+10^\circ$
$=20^\circ$
Hence,
$\cos^{-1}(\cos350^\circ)-\sin^{-1}(\sin350^\circ)=20^\circ$
View full question & answer→Question 123 Marks
Prove the following results
$\sin\Big(\cos^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)=\frac{63}{65}$
Answer$\text{L.H.S}=\sin\Big(\cos^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)=\frac{63}{65}$
$=\sin\Bigg[\sin6{-1}\sqrt{1-\Big(\frac{3}{5}\Big)^2}+\sin^{-1}\frac{5}{13}\Bigg]$
$=\sin\Big[\sin^{-1}\frac{4}{5}+\sin^{-1}\frac{5}{13}\Big]$
$=\sin\Bigg\{\sin^{-1}\Bigg[\frac{4}{5}\times\sqrt{1-\Big(\frac{5}{13}\Big)^2}+\frac{5}{13}\times\sqrt{1-\Big(\frac{4}{5}\Big)^2}\Bigg]\Bigg\}$
$=\sin\Big[\sin^{-1}\Big(\frac{48}{65}+\frac{15}{65}\Big)\Big]$
$=\sin\Big(\sin^{-1}\frac{63}{65}\Big)$
$=\frac{63}{65}=\text{R.H.S}$
View full question & answer→Question 133 Marks
Solve the following equation for x:
$3\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}-4\cos^{-1}\frac{1-\text{x}^2}{1+\text{x}^2}+2\tan^{-1}\frac{2\text{x}}{1-\text{x}^2}=\frac{\pi}{3}$
Answer$3\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}-4\cos^{-1}\frac{1-\text{x}^2}{1+\text{x}^2}+2\tan^{-1}\frac{2\text{x}}{1-\text{x}^2}=\frac{\pi}{3}$
$\Rightarrow3\big(2\tan^{-1}\text{x}\big)-4\big(2\tan^{-1}\text{x}\big)+2\big(2\tan^{-1}\text{x}\big)=\frac{\pi}{3}$
$\Big\{\text{Since},2\tan^{-1}\text{x}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}=\cos^{-1}\frac{1-\text{x}^2}{1+\text{x}^2}\Big\}$
$\Rightarrow6\tan^{-1}\text{x}-8\tan^{-1}\text{x}+4\tan^{-1}\text{x}=\frac{\pi}{3}$
$\Rightarrow2\tan^{-1}\text{x}=\frac{\pi}{3}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}$
View full question & answer→Question 143 Marks
Prove the following results:
$\sin^{-1}\frac{63}{65}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
Answer$\text{R.H.S}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
$=\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{4}{5}$ $\Big[\because\ \cos^{-1}\text{x}=\sin^{-1}\sqrt{1-\text{x}^2}\Big]$
$=\sin^{-1}\bigg\{\frac{5}{13}\sqrt{1-\Big(\frac{4}{5}\Big)^2}+\frac{4}{5}\sqrt{1-\Big(\frac{5}{13}\Big)^2}\bigg\}$
$=\sin^{-1}\Big\{\frac{5}{13}\times\frac{3}{5}+\frac{4}{5}\times\frac{12}{13}\Big\}$
$=\sin^{-1}\Big\{\frac{15}{65}+\frac{48}{65}\Big\}$
$=\sin^{-1}\frac{63}{65}=\text{L.H.S}$
View full question & answer→Question 153 Marks
For the principal values, evaluate the following:
$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)-2\sec^{-1}\Big(2\tan\frac{\pi}{6}\Big)$
Answer$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)-2\sec^{-1}\Big(2\tan\frac{\pi}{6}\Big)$
$=-\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)-2\sec^{-1}\Big(2\times\frac{1}{\sqrt3}\Big)$
$=-\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)-2\sec^{-1}\Big(\frac{2}{\sqrt3}\Big)$
$=-\sin^{-1}\Big(\sin\frac{\pi}{3}\Big)-2\sec^{-1}\Big(\sec\frac{\pi}{6}\Big)$
$=-\frac{\pi}{3}-\frac{\pi}{3}$
$=-\frac{2\pi}{3}$
View full question & answer→Question 163 Marks
Solve:
$\sin\Big(\sin^{-1}\frac{1}{5}+\cos^{-1}\text{x}\Big)=1$
Answer$\sin\Big(\sin^{-1}\frac{1}{5}+\cos^{-1}\text{x}\Big)=1$
$\Rightarrow\sin^{-1}\frac{1}{5}+\cos^{-1}\text{x}=\sin^{-1}1$
$\Rightarrow\sin^{-1}\frac{1}{5}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\sin^{-1}\frac{1}{5}=\frac{\pi}{2}-\cos^{-1}\text{x}$
$\Rightarrow\sin^{-1}\frac{1}{5}=\sin^{-1}\text{x}$ $\Big[\because\ \sin^{-1}\text{x}=\frac{\pi}{2}-\cos^{-1}\text{x}\Big]$
$\Rightarrow\text{x}=\frac{1}{5}$
View full question & answer→Question 173 Marks
Solve the following:
$\cos^{-1}\text{x}+\sin^{-1}\frac{\text{x}}{2}=\frac{\pi}{6}$
Answer$\cos^{-1}\text{x}+\sin^{-1}\frac{\text{x}}{2}=\frac{\pi}{6}$
$\Rightarrow\sin^{-1}\frac{\text{x}}{2}=\sin^{-1}\Big(\frac{1}{2}\Big)-\sin^{-1}\Big(\sqrt{1-\text{x}^2}\Big)$
$\Rightarrow\sin^{-1}\frac{\text{x}}{2}=\sin^{-1}\Big[\frac{1}{2}\sqrt{1-1+\text{x}^2}-\sqrt{1-\text{x}^2}\sqrt{1-\frac{1}{4}}\Big]$
$\Rightarrow\frac{\text{x}}{2}=\frac{\text{x}}{2}-\frac{\sqrt3\sqrt{1-\text{x}^2}}{2}$
$\Rightarrow\frac{\sqrt3\sqrt{1-\text{x}^2}}{2}=0$
$\Rightarrow\sqrt{1-\text{x}^2}=0$
$\Rightarrow\text{x}=\pm\frac{1}{2}$
View full question & answer→Question 183 Marks
Find the principal values of the following:
$\text{cosec}^{-1}\Big(2\cos\frac{2\pi}{3}\Big)$
AnswerLet $\text{cosec}^{-1}\Big(2\cos\frac{2\pi}{3}\Big)=\text{y}$ Then, $\text{cosec y}=2\cos\frac{2\pi}{3}$ We know that the range of the principal value branch is $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]-\{0\}$Thus,
$\text{cosec y}=2\cos\frac{2\pi}{3}=2\times\frac{-1}{2}=-1=\text{cosec}\Big(-\frac{\pi}{2}\Big)$ $\Rightarrow\text{y}=-\frac{\pi}{2}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big],\text{y}\neq0$ Hence, the principal value of $\text{cosec}^{-1}\Big(2\cos\frac{2\pi}{3}\Big)$ is $-\frac{\pi}{2}.$
View full question & answer→Question 193 Marks
Prove the following results:
$\tan^{-1}\frac{2}{3}=\frac{1}{2}\tan^{-1}\frac{12}{5}$
Answer$\text{L.H.S}=\tan^{-1}\frac{2}{3}$
$=\frac{1}{2}\tan^{-1}\begin{Bmatrix}\frac{2\times\frac{2}{3}}{1-\Big(\frac{2}{3}\Big)^2}\end{Bmatrix}$ $\Big[\because\ \tan^{-1}\text{x}=\frac{1}{2}\tan^{-1}\Big\{\frac{2\text{x}}{1-\text{x}^2}\Big\}\Big]$
$=\frac{1}{2}\tan^{-1}\Bigg\{\frac{\frac{4}{3}}{\frac{5}{9}}\Bigg\}$
$=\frac{1}{2}\tan^{-1}\frac{12}{5}=\text{R.H.S}$
View full question & answer→Question 203 Marks
Evaluate the following:
$\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)+\tan^{-1}\Big(-\sqrt3\Big)+\tan^{-1}\Big(\sin\Big(-\frac{\pi}{2}\Big)\Big)$
Answer$\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)+\tan^{-1}\Big(-\sqrt3\Big)+\tan^{-1}\Big(\sin\Big(-\frac{\pi}{2}\Big)\Big)$
$=\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)+\tan^{-1}\Big(-\sqrt3\Big)+\tan^{-1}\Big(-\sin\Big(\frac{\pi}{2}\Big)\Big)$
$=\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)+\tan^{-1}\Big(-\sqrt3\Big)+\tan^{-1}(-1)$
$=-\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)-\tan^{-1}\Big(-\sqrt3\Big)-\tan^{-1}(1)$
$=-\tan^{-1}\Big(\tan\frac{\pi}{6}\Big)-\tan^{-1}\Big(\frac{\pi}{3}\Big)-\tan^{-1}\Big(\frac{\pi}{4}\Big)$
$=-\frac{\pi}{6}-\frac{\pi}{3}-\frac{\pi}{4}$
$=-\frac{3\pi}{4}$
View full question & answer→Question 213 Marks
Find the principal value of the following:
$\sec^{-1}\Big(2\tan\frac{3\pi}{4}\Big)$
AnswerLet $\sec^{-1}\Big(2\tan\frac{3\pi}{4}\Big)=\text{y}$Then,
$\sec\text{y}=2\tan\frac{3\pi}{4}$ We know that the range of the principal value branch is $[0,\pi]-\Big\{\frac{\pi}{2}\Big\}.$ Thus, $\sec\text{y}=2\tan\frac{3\pi}{4}=2\times(-1)$ $=-2=\sec\Big(\frac{2\pi}{3}\Big)$ $\Rightarrow\text{y}=\frac{2\pi}{3}\in[0,\pi]$ Hence, the principal value of $\sec^{-1}\Big(2\tan\frac{3\pi}{4}\Big)$ is $\frac{2\pi}{3}.$
View full question & answer→Question 223 Marks
Prove the following results
$\tan\Big(\sin^{-1}\frac{15}{13}+\cos^{-1}\frac{3}{5}\Big)=\frac{63}{16}$
Answer$\tan\Big(\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}\Big)$
$\tan\Big(\tan^{-1}\frac{5}{12}+\tan^{-1}\frac{4}{3}\Big)\\\dots\dots\begin{bmatrix}\sin^{-1}\Big(\frac{\text{p}}{\text{h}}\Big)=\tan^{-1}\Big(\frac{\text{p}}{\text{b}}\Big)\\\cos^{-1}\Big(\frac{\text{b}}{\text{h}}\Big)=\tan^{-1}\Big(\frac{\text{p}}{\text{b}}\Big)\end{bmatrix}$
$=\tan\Bigg(\tan^{-1}\bigg(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times\frac{4}{3}}\bigg)\Bigg)\dots\dots\\\Big[\tan^{-1}(\text{x})+\tan^{-1}(\text{y})=\tan^{-1}\Big(\frac{\text{x+y}}{1-\text{xy}}\Big)\Big]$
$$$=\tan\bigg(\tan^{-1}\bigg(\frac{\frac{21}{12}}{\frac{4}{9}}\bigg)\bigg)$
$=\tan\Big(\tan^{-1}\Big(\frac{63}{16}\Big)\Big)$
$=\frac{63}{16}$
View full question & answer→Question 233 Marks
Write the value of $\tan^{-1}\text{x}+\tan^{-1}\Big(\frac{1}{\text{x}}\Big)$ for x > 0.
Answer$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big),\text{xy}<1$
$\therefore\ \tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}=\tan^{-1}\bigg(\frac{\text{x}+\frac{1}{\text{x}}}{1-\text{x}\frac{1}{\text{x}}}\bigg),\text{x}>0$
$=\tan^{-1}\Big(\frac{\text{x}^2+1}{0}\Big)$
$=\tan^{-1}(\infty)$
$=\tan^{-1}\Big(\tan\frac{\pi}{2}\Big)$
$=\frac{\pi}{2}$
$\therefore\ \tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{2}$
View full question & answer→Question 243 Marks
Prove the following results:
$4\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}=\frac{\pi}{4}$
Answer$4\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}$
$=\tan^{-1}\begin{bmatrix}\frac{4\big(\frac{1}{5}\big)-4\big(\frac{1}{5}\big)^3}{1-6\big(\frac{1}{5}\big)^2+\big(\frac{1}{5}\big)^4}\end{bmatrix}-\tan^{-1}\frac{1}{239}$ $\Big[4\tan^{-1}(\text{x})=\tan^{-1}\Big(\frac{4\text{x}-4\text{x}^3}{1-6\text{x}^2+\text{x}^4}\Big)\Big]$
$=\tan^{-1}\Big[\frac{120}{119}\Big]-\tan^{-1}\frac{1}{239}$
$=\tan^{-1}\Big(\frac{120\times239-119}{119\times239+120}\Big)$ $\Big[\tan^{-1}(\text{x})-\tan^{-1}(\text{y})=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$
$=\tan^{-1}\Big(\frac{28561}{28561}\Big)$
$=\tan^{-1}(1)$
$=\frac{\pi}{4}$
View full question & answer→Question 253 Marks
Find the domain of $\text{f(x)}=2\cos^{-1}2\text{x}=\sin^{-1}\text{x}.$
AnswerDomain of $\cos^{-1}\text{x}$ lies in the interval [-1, 1]
$\therefore$ Domain of $\cos^{-1}(2\text{x})$ lies in the interval [-1, 1]
$\Rightarrow-1\leq2\text{x}\leq1$
$\Rightarrow\frac{-1}{2}\leq\text{x}\leq\frac{1}{2}$
Domain of $\cos^{-1}(2\text{x})$ is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$
Domain of $\cos^{-1}\text{x}$ lies in the interval [-1, 1]
$\therefore$ Domain of $\cos^{-1}(2\text{x})+\sin^{-1}\text{x}$ lies in the interval $\Big[-\frac{1}{2},\frac{1}{2}\Big].$
View full question & answer→Question 263 Marks
Find the principal values of the following:
$\text{cosec}^{-1}\Big(\frac{2}{\sqrt3}\Big)$
AnswerLet $\text{cosec}^{-1}\Big(\frac{2}{\sqrt3}\Big)=\text{y}$ Then, $\text{cosec y}=\frac{2}{\sqrt3}$ We know that the range of the principal value branch is $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]-\{0\}$Thus,
$\text{cosec y}=\frac{2}{\sqrt3}=\text{cosec}\Big(\frac{\pi}{3}\Big)$ $\Rightarrow\text{y}=\frac{\pi}{3}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big],\text{y}\neq0$ Hence, the principal value of $\text{cosec}^{-1}\Big(\frac{2}{\sqrt3}\Big)$ is $\frac{\pi}{3}.$
View full question & answer→Question 273 Marks
If x < 1, then write the value of $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$ in terms of $\tan^{-1}\text{x.}$
AnswerHence,
$\text{x}<0$
$\Rightarrow-\infty<\text{x}<0$
Let, $\text{x}=\tan\theta$
$\Rightarrow-\infty<\tan\theta<0$
$\Rightarrow-\frac{\pi}{2}<\theta<0$
Multiplying by (-2),
We know that
$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$\cos(-2\theta)=\frac{1-\text{x}^2}{1+\text{x}^2}$
$-2\theta=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$-2\tan^{-1}\text{x}=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$ $\{\text{Since},\text{x}=\tan\theta=\tan^{-1}\text{x}\}$
So,
$\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)=-2\tan^{-1}\text{x}$
View full question & answer→Question 283 Marks
Write the value of $\cos^{-1}\Big(\frac{1}{2}\Big)+2\sin^{-1}\Big(\frac{1}{2}\Big).$
Answer$\cos^{-1}\Big(\frac{1}{2}\Big)+2\sin^{-1}\Big(\frac{1}{2}\Big)$
$=\frac{\pi}{3}+2\times\frac{\pi}{6}$
$\begin{Bmatrix}\text{Since},\cos^{-1}\text{x}=\text{An angle in }[0,\pi]\text{ whose cosin is x}\\\sin^{-1}\text{x}=\text{An angle in }\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\text{ whose sine is x}\end{Bmatrix}$
$=\frac{\pi}{3}+\frac{\pi}{3}$
$=\frac{2\pi}{3}$
So,
$\cos^{-1}\Big(\frac{1}{2}\Big)+2\sin^{-1}\Big(\frac{1}{2}\Big)=\frac{2\pi}{3}.$
View full question & answer→Question 293 Marks
Evaluate $\sin\Big(\tan^{-1}\frac{3}{4}\Big).$
AnswerWe know that
$\tan^{-1}\text{x}=\sin^{-1}\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
$\therefore\ \sin\Big(\tan^{-1}\frac{3}{4}\Big)=\sin\Bigg\{\sin^{-1}\Bigg(\frac{\frac{3}{4}}{\sqrt{1+\frac{9}{10}}}\Bigg)\Bigg\}$
$=\sin\Bigg\{\sin^{-1}\Bigg(\frac{\frac{3}{4}}{\frac{5}{4}}\Bigg)\Bigg\}$
$=\sin\Big(\sin^{-1}\frac{3}{5}\Big)$
$=\frac{3}{5}$ $\big[\because\ \sin\big(\sin^{-1}\text{x}\big)=\text{x}\big]$
$\therefore\ \Big(\tan^{-1}\frac{3}{4}\Big)=\frac{3}{5}$
View full question & answer→Question 303 Marks
Write the value of $\sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)+\cos^{-1}\Big(\frac{-1}{2}\Big)$
Answer$\sin^{-1}(-\text{x})=\sin^{-1}\text{x},\text{x}\in[-1,1]$
$\cos^{-1}(-\text{x})=\pi-\cos^{-1}\text{x},\text{x}\in[-1,1]$
$\therefore\ \sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)+\cos^{-1}\Big(\frac{-1}{2}\Big)$
$=-\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)+\pi-\cos^{-1}\Big(\frac{1}{2}\Big)$
$=-\sin^{-1}\Big(\sin\frac{\pi}{3}\Big)+\pi-\cos^{-1}\Big(\cos\frac{\pi}{3}\Big)$
$=-\frac{\pi}{3}+\pi-\frac{\pi}{3}$
$=\frac{\pi}{3}$
$\therefore\ \sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)+\cos^{-1}\Big(\frac{-1}{2}\Big)=\frac{\pi}{3}$
View full question & answer→Question 313 Marks
Prove that
$\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+\cot^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)=\frac{\pi}{2}$
Answer$\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+\cot^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)=\frac{\pi}{2}$
$\text{L.H.S}=\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+\cot^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)$
$=\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+\frac{\pi}{2}-\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)$
$\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=\frac{\pi}{2}=\text{R..H.S}$
View full question & answer→Question 323 Marks
Write the following in the simplest form:
$\sin^{-1}\Big\{\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\Big\},-1<\text{x}<1$
AnswerLet, $\text{x}=\text{a}\sin\theta$
Now,
$\sin^{-1}\Big\{\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\Big\}=\sin^{-1}\Big\{\frac{\sin\theta+\sqrt{1-\sin^2\theta}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\cos\theta}{\sqrt2}\Big\}$
$=\sin^{-1}\Big\{\frac{1}{\sqrt2}\sin\theta=\frac{1}{\sqrt2}\cos\theta\Big\}$
$=\sin^{-1}\Big\{\cos\frac{\pi}{4}\sin\theta+\sin\frac{\pi}{4}\cos\theta\Big\}$
$=\sin^{-1}\Big\{\sin\Big(\theta+\frac{\pi}{4}\Big)\Big\}$
$=\theta+\frac{\pi}{4}$
$=\frac{\pi}{4}=\sin^{-1}\text{x}$
$\therefore\ \sin^{-1}\bigg\{\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\bigg\}=\cos^{-1}\text{x}+\frac{\pi}{4}$
View full question & answer→Question 333 Marks
Prove the following results:
$2\tan^{-1}\frac{3}{4}-\tan^{-1}\frac{17}{31}=\frac{\pi}{4}$
Answer$\text{L.H.S}=2\tan^{-1}\frac{3}{4}-\tan^{-1}\frac{17}{31}$ $=\tan^{-1}\Bigg(\frac{2\times\frac{3}{4}}{1-\big(\frac{3}{4}\big)^2}\Bigg)+\tan^{-1}\Big(\frac{17}{31}\Big)$ $\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2}\Big]$ $=\tan^{-1}\Bigg\{\frac{\frac{3}{2}}{\frac{7}{16}}\Bigg\}-\tan^{-1}\Big(\frac{17}{31}\Big)$ $=\tan^{-1}\Big(\frac{24}{7}\Big)+\tan^{-1}\Big(\frac{17}{31}\Big)$ $=\tan^{-1}\Bigg(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}\times\frac{17}{31}}\Bigg)$ $\Big[\text{Since }\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$$=\tan^{-1}\Bigg(\frac{\frac{625}{217}}{\frac{625}{217}}\Bigg)$
$=\tan^{-1}(1)=\frac{\pi}{4}=\text{R.H.S}$
View full question & answer→Question 343 Marks
Find the principal values of each of the following:
$\text{cosec}^{-1}\big(-\sqrt2\big)$
AnswerLet $\text{cosec}^{-1}\big(-\sqrt2\big)=\text{y}$ Then, $\text{cosec}\text{y}=-\sqrt{2}$We know that the range of the principal value branch is $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]-\{0\}$
Thus, $\text{cosec}\text{y}=-\sqrt{2}=\text{cosec}\Big(-\frac{\pi}{4}\Big)$ $\text{y}=-\frac{\pi}{4}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big],\text{y}\neq0$ Hence, the principal value of $\text{cosec}^{-1}\big(-\sqrt2\big)$ is $-\frac{\pi}{4}.$
View full question & answer→Question 353 Marks
Write the value of $\sin^{-1}(\sin(-600^\circ))\sin(-600^\circ).$
Answer$\sin^{-1}\{\sin(-600^\circ)\}$
$=\sin^{-1}\{\sin(-600^\circ+360\times2)\}$ $\{\text{Since},\sin(2\text{n}\pi+\theta)=\sin\theta\}$
$=\sin^{-1}\{\sin120^\circ\}$
$=180^\circ-120^\circ$
$\begin{Bmatrix}\text{Since},\sin^{-1}(\sin\theta)=\begin{cases}-\pi-\theta,&\text{ if }\theta\in\Big[\frac{-3\pi}{2},\frac{-\pi}{2}\Big]\\\theta,&\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\\\pi}{2}\Big]\\\pi-\theta,&\text{ if }\theta\in\Big[\frac{\pi}{2},\frac{3\pi}{2}\Big]\\\pi-\theta,&\text{ if }\theta\in\Big[\frac{3\pi}{2},\frac{5\pi}{2}\Big]\end{cases}\end{Bmatrix}$
$=60^\circ$
Hence,
$\sin^{-1}\{\sin(-600^\circ)\}=60^\circ$
View full question & answer→Question 363 Marks
Write the value of $\cos\Big(2\sin^{-1}\frac{1}{3}\Big).$
AnswerLet $\text{y}=\sin^{-1}\frac{1}{3}$
Then, $\sin\text{y}=\frac{1}{3}$
Now, $\cos\text{y}=\sqrt{1-\sin^2\text{y}}$
$\Rightarrow\cos\text{y}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt2}{3}$
$\cos\Big(2\sin^{-1}\frac{1}{3}\Big)=\cos(2\text{y})$
$=\cos^2\text{y}-\sin^2\text{y}$ $\big[\because\ \cos2\text{x}=\cos^2\text{x}-\sin^2\text{x}\big]$
$=\Big(\frac{2\sqrt2}{3}\Big)^2-\Big(\frac{1}{3}\Big)^2$
$=\frac{8}{9}-\frac{1}{9}$
$=\frac{7}{9}$
$\because\ \cos\Big(2\sin^{-1}\frac{1}{3}\Big)=\frac{7}{9}$
View full question & answer→Question 373 Marks
Prove the following results:
$\frac{9\pi}{4}-\frac{9}{4}\sin^{-1}\frac{1}{3}=\frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}$
Answer$\text{L.H.S}=\frac{9\pi}{4}-\frac{9}{4}\sin^{-1}\frac{1}{3}$
$=\frac{9}{4}\Big(\frac{\pi}{2}-\sin^{-1}\frac{1}{3}\Big)...(1)$ $\Big[\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
Now, let $\cos^{-1}\frac{1}{3}=\text{x}.$ Then,
$\cos\text{x}=\frac{1}{3}\Rightarrow\sin\text{x}=\sqrt{1-\Big(\frac{1}{3}\Big)^2}=\frac{2\sqrt2}{3}.$
$\therefore\ \text{x}=\sin^{-1}\frac{2\sqrt2}{3}\Rightarrow\cos^{-1}\frac{1}{3}=\frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}$
$\therefore\ \text{L.H.S}=\frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}=\text{R.H.S}$
View full question & answer→Question 383 Marks
Evaluate the following:
$\cos^{-1}\Big\{\cos\Big(-\frac{\pi}{4}\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}$
$\therefore\ \cos^{-1}\Big\{\cos\Big(-\frac{\pi}{4}\Big)\Big\}$
$=-\Big(-\frac{\pi}{4}\Big)$ $\Big\{\because-\frac{\pi}{4}\in[-\pi,0]\Big\}$
$=\frac{\pi}{4}$
Hence,
$\therefore\ \cos^{-1}\Big\{\cos\Big(-\frac{\pi}{4}\Big)\Big\}=\frac{\pi}{4}$
View full question & answer→Question 393 Marks
For the principal values, evaluate the following:
$\sin^{-1}[\cos\{2\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 c}\Big(-\frac{\pi}{6}\Big)$ $\Rightarrow\text{x}=-\frac{\pi}{6}$ $\sin^{-1}[\cos\{2\text{cosec}^{-1}(-2)\}]=\sin^{-1}\Big[\cos\Big\{2\times\Big(-\frac{\pi}{6}\Big)\Big\}\Big]$ $=\sin^{-1}\Big[\cos\Big(-\frac{\pi}{3}\Big)\Big]=\sin^{-1}\Big[\frac{1}{2}\Big]$ $\sin^{-1}\text{x}$ represents an angle in $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$ whose sin is x.Let $\text{x}=\sin^{-1}\Big[\frac{1}{2}\Big]$
$\Rightarrow\sin\text{x}=\frac{1}{2}=\sin\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=\frac{\pi}{6}$
View full question & answer→Question 403 Marks
Evaluate the following:
$\tan^{-1}\Big(\tan\frac{5\pi}{6}\Big)+\cos^{-1}\Big\{\cos\Big(\frac{13\pi}{6}\Big)\Big\}$
Answer$\tan^{-1}\Big(\tan\frac{5\pi}{6}\Big)+\cos^{-1}\Big\{\cos\Big(\frac{13\pi}{6}\Big)\Big\}$
$=\tan^{-1}\Big\{\tan\Big(\pi-\frac{5\pi}{6}\Big)\Big\}+\cos^{-1}\Big\{\cos\Big(2\pi+\frac{\pi}{6}\Big)\Big\}$
$=\tan^{-1}\Big\{-\tan\Big(\frac{\pi}{6}\Big)\Big\}+\cos^{-1}\Big\{\cos\Big(\frac{\pi}{6}\Big)\Big\}$
$=-\tan^{-1}\Big\{\tan\Big(\frac{\pi}{6}\Big)\Big\}+\cos^{-1}\Big\{\cos\Big(\frac{\pi}{6}\Big)\Big\}$
$=-\frac{\pi}{6}+\frac{\pi}{6}$
$=0$
View full question & answer→Question 413 Marks
Evaluate: $\sin^{-1}\Big(\sin\frac{3\pi}{5}\Big).$
Answer$\sin^{-1}\Big(\sin\frac{3\pi}{5}\Big).$
$ =\pi-\frac{3\pi}{5}$
$\begin{Bmatrix}\text{Since},\sin^{-1}(\sin\theta)=\begin{cases}-\pi-\theta,&\text{if }\theta\in\Big[-\frac{3\pi}{2},-\frac{\pi}{2}\Big]\\\theta,&\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\\\pi-\theta,&\text{if }\theta\in\Big[\frac{\pi}{2},\frac{\\\pi}{2}\Big]\\-2\pi+\theta,&\text{if }\theta\in\Big[\frac{3\pi}{2},\frac{5\pi}{2}\Big]\end{cases}\end{Bmatrix}$
$=\frac{2\pi}{5}$
View full question & answer→Question 423 Marks
Write the difference between maximum and minimum values of $\sin^{-1}\text{x}$ for $\text{x}\in[-1,1].$
AnswerWe have to find the difference between maximum and minimum values of $\sin^{-1}\text{x}$ for $\text{x}\in[-1,1]$ We know that, $\sin^{-1}\text{x}=$ An angle in $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$ whose sin is x. So, minimum value of $\sin^{-1}\text{x}=-\frac{\pi}{2}$ maximum value of $\sin^{-1}\text{x}=\frac{\pi}{2}$Difference between maximum and minimum values of
$\sin^{-1}\text{x}=\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)$ $=\frac{\pi}{2}+\frac{\pi}{2}$ $=\pi$ The required difference $=\pi.$
View full question & answer→Question 433 Marks
Write the value of $\sin\Big\{\frac{\pi}{3}-\sin^{-1}\Big(-\frac{1}{2}\Big)\Big\}.$
Answer$\sin\Big\{\frac{\pi}{3}-\sin^{-1}\Big(-\frac{1}{2}\Big)\Big\}$
$=\sin\Big\{\frac{\pi}{3}+\sin^{-1}\Big(\frac{1}{2}\Big)\Big\}$ $\big\{\text{Since},\sin^{-1}(-\theta)=-\sin^{-1}\theta\big\}$
$=\sin\Big\{\frac{\pi}{3}+\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\}$
$=\sin\Big(\frac{\pi}{2}\Big)$
$=1$
Hence,
$\sin\Big\{\frac{\pi}{3}-\sin^{-1}\Big(-\frac{1}{2}\Big)\Big\}=1$
View full question & answer→Question 443 Marks
If $\big(\sin^{-1}\text{x}\big)^2+\big(\sin^{-1}\text{y}\big)^2+\big(\sin^{-1}\text{z}\big)^2=\frac{3}{4}\pi^2,$ find the value of $x^2 + y^2 + z^2$
AnswerRange of $\sin^{-1}$ is $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big].$
Given that $\big(\sin^{-1}\text{x}\big)^2+\big(\sin^{-1}\text{y}\big)^2+\big(\sin^{-1}\text{z}\big)^2=\frac{3}{4}\pi^2$
⇒ Each of $\sin^{-1}\text{x},\sin^{-1}\text{y}$ and $\sin^{-1}\text{z}$ takes value of $\frac{\pi}{2}.$
$\Rightarrow x = 1, y = 1$ and $z = 1$
$x^2 + y^2 + z^2 = 1 + 1 + 1 = 3$
View full question & answer→Question 453 Marks
For the principal values, evaluate the following:
$\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)+\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)$
Answer$\sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)+\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)$
$=\sin^{-1}\Big\{\sin\Big(-\frac{\pi}{3}\Big)\Big\}=\cos^{-1}\Big(\cos\frac{\pi}{6}\Big)$
$=-\frac{\pi}{3}+\frac{\pi}{6}$ $\begin{bmatrix}\because\text{Range of sine is }\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big];-\frac{\pi}{3}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\\\ \\\text{and range of cosine is }[0,\pi];\frac{\pi}{6}\in[0,\pi]\end{bmatrix}$
$=-\frac{\pi}{6}$
$\therefore\sin^{-1}\Big(-\frac{\sqrt3}{2}\Big)+\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)=-\frac{\pi}{6}$
View full question & answer→Question 463 Marks
Write the following in the simplest form:
$\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}$
Answer$\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}$
$=\sin\begin{Bmatrix}\sin^{-1}\begin{pmatrix}\frac{2\sqrt{\frac{1-\text{x}}{1+\text{x}}}}{1+\Big(\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big)^2}\end{pmatrix}\end{Bmatrix}$ $\Big\{\text{Since},2\tan^{-1}\text{x}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}\Big\}$
$=\sin\begin{Bmatrix}\sin^{-1}\begin{pmatrix}\frac{2\sqrt{\frac{1-\text{x}}{1+\text{x}}}}{\frac{1+\text{x}+1-\text{x}}{1+\text{x}}}\end{pmatrix}\end{Bmatrix}$
$=2\sqrt{\frac{1-\text{x}}{1+\text{x}}}\times\frac{1+\text{x}}{2}$
$=\sqrt{1-\text{x}}\sqrt{1+\text{x}}$
$=\sqrt{1-\text{x}^2}$
Hence, $\sin\Big\{2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}=\sqrt{1-\text{x}^2}$
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If $\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}+\sin^{-1}\frac{2\text{b}}{1+\text{b}^2}=2\tan^{-1}\text{x},$ prove that $\text{x}=\frac{\text{a}+\text{b}}{1-\text{ab}}.$
AnswerLet: $\text{a}=\tan\text{z}$ $\text{b}=\tan\text{y}$ Then, $\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}+\sin^{-1}\frac{2\text{b}}{1+\text{b}^2}=2\tan^{-1}\text{x}$$\Rightarrow\sin^{-1}\frac{2\tan\text{z}}{1+\tan^2\text{z}}+\sin^{-1}\frac{2\tan\text{y}}{1+\tan^2\text{y}}=2\tan^{-1}\text{x}$
$\Rightarrow\sin^{-1}(\sin2\text{z})+\sin^{-1}(\sin2\text{y})=2\tan^{-1}\text{x}$ $\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big]$ $\Rightarrow2\text{z}+2\text{y}=2\tan^{-1}\text{x}$ $\Rightarrow\tan^{-1}\text{a}+\tan^{-1}\text{b}=\tan^{-1}\text{x}$ $[\because\ \text{a}=\tan\text{z}\text{ and }\text{b}=\tan\text{y}]$ $\Rightarrow\tan^{-1}\frac{\text{a}+\text{b}}{1-\text{ab}}=\tan^{-1}\text{x}$ $\Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big]$ $\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{1-\text{ab}}$
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Solve the following equation for x:
$\cos^{-1}\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)+\frac{1}{2}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{2\pi}{3}$
Answer$\cos^{-1}\Big(\frac{\text{x}^2-1}{\text{x}^2+1}\Big)+\frac{1}{2}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{2\pi}{3}$
$\Rightarrow\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\frac{1}{2}\times2\tan^{-1}\text{x}=\frac{2\pi}{3}$ $\Big[\because\ \tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=2\tan^{-1}\text{x}\Big]$
$\Rightarrow2\tan^{-1}\text{x}+\tan^{-1}\text{x}=\frac{2\pi}{3}$ $\Big[\because\ \cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)=2\tan^{-1}\text{x}\Big]$
$\Rightarrow3\tan^{-1}\text{x}=\frac{2\pi}{3}$
$\Rightarrow\tan^{-1}\text{x}=\frac{2\pi}{9}$
$\Rightarrow\text{x}=\tan\Big(\frac{2\pi}{9}\Big)$
View full question & answer→Question 493 Marks
Find the principal values of the following:
$\tan^{-1}\Big(\cos\frac{\pi}{2}\Big)$
AnswerLet $\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)=\text{y}$Then,
$\tan\text{y}=\cos\frac{\pi}{2}$
We know that the range of the principal value branch is $\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big).$
Thus,
$\tan\text{y}=\cos\frac{\pi}{2}=0=\tan(0)$
$\Rightarrow\text{y}=0\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$
Hence, the principal value of $\tan^{-1}\Big(\cos\frac{\pi}{2}\Big)$ is 0.
View full question & answer→Question 503 Marks
Write the value of $\cos^2\Big(\frac{1}{2}\cos^{-1}\frac{3}{5}\Big).$
AnswerLet $\text{y}=\cos^{-1}\Big(\frac{3}{5}\Big)$
$\Rightarrow\cos\text{y}=\frac{3}{5}$
Now,
$\cos^2\Big(\frac{1}{2}\cos^{-1}\frac{3}{5}\Big)=\cos^{2}\Big(\frac{1}{2}\text{y}\Big)$
$=\frac{\cos\text{y}+1}{2}$ $\big[\because\ \cos2\text{x}=\cos^2\text{x}-1\big]$
$=\frac{\frac{3}{5}+1}{2}$
$=\frac{\frac{8}{5}}{2}$
$=\frac{4}{5}$
$\therefore\ \cos^2\Big(\frac{1}{2}\cos^{-1}\frac{3}{5}\Big)=\frac{4}{5}$
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