Question 13 Marks
Write the following in the simplest form:
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\},0<\text{x}<1$
AnswerLet, $\text{x}=\cos\theta$
Now,
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{1+\cos\theta}+\sqrt{1-\cos\theta}}{2}\Big\}$
$=\sin^{-1}\Bigg\{\frac{\sqrt{2\cos^2\frac{\theta}{2}}+\sqrt{2\sin^2\frac{\theta}{2}}}{2}\Bigg\}$
$=\sin^{-1}\bigg\{\frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}{\sqrt3}\bigg\}$
$=\sin^{-1}\Big\{\frac{1}{\sqrt2}\sin\frac{\theta}{2}+\frac{1}{\sqrt2}\cos\frac{\theta}{2}\Big\}$
$=\sin^{-1}\Big\{\sin\Big(\frac{\theta}{2}+\frac{\pi}{4}\Big)\Big\}$
$=\frac{\theta}{2}+\frac{\pi}{4}$
$=\frac{\cos^{-1}\text{x}}{2}+\frac{\pi}{4}$
$\therefore\ \sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\}=\frac{\cos^{-1}\text{x}}{2}+\frac{\pi}{4}$
View full question & answer→Question 23 Marks
Find the principal value of the following:
$\sec^{-1}\Big(2\sin\frac{3\pi}{4}\Big)$
AnswerLet $\sec^{-1}\Big(2\sin\frac{3\pi}{4}\Big)=\text{y}$Then,
$\sec\text{y}=2\sin\frac{3\pi}{4}$ We know that the range of the principal value branch is $[0,\pi]-\Big\{\frac{\pi}{2}\Big\}.$ Thus, $\sec\text{y}=2\sin\frac{3\pi}{4}=2\times\frac{1}{\sqrt2}$ $=\sqrt2=\sec\frac{\pi}{4}$ $\Rightarrow\text{y}=\frac{\pi}{4}\in[0,\pi]$
View full question & answer→Question 33 Marks
Sum the following siries:
$\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{2}{9}+\tan^{-1}\frac{4}{33}+...+\tan^{-1}\frac{2^{\text{n}-1}}{1+2^{2\text{n}-1}}$
Answer$\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{2}{9}+\tan^{-1}\frac{4}{33}+...+\tan^{-1}\frac{2^{\text{n}-1}}{1+2^{2\text{n}-1}}$
$\Rightarrow\tan^{-1}\Big(\frac{2-1}{1+2\times1}\Big)+\tan^{-1}\Big(\frac{4-2}{1+4\times2}\Big)+\tan^{-1}\Big(\frac{8-4}{1+8\times4}\Big)+...\tan^{-1}\Big(\frac{2^\text{n}-2^{\text{n}-1}}{1+2^\text{n}.2^{\text{n}-1}}\Big)$
$\Rightarrow\big(\tan^{-1}2-\tan^{-1}1\big)+(\tan^{-1}4-\tan^{-1}2\big)+(\tan^{-1}8-\tan^{-1}4\big)+\\...+\big(\tan^{-1}2^{\text{n}-1}-\tan^{-1}2^{\text{n}-2}\big)+\big(\tan^{-1}2^{\text{n}}-\tan^{-1}2^{\text{n}-1}\big)$
$\Rightarrow\tan^{-1}2^{\text{n}}-\tan^{-1}1$
$\Rightarrow\tan^{-1}2^{\text{n}}-\frac{\pi}{4}$
View full question & answer→Question 43 Marks
Find the principal values of the following:
$\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)$
AnswerWe have $\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)=-\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)$$[\because\tan^{-1}(-\text{x})=-\tan^{-1}\text{x}]$Let $\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)=\text{y}$
Then,
$\tan\text{y}=\frac{1}{\sqrt3}$ We know that the range of the principal value branch is $\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big).$Thus,
$\tan\text{y}=\frac{1}{\sqrt3}=\tan\Big(\frac{\pi}{6}\Big)$ $\Rightarrow\text{y}=\frac{\pi}{6}$ $\therefore\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)=-\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)$ $=-\text{y}$ $=-\frac{\pi}{6}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$ Hence, the principal value of $\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)$ is $-\frac{\pi}{6}.$
View full question & answer→Question 53 Marks
Prove the following results
$\cos\Big(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\Big)=\frac{6}{5\sqrt{13}}$
Answer$\cos\Big(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\Big)$
$=\cos\Big(\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{2}{3}\Big)\\\dots\dots\begin{bmatrix}\sin^{-1}\Big(\frac{\text{p}}{\text{h}}\Big)=\tan^{-1}\Big(\frac{\text{p}}{\text{b}}\Big)\\\cot^{-1}\Big(\frac{\text{b}}{\text{p}}\Big)=\tan^{-1}\Big(\frac{\text{p}}{\text{b}}\Big)\end{bmatrix}$
$=\cos\Bigg(\tan^{-1}\bigg(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times\frac{2}{3}}\bigg)\Bigg)\\\dots\dots\Big[\tan^{-1}(\text{x})+\tan^{-1}(\text{y})=\tan^{-1}\Big(\frac{\text{x+y}}{1-\text{xy}}\Big)\Big]$
$$$=\cos\bigg(\tan^{-1}\bigg(\frac{\frac{17}{12}}{\frac{1}{2}}\bigg)\bigg)$
$=\cos\Big(\tan^{-1}\Big(\frac{17}{6}\Big)\Big)$
$=\cos\Big(\cos^{-1}\Big(\frac{6}{5\sqrt{13}}\Big)\Big)$ $\dots\dots\Big[\tan^{-1}\Big(\frac{\text{p}}{\text{b}}\Big)=\cos^{-1}\Big(\frac{\text{b}}{\text{h}}\Big)\Big]$
$=\frac{6}{5\sqrt3}$
View full question & answer→Question 63 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 73 Marks
Evaluate $\sin\Big(\frac{1}{2}\sin^{-1}\frac{4}{5}\Big).$
Answer$\sin\Big(\frac{1}{2}\sin^{-1}\frac{4}{5}\Big).$
$=\sin\Bigg(\frac{1}{2}\times2\tan^{-1}\sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}}\Bigg)$ $\Big\{\text{Since},\cos^{-1}\text{x}=2\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}$
$=\sin\Big(\tan^{-1}\frac{1}{3}\Big)$
$=\sin\begin{pmatrix}\sin^{-1}\frac{\frac{1}{3}}{\sqrt{1+\big(\frac{1}{3}\big)^2}}\end{pmatrix}$ $\Big\{\text{Since},\tan^{-1}\text{x}=\sin^{-1}\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big\}$
$=\frac{\frac{1}{3}}{\frac{\sqrt{10}}{3}}$
$=\frac{1}{\sqrt{10}}$
$\sin\Big(\frac{1}{2}\cos^{-1}\frac{4}{5}\Big)=\frac{1}{\sqrt{10}}$
View full question & answer→Question 83 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 93 Marks
If $x < 0, y < 0$ such that $xy = 1,$ then write the value of $\tan^{-1}x + \tan^{-1}y.$
AnswerWe know,$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$x < 0, y < 0$ such that
$xy = 1$
Let $x = -a$ and $y = -b$ where both $a$ and $b$ are positive.
$\therefore\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{-\text{a}-\text{a}}{1-1}\Big)$
$=\tan^{-1}(-\infty)$
$=\tan^{-1}\Big\{\tan\Big(-\frac{\pi}{2}\Big)\Big\}$
$=-\frac{\pi}{2}$
View full question & answer→Question 103 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 113 Marks
If $\sin^{-1}\text{x}+\sin^{-1}\text{y}+\sin^{-1}\text{z}=\frac{3\pi}{2},$ then write the values of x + y + z.
AnswerGiven,
$\sin^{-1}\text{x}+\sin^{-1}\text{y}+\sin^{-1}\text{z}=\frac{3\pi}{2}$
We know that maximum and minimum values of $\sin^{-1}\text{x}$ are $\frac{\pi}{2}$ and $-\frac{\pi}{2}$ respectively.
$\sin^{-1}\text{x}+\sin^{-1}\text{y}+\sin^{-1}\text{z}=\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}$
$\Rightarrow\sin^{-1}\text{x}=\frac{\pi}{2},\sin^{-1}\text{y}=\frac{\pi}{2},\sin^{-1}\text{z}=\frac{\pi}{2}$
⇒ x = 1, y = 1, z = 1
So,
x + y + z = 1 + 1 + 1
= 3
Hence,
x + y + z = 3
View full question & answer→Question 123 Marks
Write the value of $\cos^{-1}\Big(\tan\frac{3\pi}{4}\Big).$
AnswerWe have
$\cos^{-1}\Big(\tan\frac{3\pi}{4}\Big)=\cos^{-1}\Big\{-\tan\Big(-\pi-\frac{3\pi}{4}\Big)\Big\}$ $[\because\ \tan(\pi-\text{x}=-\tan\text{x})]$
$=\cos^{-1}\Big\{\tan\Big(-\frac{\pi}{4}\Big)\Big\}$
$=\cos^{-1}\Big\{-\tan\Big(\frac{\pi}{4}\Big)\Big\}$
$=\cos^{-1}(-1)$
$=\cos^{-1}(\cos\pi)$ $[\therefore\ \cos\pi=-1]$
$=\pi$
$\therefore\ \cos^{-1}\Big(\tan\frac{3\pi}{4}\Big)=\pi$
View full question & answer→Question 133 Marks
Solve:
$\tan^{-1}\text{x}+2\cot^{-1}\text{x}=\frac{2\pi}{3}$
Answer$\tan^{-1}\text{x}+2\cot^{-1}\text{x}=\frac{2\pi}{3}$
$\Rightarrow\tan^{-1}\text{x}+2\Big(\frac{\pi}{2}-\tan^{-1}\text{x}\Big)=\frac{2\pi}{3}$
$\Big[\because\ \cot^{-1}\text{x}=\frac{\pi}{2}-\tan^{-1}\text{x}\Big]$
$\Rightarrow\tan^{-1}\text{x}+\pi-2\tan^{-1}\text{x}=\frac{2\pi}{3}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{3}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{3}$
$\Rightarrow\text{x}=\tan\frac{\pi}{3}=\sqrt3$
View full question & answer→Question 143 Marks
Solve the following equation for x:
$\tan^{-1}(2+\text{x})+\tan^{-1}(2-\text{x})=\tan^{-1}\frac{2}{3},$ where $\text{x}<-\sqrt3$ or, $\text{x}>\sqrt3$
AnswerWe know
$ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}(2+\text{x})+\tan^{-1}(2-\text{x})=\tan^{-1}\frac{2}{3},$
$\Rightarrow\tan^{-1}\Big(\frac{2+\text{x}+2-\text{x}}{1-(2+\text{x})(2-\text{x})}\Big)=\tan^{-1}\frac{2}{3}$
$\Rightarrow\frac{4}{1-4+\text{x}^2}=\tan\frac{2}{3}$
$\Rightarrow-6+2\text{x}^2=12$
$\Rightarrow2\text{x}^2=18$
$\Rightarrow\text{x}^2=9$
$\Rightarrow\text{x}=\pm3$
View full question & answer→Question 153 Marks
Prove the following results:
$\sin^{-1}\frac{4}{5}+2\tan^{-1}\frac{1}{3}=\frac{\pi}{2}$
Answer$\text{L.H.S}=\sin^{-1}\frac{4}{5}+2\tan^{-1}\frac{1}{3}$
$=\sin^{-1}\frac{4}{5}+\tan^{-1}\Bigg\{\frac{2\times\frac{1}{2}}{1-\big(\frac{1}{3}\big)^2}\Bigg\}$ $\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big\{\frac{2\text{x}}{1-\text{x}^2}\Big\}\Big]$
$=\sin^{-1}\frac{4}{5}+\tan^{-1}\Bigg\{\frac{\frac{2}{3}}{\frac{8}{9}}\Bigg\}$
$=\sin^{-1}\frac{4}{5}+\tan^{-1}\frac{3}{4}$
$=\sin^{-1}\frac{4}{5}+\cos^{-1}\frac{1}{\sqrt{1+\frac{9}{16}}}$ $\Big[\because\ \tan^{-1}\text{x}=\cos^{-1}\frac{1}{\sqrt{1+\text{x}^2}}\Big]$
$=\sin^{-1}\frac{4}{5}+\cos^{-1}\frac{1}{\frac{5}{4}}$
$=\sin^{-1}\frac{4}{5}+\cos^{-1}\frac{4}{5}$
$=\frac{\pi}{2}=\text{R.H.S}$
View full question & answer→Question 163 Marks
Solve the following equation for x:
$\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\cot^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)=\frac{2\pi}{3},\text{x}>0$
AnswerWe know,
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\cot^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)=\frac{2\pi}{3}$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{2\pi}{3}$ $\Big[\because\ \cot^{-1}\text{x}=\tan^{-1}\frac{1}{\text{x}}\Big]$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{\pi}{3}$
$\Rightarrow2\tan^{-1}\text{x}=\frac{\pi}{3}$ $\Big[\because\ 2\tan^{-1}\text{x}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\tan\frac{\pi}{6}$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}$
View full question & answer→Question 173 Marks
Find the principal value of the following:
$\sec^{-1}\Big(-\sqrt2\Big)$
AnswerLet $\sec^{-1}\Big(-\sqrt2\Big)=\text{y}$
Then,
$\sec\text{y}=-\sqrt2$
We know that the range of the principal value branch is $[0,\pi]-\Big\{\frac{\pi}{2}\Big\}.$
Thus,
$\sec\text{y}=-\sqrt2=\sec\Big(\frac{3\pi}{4}\Big)$
$\Rightarrow\text{y}=\frac{3\pi}{4}\in[0,\pi],\text{y}\neq\frac{\pi}{2}$
Hence, the principal value of $\sec^{-1}\Big(-\sqrt2\Big)$ is $\frac{3\pi}{4}.$
View full question & answer→Question 183 Marks
Solve the following equation for x:
$\tan^{-1}(\text{x}-1)+\tan^{-1}\text{x}+\tan^{-1}(\text{x}+1)=\tan^{-1}3\text{x}$
AnswerWe know
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$ and
$\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)$
$\therefore\ \tan^{-1}(\text{x}+1)+\tan^{-1}(\text{x}-1)+\tan^{-1}\text{x}=\tan^{-1}3\text{x}$
$\Rightarrow\tan^{-1}\Big\{\frac{\text{x}+1+\text{x}-1}{1-(\text{x}+1)\times(\text{x} +1)}\Big\}=\tan^{-1}3\text{x}-\tan^{-1}\text{x}$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}}{2-\text{x}^2}\Big)=\tan^{-1}\Big(\frac{3\text{x}-\text{x}}{1+3\text{x}^2}\Big)$
$\Rightarrow\frac{2\text{x}}{2-\text{x}^2}=\frac{2\text{x}}{1+3\text{x}^2}$
$\Rightarrow2-\text{x}^2=1+3\text{x}^2$
$\Rightarrow4\text{x}^2-1=0$
$\Rightarrow\text{x}^2=\frac{1}{4}$
$\Rightarrow\text{x}=\pm\frac{1}{2}$
View full question & answer→Question 193 Marks
Write the following in the simplest form:
$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}+1}{\text{x}}\Big\},\text{x}\neq0$
AnswerLet $\text{x}=\tan\theta$Now,
$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}+1}{\text{x}}\Big\}$ $=\tan^{-1}\Big\{\frac{\sqrt{1+\tan^2\theta}+1}{\tan\theta}\Big\}$ $=\tan^{-1}\Big\{\frac{\sqrt{1+\sec^2\theta}+1}{\tan\theta}\Big\}$ $=\tan^{-1}\Big\{\frac{\sec\theta+1}{\tan\theta}\Big\}$ $=\tan^{-1}\Big\{\frac{\cos\theta+1}{\sin\theta}\Big\}$ $=\tan^{-1}\Bigg\{\frac{2\cos^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\Bigg\}$ $=\tan^{-1}\Big(\frac{\cot\theta}{2}\Big)$ $=\tan^{-1}\Big\{\tan\Big(\frac{\pi}{2}-\frac{\theta}{2}\Big)\Big\}$ $=\Big(\frac{\pi}{2}-\frac{\theta}{2}\Big)$ $=\frac{\pi}{2}-\frac{\tan^{-1}\text{x}}{2}$
View full question & answer→Question 203 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 213 Marks
Solve the following equation for x:
$\tan^{-1}(\text{x}+1)+\tan^{-1}(\text{x}-1)=\tan^{-1}\frac{8}{31}$
AnswerWe know
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}(\text{x}+1)+\tan^{-1}(\text{x}-1)=\tan^{-1}\frac{8}{31}$
$\Rightarrow\tan^{-1}\Big\{\frac{\text{x}+1+\text{x}-1}{1-(\text{x}+1)\times(\text{x} -1)}\Big\}=\tan^{-1}\frac{8}{31}$
$\Rightarrow\frac{2\text{x}}{1-\text{x}^2+1}=\frac{8}{31}$
$\Rightarrow\frac{2\text{x}}{2-\text{x}^2}=\frac{8}{31}$
$\Rightarrow31\text{x}=8-4\text{x}^2$
$\Rightarrow4\text{x}^2+31\text{x}-8=0$
$\Rightarrow4\text{x}^2+32\text{x}-\text{x}-8=0$
$\Rightarrow(4\text{x}-1)(\text{x}+8)=0$
$\Rightarrow\text{x}=\frac{1}{4}$ [As x =-8 is not satisfying the equation]
View full question & answer→Question 223 Marks
Write the value of $\tan^{-1}\frac{\text{a}}{\text{b}}-\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big).$
Answer$\tan^{-1}\frac{\text{a}}{\text{b}}-\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)$
$=\tan^{-1}\begin{bmatrix}\frac{\frac{\text{a}}{\text{b}}-\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}{1+\Big(\frac{\text{a}}{\text{b}}\Big)-\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)}\end{bmatrix}$
$\Big\{\text{Since},\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big\}$
$=\tan^{-1}\begin{bmatrix}\frac{\frac{\text{a}^2+\text{ab}-\text{ab}+\text{b}^2}{\text{b}(\text{a}+\text{b})}}{\frac{\text{ba}+\text{b}^2+\text{a}^2-\text{ab}}{{\text{b}(\text{a}+\text{b})}}}\end{bmatrix}$
$=\tan^{-1}\Big[\frac{\text{a}^2+\text{b}^2}{\text{a}^2+\text{b}^2}\Big]$
$=\tan^{-1}(1)$
$=\frac{\pi}{4}$
Hence,
$\tan^{-1}\frac{\text{a}}{\text{b}}-\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)=\frac{\pi}{4}$
View full question & answer→Question 233 Marks
Write the following in the simplest form:
$\cot^{-1}\frac{\text{a}}{\sqrt{\text{x}^2-\text{a}^2}},|\text{x}|>\text{a}$
Answer$\cot^{-1}\frac{\text{a}}{\sqrt{\text{x}^2-\text{a}^2}},|\text{x}|>\text{a}$ Let, $\text{x}=\text{a}\sec\theta$ $\cot^{-1}\bigg(\frac{\text{a}}{\sqrt{\text{a}^2\sec^{2}\theta-\text{a}^2}}\bigg)$ $=\cot^{-1}\begin{pmatrix}\frac{\text{a}}{\sqrt{\text{a}^2\big(\sec^{2}\theta-1\big)}}\end{pmatrix}$ $=\cot^{-1}\frac{1}{\sqrt{\tan^2\theta}}$ $\{\text{since},\sec^2\theta-1=\tan^2\theta\}$ $=\cot^{-1}(\cot\theta)$ $=\theta$ $=\sec^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$Hence,
$\cot^{-1}\frac{a}{\sqrt{\text{x}^2-\text{a}^2}}=\sec^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$
View full question & answer→Question 243 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 253 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 263 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 273 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 283 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 293 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 303 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 313 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 323 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 333 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 343 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 353 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 363 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 373 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 383 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 393 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 403 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 413 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 423 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 433 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 443 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 453 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 463 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 473 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 483 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 493 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 503 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→