Question 513 Marks
Solve: $\cos\big(\sin^{-1}\text{x}\big)=\frac{1}{6}$
Answer$\cos\big(\sin^{-1}\text{x}\big)=\frac{1}{6}$
$\Rightarrow\cos\Big(\cos^{-1}\sqrt{1-\text{x}^2}\Big)=\frac{1}{6}$
$\Rightarrow\sqrt{1-\text{x}^2}=\frac{1}{6}$
$\Rightarrow1-\text{x}^2=\frac{1}{36}$
$\Rightarrow1-\frac{1}{36}=\text{x}^2$
$\Rightarrow\text{x}^2=\frac{35}{36}$
$\Rightarrow\text{x}=\pm\frac{\sqrt{35}}{6}$
View full question & answer→Question 523 Marks
Evaluate:
$\cos\Big(\tan^{-1}\frac{3}{4}\Big)$
AnswerWe have
$\cos\Big(\tan^{-1}\frac{3}{4}\Big)$
$=\cos\Bigg[\frac{1}{2}\cos^{-1}\Bigg(\frac{1-\big(\frac{3}{4}\big)^2}{1+\big(\frac{3}{4}\big)^2}\Bigg)\Bigg]$ $\Big[\because\ 2\tan^{-1}\text{x}=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\Big]$
$=\cos\Big[\frac{1}{2}\cos^{-1}\Big(\frac{7}{25}\Big)\Big]$
Let
$\text{y}=\cos^{-1}\Big(\frac{7}{25}\Big)$
$\Rightarrow\cos\text{y}=\frac{7}{25}$
Now,
$=\cos\Big[\frac{1}{2}\cos^{-1}\Big(\frac{7}{25}\Big)\Big]=\cos\Big[\frac{1}{2}\text{y}\Big]$
$=\sqrt{ \frac{\cos\text{y}+1}{2}}$ $\big[\because\ \cos2\text{x}=2\cos^2\text{x}-1\big]$
$=\sqrt{\frac{\frac{7}{25}+1}{2}}$
$=\sqrt{\frac{32}{50}}$
$=\frac{4}{5}$
$\therefore\ \cos\Big[\tan^{-1}\Big(\frac{3}{4}\Big)\Big]=\frac{4}{5}$
View full question & answer→Question 533 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 543 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 553 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 563 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 573 Marks
Prove the following results:
$\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}=\sin^{-1}\frac{1}{\sqrt5}$
Answer$\text{L.H.S}=\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}$
$=\tan^{-1}\Bigg(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\times\frac{2}{9}}\Bigg)$ $\Big[\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x+y}}{1-\text{xy}}\Big)\Big]$
$=\tan^{-1}\Bigg(\frac{\frac{17}{36}}{\frac{34}{36}}\Bigg)$
$=\tan^{-1}\frac{1}{2}$
$=\sin^{-1}\frac{\frac{1}{2}}{\sqrt{1+\big(\frac{1}{2}\big)^2}}$
$=\sin^{-1}\frac{1}{\sqrt5}=\text{R.H.S}$
View full question & answer→Question 583 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 593 Marks
If $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{\pi}{4},$ then write the value of x + y + xy.
AnswerGiven,
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)=\frac{\pi}{4}$ $\Big\{\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big\}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)=\tan^{-1}(1)$
$\Rightarrow\frac{\text{x}+\text{y}}{1-\text{xy}}=1$
$\Rightarrow\text{x}+\text{y}=1-\text{xy}$
$\Rightarrow\text{x}+\text{y}+\text{xy}=1$
So,
$\Rightarrow\text{x}+\text{y}+\text{xy}=1$
View full question & answer→Question 603 Marks
Write the value of $\sin\big(\cot^{-1}\text{x}\big).$
AnswerWe know$\cot^{-1}\text{x}=\tan^{-1}\frac{1}{\text{x}}$
Now, we have $\sin\big(\cot^{-1}\text{x}\big)=\sin\Big(\tan^{-1}\frac{1}{\text{x}}\Big)$ $=\sin\Bigg[\sin^{-1}\Bigg(\frac{\frac{1}{\text{x}}}{\sqrt{1+\frac{1}{\text{x}^2}}}\Bigg)\Bigg]$ $\Big[\because\ \tan^{-1}\text{x}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}}}\Big)\Big]$ $=\sin\Bigg[\sin^{-1}\Bigg(\frac{\frac{1}{\text{x}}}{\frac{\sqrt{\text{x}^2+1}}{\text{x}}}\Bigg)\Bigg]$ $=\sin\bigg(\sin^{-1}\frac{1}{\sqrt{\text{x}^2+1}}\bigg)$ $=\frac{1}{\sqrt{\text{x}^2+1}}$ $\big[\because\ \sin\big(\sin^{-1}\text{x}=\text{x}\big)\big]$ Hence, $\sin\big(\cot^{-1}\text{x}\big)=\frac{1}{\sqrt{\text{x}^2-1}}.$
View full question & answer→Question 613 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 623 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 633 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 643 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 653 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 663 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 673 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 683 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 693 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 703 Marks
If $4\sin^{-1}\text{x}+\cos^{-1}\text{x}=\pi,$ then what is the value of x?
AnswerGiven,
$4\sin^{-1}\text{x}+\cos^{-1}\text{x}=\pi$
$\Rightarrow4\sin^{-1}\text{x}+\frac{\pi}{2}-\sin^{-1}\text{x}=\pi$ $\Big\{\text{Since},\cos^{-1}\text{x}=\frac{\pi}{2}-\sin^{-1}\text{x}\Big\}$
$\Rightarrow3\sin^{-1}\text{x}=\pi-\frac{\pi}{2}$
$\Rightarrow3\sin^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\sin^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\sin^{-1}\text{x}=\sin^{-1}\Big(\frac{1}{2}\Big)$
$\Rightarrow\text{x}=\frac{1}{2}$
Hence,
$\text{x}=\frac{1}{2}$
View full question & answer→Question 713 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 723 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 733 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 743 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 753 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→