Class 12: Inverse Trigonometric Functions – Miscellaneous Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1. } \text{What is the value of }\cot\left(\frac{\pi}{2}+\tan^{-1}\frac{1}{\sqrt{2}}\right)\text{?} \ \$
$\displaystyle \text{(a) }\frac{1}{\sqrt{2}} \ \$
$\displaystyle \text{(b) }\frac{\pi}{2} \ \$
$\displaystyle \text{(c) }-1 \ \$
$\displaystyle \text{(d) }\frac{-1}{\sqrt{2}} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{ (d) } \cot\left(\frac{\pi}{2}+\tan^{-1}\frac{1}{\sqrt{2}}\right)$
$\displaystyle = -\tan\left(\tan^{-1}\frac{1}{\sqrt{2}}\right)$
$\displaystyle \left[\because \cot\left(\frac{\pi}{2}+\theta\right)=-\tan\theta\right]$
$\displaystyle = -\frac{1}{\sqrt{2}}$
$\\$

$\displaystyle \textbf{Question 2. } \text{The value of }\tan^{-1}\sqrt{3}-\sec^{-1}(-2)\text{ is equal to } \ \$
$\displaystyle \text{(a) }\frac{\pi}{3} \ \$
$\displaystyle \text{(b) }\frac{2\pi}{3} \ \$
$\displaystyle \text{(c) }\frac{-\pi}{3} \ \$
$\displaystyle \text{(d) }\frac{\pi}{4} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Let } \tan^{-1}\sqrt{3}=x \Rightarrow \tan x=\sqrt{3}=\tan\frac{\pi}{3}$
$\displaystyle \text{Since, the principal value branch of } \tan^{-1} \text{ is } \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
$\displaystyle \text{and } x=\frac{\pi}{3}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
$\displaystyle \text{So, the principal value of } \tan^{-1}\sqrt{3} \text{ is } \frac{\pi}{3}$
$\displaystyle \text{Again, let } \sec^{-1}(-2)=y$
$\displaystyle \Rightarrow \sec y=-2=-\sec\frac{\pi}{3}$
$\displaystyle \Rightarrow \sec y=\sec\left(\pi-\frac{\pi}{3}\right)=\sec\frac{2\pi}{3}$
$\displaystyle \text{Since, the principal value branch of } \sec^{-1} \text{ is } [0,\pi]-\left\{\frac{\pi}{2}\right\}$
$\displaystyle \text{and } x=\frac{2\pi}{3}\in[0,\pi]-\left\{\frac{\pi}{2}\right\}$
$\displaystyle \text{So, the principal value of } \sec^{-1}(-2) \text{ is } \frac{2\pi}{3}$
$\displaystyle \text{Hence, } \tan^{-1}\sqrt{3}-\sec^{-1}(-2)=\frac{\pi}{3}-\frac{2\pi}{3}=-\frac{\pi}{3}$
$\\$

$\displaystyle \textbf{Question 3. } \text{What will be the principal value of }\mathrm{cosec}^{-1}(-\sqrt{2})\text{? } \ \$
$\displaystyle \text{(a) }\frac{3\pi}{4} \ \$
$\displaystyle \text{(b) }\frac{-\pi}{6} \ \$
$\displaystyle \text{(c) }\frac{\pi}{4} \ \$
$\displaystyle \text{(d) }\frac{-\pi}{4} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) Let } y=\mathrm{cosec}^{-1}(-\sqrt{2})$
$\displaystyle \Rightarrow \mathrm{cosec}\ y=-\sqrt{2}$
$\displaystyle \Rightarrow \mathrm{cosec}\ y=\mathrm{cosec}\left(-\frac{\pi}{4}\right)$
$\displaystyle \text{We know that the range of the principal value branch of } \mathrm{cosec}^{-1} x \text{ is } \left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}$
$\displaystyle \therefore y=-\frac{\pi}{4}$
$\displaystyle \text{Therefore, the principal value of } \mathrm{cosec}^{-1}(-\sqrt{2}) \text{ is } -\frac{\pi}{4}$
$\\$

$\displaystyle \textbf{Question 4. } \text{Simplified value of }\sin\left[\frac{\pi}{2}-\sin^{-1}\left(\frac{-\sqrt{3}}{2}\right)\right]\text{ is } \ \$
$\displaystyle \text{(a) }\frac{1}{2} \ \$
$\displaystyle \text{(b) }\frac{1}{\sqrt{2}} \ \$
$\displaystyle \text{(c) }\frac{\sqrt{3}}{2} \ \$
$\displaystyle \text{(d) }\frac{-\sqrt{3}}{2} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) } \sin\left[\frac{\pi}{2}-\sin^{-1}\left(\frac{-\sqrt{3}}{2}\right)\right]$
$\displaystyle = \sin\left[\frac{\pi}{2}-\left(-\sin^{-1}\frac{\sqrt{3}}{2}\right)\right]$
$\displaystyle = \sin\left[\frac{\pi}{2}+\sin^{-1}\frac{\sqrt{3}}{2}\right]$
$\displaystyle = \sin\left[\frac{\pi}{2}+\frac{\pi}{3}\right]$
$\displaystyle = \cos\frac{\pi}{3}=\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 5. } \text{Find the domain of the function }\cos^{-1}x+\sin x\text{. } \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } f(x)=\cos^{-1}x+\sin x$
$\displaystyle \text{We have, domain of } \cos^{-1}x \text{ as } [-1,1] \text{ and domain of } \sin x \text{ as } R$
$\displaystyle \therefore \text{Domain of } f(x) \text{ will be common value of domain of } \cos^{-1}x \text{ and } \sin x$
$\displaystyle \therefore \text{Domain of } f(x) \text{ is } [-1,1]$
$\\$

$\displaystyle \textbf{Question 6. } \text{Solve }\sin^{-1}\cos\left(\sin^{-1}x\right)=\frac{\pi}{3}\text{. } \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \sin^{-1}\{\cos(\sin^{-1}x)\}=\frac{\pi}{3}$
$\displaystyle \Rightarrow \cos(\sin^{-1}x)=\sin\frac{\pi}{3}$
$\displaystyle \Rightarrow \cos(\sin^{-1}x)=\frac{\sqrt{3}}{2}$
$\displaystyle \Rightarrow \sin^{-1}x=\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$\displaystyle \Rightarrow \sin^{-1}x=\frac{\pi}{6}$
$\displaystyle \Rightarrow x=\sin\frac{\pi}{6}=\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 7. } \text{Prove that }\cos\left[\tan^{-1}\left\{\cot(\sin^{-1}x)\right\}\right]=x\text{. } \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{ We have, } \cos[\tan^{-1}\{\cot(\sin^{-1}x)\}]$
$\displaystyle \text{Let } \sin^{-1}x=\theta \Rightarrow \sin\theta=x$
$\displaystyle =\cos\left[\tan^{-1}\{\cot\theta\}\right]$
$\displaystyle =\cos\left[\tan^{-1}\left\{\tan\left(\frac{\pi}{2}-\theta\right)\right\}\right]$
$\displaystyle =\cos\left(\frac{\pi}{2}-\theta\right)$
$\displaystyle =\sin\theta=x\ \ \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 8. }\text{If }\cos^{-1}\left(\frac{x}{a}\right)+\cos^{-1}\left(\frac{y}{b}\right)=\alpha,\text{ prove that} \ \$
$\displaystyle \frac{x^{2}}{a^{2}}-\frac{2xy}{ab}\cos\alpha+\frac{y^{2}}{b^{2}}=\sin^{2}\alpha.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \cos^{-1}\frac{x}{a} + \cos^{-1}\frac{y}{b} = \alpha$
$\displaystyle \text{We know that } \cos^{-1}u + \cos^{-1}v = \cos^{-1}(uv - \sqrt{(1-u^{2})(1-v^{2})})$
$\displaystyle \Rightarrow \cos^{-1}\left(\frac{x}{a}\cdot\frac{y}{b} - \sqrt{\left(1 - \frac{x^{2}}{a^{2}}\right)\left(1 - \frac{y^{2}}{b^{2}}\right)}\right) = \alpha$
$\displaystyle \Rightarrow \cos\alpha = \frac{xy}{ab} - \sqrt{\left(1 - \frac{x^{2}}{a^{2}}\right)\left(1 - \frac{y^{2}}{b^{2}}\right)}$
$\displaystyle \Rightarrow \sqrt{\left(1 - \frac{x^{2}}{a^{2}}\right)\left(1 - \frac{y^{2}}{b^{2}}\right)} = \frac{xy}{ab} - \cos\alpha$
$\displaystyle \text{On squaring both sides, we get}$
$\displaystyle \left(1 - \frac{x^{2}}{a^{2}}\right)\left(1 - \frac{y^{2}}{b^{2}}\right) = \left(\frac{xy}{ab}\right)^{2} - \frac{2xy}{ab}\cos\alpha + \cos^{2}\alpha$
$\displaystyle \Rightarrow 1 - \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} + \left(\frac{xy}{ab}\right)^{2} = \left(\frac{xy}{ab}\right)^{2} - \frac{2xy}{ab}\cos\alpha + \cos^{2}\alpha$
$\displaystyle \Rightarrow 1 - \cos^{2}\alpha = \frac{x^{2}}{a^{2}} - \frac{2xy}{ab}\cos\alpha + \frac{y^{2}}{b^{2}}$
$\displaystyle \Rightarrow \sin^{2}\alpha = \frac{x^{2}}{a^{2}} - \frac{2xy}{ab}\cos\alpha + \frac{y^{2}}{b^{2}}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 9. }\text{If }\alpha\leq 2\sin^{-1}x+\cos^{-1}x\leq \beta,\text{ then }(\alpha,\beta)\text{ is } \ \$
$\displaystyle \text{(a) }(0,\pi) \ \$
$\displaystyle \text{(b) }\left(\frac{-\pi}{2},\frac{\pi}{2}\right) \ \$
$\displaystyle \text{(c) }\left(\frac{-3\pi}{2},\frac{\pi}{2}\right) \ \$
$\displaystyle \text{(d) None of these} \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) } \text{We know that } -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2}$
$\displaystyle \text{On adding } \frac{\pi}{2} \text{ in each term of the inequality }$
$\displaystyle -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2}, \text{ we have }$
$\displaystyle -\frac{\pi}{2} + \frac{\pi}{2} \leq \sin^{-1}x + \frac{\pi}{2} \leq \frac{\pi}{2} + \frac{\pi}{2}$
$\displaystyle \Rightarrow 0 \leq \sin^{-1}x + \frac{\pi}{2} \leq \pi$
$\displaystyle \Rightarrow 0 \leq \sin^{-1}x + (\sin^{-1}x + \cos^{-1}x) \leq \pi$
$\displaystyle \text{Since } \sin^{-1}\theta + \cos^{-1}\theta = \frac{\pi}{2}$
$\displaystyle \therefore 0 \leq 2\sin^{-1}x + \cos^{-1}x \leq \pi$
$\displaystyle \therefore \alpha = 0, \beta = \pi \text{ or } (\alpha, \beta) = (0, \pi)$
$\\$

$\displaystyle \textbf{Question 10. }\text{For all }x\in R,\ \cot^{-1}(-x)\text{ is equal to } \ \$
$\displaystyle \text{(a) }\pi-\cot^{-1}x \ \$
$\displaystyle \text{(b) }-\tan^{-1}x \ \$
$\displaystyle \text{(c) }-\cot^{-1}x \ \$
$\displaystyle \text{(d) }\pi+\cot^{-1}x \ \$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Let } \cot^{-1}(-x) = \theta$
$\displaystyle \Rightarrow \cot\theta = -x$
$\displaystyle \Rightarrow \cot(\pi - \theta) = x$
$\displaystyle \Rightarrow \pi - \theta = \cot^{-1}x$
$\displaystyle \Rightarrow \theta = \pi - \cot^{-1}x$
$\displaystyle \therefore \cot^{-1}(-x) = \pi - \cot^{-1}x \ \forall x \in R$
$\\$

$\displaystyle \textbf{Question 11. }\text{Prove that: } \\ \sin^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)+\cos^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2x+2}}\right)=\tan^{-1}(x^{2}+x+1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS } = \sin^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right) + \cos^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2x+2}}\right)$
$\displaystyle = \tan^{-1}\left(\frac{\frac{x}{\sqrt{1+x^{2}}}}{\sqrt{1-\frac{x^{2}}{1+x^{2}}}}\right) + \tan^{-1}\left(\frac{\sqrt{1-\frac{(x+1)^{2}}{x^{2}+2x+2}}}{\frac{x+1}{\sqrt{x^{2}+2x+2}}}\right)$
$\displaystyle \text{Since } \sin^{-1}y = \tan^{-1}\left(\frac{y}{\sqrt{1-y^{2}}}\right) \text{ and } \cos^{-1}y = \tan^{-1}\left(\frac{\sqrt{1-y^{2}}}{y}\right)$
$\displaystyle = \tan^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\times\frac{\sqrt{1+x^{2}}}{\sqrt{1}}\right) + \tan^{-1}\left(\frac{\sqrt{1}}{\sqrt{x^{2}+2x+2}}\times\frac{\sqrt{x^{2}+2x+2}}{x+1}\right)$
$\displaystyle = \tan^{-1}\left(\frac{x}{1}\right) + \tan^{-1}\left(\frac{1}{x+1}\right)$
$\displaystyle = \tan^{-1}\left(\frac{x+\frac{1}{x+1}}{1-x\cdot\frac{1}{x+1}}\right)$
$\displaystyle \text{Since } \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$
$\displaystyle = \tan^{-1}\left(\frac{\frac{x^{2}+x+1}{x+1}}{\frac{1}{x+1}}\right)$
$\displaystyle = \tan^{-1}(x^{2}+x+1)$
$\displaystyle = \text{RHS}$
$\displaystyle \text{Hence proved.}$
$\\$