\displaystyle \textbf{Question 1. }\text{The domain in which sine function will be one-one, is}
\displaystyle \text{(a) }\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\qquad \text{(b) }\left[\frac{\pi}{2},\frac{3\pi}{2}\right]
\displaystyle \text{(c) }[0,\pi]\qquad \text{(d) Both (a) and (b)}
\displaystyle \text{Answer:}
\displaystyle \sin x\text{ is one-one in }\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{ and }\left[\frac{\pi}{2},\frac{3\pi}{2}\right].
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 2. }\text{If }\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\frac{3\pi}{2},\text{ then the value of } \\ x^{100}+y^{100}+z^{100}-\frac{9}{x^{101}+y^{101}+z^{101}}\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }2\qquad \text{(d) }3
\displaystyle \text{Answer:}
\displaystyle \sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\frac{3\pi}{2}
\displaystyle \text{Since maximum value of }\sin^{-1}x\text{ is }\frac{\pi}{2},\text{ we get }x=y=z=1.
\displaystyle x^{100}+y^{100}+z^{100}-\frac{9}{x^{101}+y^{101}+z^{101}}
\displaystyle =1+1+1-\frac{9}{1+1+1}
\displaystyle =3-3=0
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 3. }\text{The value of }\sin^{-1}\left[\cot\left\{\sin^{-1}\frac{\sqrt{2-\sqrt{3}}}{2}+\cos^{-1}\frac{\sqrt{12}}{4}+\sec^{-1}\sqrt{2}\right\}\right]\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }\frac{\pi}{2}\qquad \text{(c) }\frac{\pi}{3}\qquad \text{(d) }\frac{\pi}{4}
\displaystyle \text{Answer:}
\displaystyle \sin^{-1}\frac{\sqrt{2-\sqrt{3}}}{2}=\sin^{-1}\left(\sin\frac{\pi}{12}\right)=\frac{\pi}{12}
\displaystyle \cos^{-1}\frac{\sqrt{12}}{4}=\cos^{-1}\frac{\sqrt{3}}{2}=\frac{\pi}{6}
\displaystyle \sec^{-1}\sqrt{2}=\frac{\pi}{4}
\displaystyle \therefore \frac{\pi}{12}+\frac{\pi}{6}+\frac{\pi}{4}=\frac{\pi+2\pi+3\pi}{12}=\frac{\pi}{2}
\displaystyle \sin^{-1}\left(\cot\frac{\pi}{2}\right)=\sin^{-1}0=0
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 4. }\text{If }0\leq x<1,\text{ then }\sin\left\{\tan^{-1}\left(\frac{1-x^{2}}{2x}\right)+\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right\}\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }-1\qquad \text{(c) }0\qquad \text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\tan\theta,\text{ where }0\leq\theta<\frac{\pi}{4}.
\displaystyle \frac{1-x^{2}}{2x}=\cot2\theta
\displaystyle \frac{1-x^{2}}{1+x^{2}}=\cos2\theta
\displaystyle \tan^{-1}(\cot2\theta)=\frac{\pi}{2}-2\theta
\displaystyle \cos^{-1}(\cos2\theta)=2\theta
\displaystyle \therefore \sin\left\{\frac{\pi}{2}-2\theta+2\theta\right\}=\sin\frac{\pi}{2}=1
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 5. }\text{The value of }\cos\left[\tan^{-1}\{\sin(\cot^{-1}x)\}\right]\text{ is}
\displaystyle \text{(a) }\frac{1}{\sqrt{x^{2}+2}}\qquad \text{(b) }\sqrt{\frac{x^{2}+2}{x^{2}+1}}
\displaystyle \text{(c) }\sqrt{\frac{x^{2}+1}{x^{2}+2}}\qquad \text{(d) }\frac{1}{\sqrt{x^{2}+1}}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\cot^{-1}x.
\displaystyle \therefore \sin\theta=\frac{1}{\sqrt{x^{2}+1}}
\displaystyle \cos\left[\tan^{-1}\{\sin(\cot^{-1}x)\}\right]=\cos\left[\tan^{-1}\left(\frac{1}{\sqrt{x^{2}+1}}\right)\right]
\displaystyle =\frac{1}{\sqrt{1+\left(\frac{1}{\sqrt{x^{2}+1}}\right)^{2}}}
\displaystyle =\sqrt{\frac{x^{2}+1}{x^{2}+2}}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 6. }\text{The principal value of }\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\text{ is}
\displaystyle \text{(a) }-\frac{2\pi}{3}\qquad \text{(b) }-\frac{\pi}{3}\qquad \text{(c) }\frac{4\pi}{3}\qquad \text{(d) }\frac{7\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 7. }\text{The principal value of }\sin^{-1}\left(\sin\frac{5\pi}{3}\right)\text{ is}
\displaystyle \text{(a) }\frac{5\pi}{3}\qquad \text{(b) }-\frac{5\pi}{3}\qquad \text{(c) }-\frac{\pi}{3}\qquad \text{(d) }\frac{4\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \sin\frac{5\pi}{3}=-\frac{\sqrt{3}}{2}
\displaystyle \therefore \sin^{-1}\left(\sin\frac{5\pi}{3}\right)=\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)
\displaystyle =-\frac{\pi}{3}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 8. }\text{If }\theta=\sin^{-1}x+\cos^{-1}x-\tan^{-1}x,\ x\geq0,\text{ then the smallest interval in which }\theta\text{ lies is}
\displaystyle \text{(a) }\frac{\pi}{2}\leq\theta\leq\frac{3\pi}{4}\qquad \text{(b) }0\leq\theta\leq\frac{\pi}{4}
\displaystyle \text{(c) }-\frac{\pi}{4}\leq\theta\leq0\qquad \text{(d) }\frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}
\displaystyle \therefore \theta=\frac{\pi}{2}-\tan^{-1}x
\displaystyle \text{Since }x\geq0,\quad 0\leq\tan^{-1}x\leq\frac{\pi}{2}
\displaystyle \therefore 0\leq\theta\leq\frac{\pi}{2}
\displaystyle \text{But }x\in[0,1]\text{ for }\sin^{-1}x\text{ and }\cos^{-1}x.
\displaystyle \therefore 0\leq\tan^{-1}x\leq\frac{\pi}{4}
\displaystyle \therefore \frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 9. }\text{The value of }\sin(\cot^{-1}x)\text{ is}
\displaystyle \text{(a) }(1+x^{2})^{\frac{3}{2}}\qquad \text{(b) }(1+x^{2})^{-\frac{3}{2}}
\displaystyle \text{(c) }(1+x^{2})^{\frac{1}{2}}\qquad \text{(d) }(1+x^{2})^{-\frac{1}{2}}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\cot^{-1}x.
\displaystyle \therefore \cot\theta=x=\frac{x}{1}
\displaystyle \therefore \sin\theta=\frac{1}{\sqrt{1+x^{2}}}
\displaystyle \therefore \sin(\cot^{-1}x)=\frac{1}{\sqrt{1+x^{2}}}=(1+x^{2})^{-\frac{1}{2}}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 10. }\text{The number of real solution of } \\ \tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^{2}+x+1}=\frac{\pi}{2}\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }2\qquad \text{(d) infinite}
\displaystyle \text{Answer:}
\displaystyle \text{For }\tan^{-1}\sqrt{x(x+1)}\text{ to be defined, }x(x+1)\geq0.
\displaystyle \text{For }\sin^{-1}\sqrt{x^{2}+x+1}\text{ to be defined, }0\leq x^{2}+x+1\leq1.
\displaystyle x^{2}+x+1\leq1
\displaystyle \Rightarrow x^{2}+x\leq0
\displaystyle \Rightarrow x(x+1)\leq0
\displaystyle \text{Combining }x(x+1)\geq0\text{ and }x(x+1)\leq0,\text{ we get }x(x+1)=0.
\displaystyle \therefore x=0\text{ or }x=-1.
\displaystyle \text{For both values, }\sqrt{x(x+1)}=0\text{ and }\sqrt{x^{2}+x+1}=1.
\displaystyle \therefore \tan^{-1}0+\sin^{-1}1=0+\frac{\pi}{2}=\frac{\pi}{2}
\displaystyle \therefore \text{Number of real solutions}=2.
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 11. }\text{Find the domain of the function }\cos^{-1}(2x-1).
\displaystyle \text{Answer:}
\displaystyle \text{For }\cos^{-1}(2x-1)\text{ to be defined, the argument must satisfy}
\displaystyle -1\leq 2x-1\leq 1
\displaystyle \therefore 0\leq 2x\leq 2
\displaystyle \therefore 0\leq x\leq 1
\displaystyle \therefore \text{The domain is }[0,1].
\\

\displaystyle \textbf{Question 12. }\text{Write the value of the following.}
\displaystyle \text{(i) }\sin^{-1}(\sin1550^\circ)\qquad \text{(ii) }\tan^{-1}\left[\tan\left(\frac{15\pi}{4}\right)\right]
\displaystyle \text{Answer:}
\displaystyle \text{(i) }1550^\circ=4(360^\circ)+110^\circ
\displaystyle \therefore \sin1550^\circ=\sin110^\circ=\sin70^\circ
\displaystyle \therefore \sin^{-1}(\sin1550^\circ)=70^\circ
\displaystyle \text{(ii) }\frac{15\pi}{4}=4\pi-\frac{\pi}{4}
\displaystyle \therefore \tan\left(\frac{15\pi}{4}\right)=\tan\left(-\frac{\pi}{4}\right)
\displaystyle \therefore \tan^{-1}\left[\tan\left(\frac{15\pi}{4}\right)\right]=-\frac{\pi}{4}
\\

\displaystyle \textbf{Question 13. }\text{If }\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5},\text{ then find }\cot^{-1}x+\cot^{-1}y.
\displaystyle \text{Answer:}
\displaystyle \cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x
\displaystyle \cot^{-1}y=\frac{\pi}{2}-\tan^{-1}y
\displaystyle \therefore \cot^{-1}x+\cot^{-1}y=\pi-(\tan^{-1}x+\tan^{-1}y)
\displaystyle =\pi-\frac{4\pi}{5}
\displaystyle =\frac{\pi}{5}
\\

\displaystyle \textbf{Question 14. }\text{If }\tan^{-1}(\sqrt{3})+\cot^{-1}x=\frac{\pi}{2},\text{ then find }x.
\displaystyle \text{Answer:}
\displaystyle \tan^{-1}(\sqrt{3})=\frac{\pi}{3}
\displaystyle \therefore \frac{\pi}{3}+\cot^{-1}x=\frac{\pi}{2}
\displaystyle \therefore \cot^{-1}x=\frac{\pi}{6}
\displaystyle \therefore x=\cot\frac{\pi}{6}=\sqrt{3}
\\

\displaystyle \textbf{Question 15. }\text{Show that }\sin^{-1}2x\sqrt{1-x^{2}}=2\sin^{-1}x,\ -\frac{1}{\sqrt{2}}\leq x\leq \frac{1}{\sqrt{2}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\sin^{-1}x.
\displaystyle \therefore x=\sin\theta
\displaystyle \text{Since }-\frac{1}{\sqrt{2}}\leq x\leq \frac{1}{\sqrt{2}},\text{ we get }-\frac{\pi}{4}\leq\theta\leq\frac{\pi}{4}
\displaystyle \therefore -\frac{\pi}{2}\leq2\theta\leq\frac{\pi}{2}
\displaystyle \text{Now, }2x\sqrt{1-x^{2}}=2\sin\theta\sqrt{1-\sin^{2}\theta}
\displaystyle =2\sin\theta\cos\theta
\displaystyle =\sin2\theta
\displaystyle \therefore \sin^{-1}(2x\sqrt{1-x^{2}})=\sin^{-1}(\sin2\theta)
\displaystyle =2\theta
\displaystyle =2\sin^{-1}x
\displaystyle \therefore \sin^{-1}2x\sqrt{1-x^{2}}=2\sin^{-1}x
\\

\displaystyle \textbf{Question 16. }\text{Find the domain of }\cos^{-1}(x^{2}-4).
\displaystyle \text{Answer:}
\displaystyle \text{For }\cos^{-1}(x^{2}-4)\text{ to be defined,}
\displaystyle -1\leq x^{2}-4\leq1
\displaystyle \therefore 3\leq x^{2}\leq5
\displaystyle \therefore x\in[-\sqrt5,-\sqrt3]\cup[\sqrt3,\sqrt5]
\displaystyle \therefore \text{Domain}=[-\sqrt5,-\sqrt3]\cup[\sqrt3,\sqrt5].
\\

\displaystyle \textbf{Question 17. }\text{Find the value of }\sin[2\sin^{-1}(0.6)].
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\sin^{-1}(0.6).
\displaystyle \therefore \sin\theta=0.6=\frac{3}{5}
\displaystyle \therefore \cos\theta=\frac{4}{5}
\displaystyle \therefore \sin[2\sin^{-1}(0.6)]=\sin2\theta
\displaystyle =2\sin\theta\cos\theta
\displaystyle =2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}
\\

\displaystyle \textbf{Question 18. }\text{Find the value of }\tan^{-1}\left(\tan\frac{5\pi}{6}\right)+\cos^{-1}\left(\cos\frac{13\pi}{6}\right).
\displaystyle \text{Answer:}
\displaystyle \tan^{-1}\left(\tan\frac{5\pi}{6}\right)=\tan^{-1}\left(\tan\left(-\frac{\pi}{6}\right)\right)=-\frac{\pi}{6}
\displaystyle \cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\cos^{-1}\left(\cos\frac{\pi}{6}\right)=\frac{\pi}{6}
\displaystyle \therefore \tan^{-1}\left(\tan\frac{5\pi}{6}\right)+\cos^{-1}\left(\cos\frac{13\pi}{6}\right)
\displaystyle =-\frac{\pi}{6}+\frac{\pi}{6}=0
\\

\displaystyle \textbf{Question 19. }\text{Find the value of }\sin\left[2\cot^{-1}\left(-\frac{5}{12}\right)\right].
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\cot^{-1}\left(-\frac{5}{12}\right).
\displaystyle \therefore \cot\theta=-\frac{5}{12}
\displaystyle \text{Since }\theta\in(0,\pi),\text{ we get }\theta\text{ in quadrant II.}
\displaystyle \therefore \sin\theta=\frac{12}{13},\quad \cos\theta=-\frac{5}{13}
\displaystyle \therefore \sin\left[2\cot^{-1}\left(-\frac{5}{12}\right)\right]=\sin2\theta
\displaystyle =2\sin\theta\cos\theta
\displaystyle =2\cdot\frac{12}{13}\cdot\left(-\frac{5}{13}\right)=-\frac{120}{169}
\\

\displaystyle \textbf{Question 20. }\text{Solve for }x,\ \tan^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan^{-1}x,\ x>0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\tan^{-1}x=2\theta.
\displaystyle \therefore \frac{1}{2}\tan^{-1}x=\theta
\displaystyle \text{Given, }\tan^{-1}\left(\frac{1-x}{1+x}\right)=\theta
\displaystyle \therefore \tan\theta=\frac{1-x}{1+x}
\displaystyle \text{Also, }x=\tan2\theta=\frac{2\tan\theta}{1-\tan^{2}\theta}
\displaystyle \therefore x=\frac{2\left(\frac{1-x}{1+x}\right)}{1-\left(\frac{1-x}{1+x}\right)^2}
\displaystyle =\frac{2(1-x)(1+x)}{(1+x)^2-(1-x)^2}
\displaystyle =\frac{2(1-x^2)}{4x}
\displaystyle \therefore 2x^2=1-x^2
\displaystyle \therefore 3x^2=1
\displaystyle \therefore x=\frac{1}{\sqrt3}\quad(\because x>0)
\\

\displaystyle \textbf{Question 21. }\text{Evaluate }\cos\left(\tan^{-1}\frac{3}{4}\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\tan^{-1}\frac{3}{4}.
\displaystyle \therefore \tan\theta=\frac{3}{4}
\displaystyle \therefore \cos\theta=\frac{4}{\sqrt{3^2+4^2}}=\frac{4}{5}
\displaystyle \therefore \cos\left(\tan^{-1}\frac{3}{4}\right)=\frac{4}{5}
\\

\displaystyle \textbf{Question 22. }\text{Show that }\sin^{-1}(3x-4x^{3})=3\sin^{-1}x,\ x\in\left[-\frac{1}{2},\frac{1}{2}\right].
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\sin^{-1}x.
\displaystyle \therefore x=\sin\theta
\displaystyle \text{Since }x\in\left[-\frac{1}{2},\frac{1}{2}\right],\text{ we get }\theta\in\left[-\frac{\pi}{6},\frac{\pi}{6}\right]
\displaystyle \therefore 3\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
\displaystyle \text{Now, }3x-4x^3=3\sin\theta-4\sin^3\theta
\displaystyle =\sin3\theta
\displaystyle \therefore \sin^{-1}(3x-4x^3)=\sin^{-1}(\sin3\theta)
\displaystyle =3\theta
\displaystyle =3\sin^{-1}x
\displaystyle \therefore \sin^{-1}(3x-4x^{3})=3\sin^{-1}x
\\

\displaystyle \textbf{Question 23. }\text{Find the value of }\sin[\cot^{-1}\{\tan(\cos^{-1}x)\}].
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\cos^{-1}x.
\displaystyle \therefore \cos\theta=x
\displaystyle \text{Let }\phi=\cot^{-1}(\tan\theta).
\displaystyle \therefore \cot\phi=\tan\theta
\displaystyle \therefore \sin\phi=\frac{1}{\sqrt{1+\cot^2\phi}}
\displaystyle =\frac{1}{\sqrt{1+\tan^2\theta}}
\displaystyle =|\cos\theta|
\displaystyle =|x|
\displaystyle \therefore \sin[\cot^{-1}\{\tan(\cos^{-1}x)\}]=|x|
\\

\displaystyle \textbf{Question 24. }\text{Solve for }x,\ \sin^{-1}\left(\frac{2\alpha}{1+\alpha^2}\right)+\sin^{-1}\left(\frac{2\beta}{1+\beta^2}\right)=2\tan^{-1}x.
\displaystyle \text{Answer:}
\displaystyle \sin^{-1}\left(\frac{2\alpha}{1+\alpha^2}\right)=2\tan^{-1}\alpha
\displaystyle \sin^{-1}\left(\frac{2\beta}{1+\beta^2}\right)=2\tan^{-1}\beta
\displaystyle \therefore 2\tan^{-1}\alpha+2\tan^{-1}\beta=2\tan^{-1}x
\displaystyle \therefore \tan^{-1}\alpha+\tan^{-1}\beta=\tan^{-1}x
\displaystyle \therefore \tan^{-1}\left(\frac{\alpha+\beta}{1-\alpha\beta}\right)=\tan^{-1}x
\displaystyle \therefore x=\frac{\alpha+\beta}{1-\alpha\beta}
\\

\displaystyle \textbf{Question 25. }\text{Prove that } \\ \tan^{-1}\left(\frac{1-x}{1+x}\right)-\tan^{-1}\left(\frac{1-y}{1+y}\right)
\displaystyle =\sin^{-1}\left(\frac{y-x}{\sqrt{1+x^2}\sqrt{1+y^2}}\right).
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S.}=\tan^{-1}\left(\frac{1-x}{1+x}\right)-\tan^{-1}\left(\frac{1-y}{1+y}\right)
\displaystyle =\left(\frac{\pi}{4}-\tan^{-1}x\right)-\left(\frac{\pi}{4}-\tan^{-1}y\right)
\displaystyle =\tan^{-1}y-\tan^{-1}x
\displaystyle \text{Let }\theta=\tan^{-1}y-\tan^{-1}x.
\displaystyle \therefore \sin\theta=\sin(\tan^{-1}y-\tan^{-1}x)
\displaystyle =\sin(\tan^{-1}y)\cos(\tan^{-1}x)-\cos(\tan^{-1}y)\sin(\tan^{-1}x)
\displaystyle =\frac{y}{\sqrt{1+y^2}}\cdot\frac{1}{\sqrt{1+x^2}}-\frac{1}{\sqrt{1+y^2}}\cdot\frac{x}{\sqrt{1+x^2}}
\displaystyle =\frac{y-x}{\sqrt{1+x^2}\sqrt{1+y^2}}
\displaystyle \therefore \theta=\sin^{-1}\left(\frac{y-x}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)
\displaystyle \therefore \tan^{-1}\left(\frac{1-x}{1+x}\right)-\tan^{-1}\left(\frac{1-y}{1+y}\right)
\displaystyle =\sin^{-1}\left(\frac{y-x}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 26. }\text{Simplify: }\tan^{-1}\left[\sqrt{\frac{1-\cos x}{1+\cos x}}\right],\ x<\pi.
\displaystyle \text{Answer:}
\displaystyle \tan^{-1}\left[\sqrt{\frac{1-\cos x}{1+\cos x}}\right]
\displaystyle =\tan^{-1}\left[\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}\right]
\displaystyle =\tan^{-1}\left(\tan\frac{x}{2}\right)
\displaystyle =\frac{x}{2}
\\

\displaystyle \textbf{Question 27. }\text{Find the minimum value of }n\text{ for which }\tan^{-1}\left(\frac{n}{\pi}\right)>\frac{\pi}{4}, \\  n\in N\text{ is valid.}
\displaystyle \text{Answer:}
\displaystyle \tan^{-1}\left(\frac{n}{\pi}\right)>\frac{\pi}{4}
\displaystyle \therefore \frac{n}{\pi}>\tan\frac{\pi}{4}
\displaystyle \therefore \frac{n}{\pi}>1
\displaystyle \therefore n>\pi
\displaystyle \text{Since }n\in N,\text{ the minimum value of }n\text{ is }4.
\\

\displaystyle \textbf{Question 28. }\text{Show that }\cos^{-1}x=\tan^{-1}\left[\frac{\sqrt{1-x^{2}}}{x}\right].
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\cos^{-1}x.
\displaystyle \therefore \cos\theta=x
\displaystyle \therefore \sin\theta=\sqrt{1-x^{2}}
\displaystyle \therefore \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\sqrt{1-x^{2}}}{x}
\displaystyle \therefore \theta=\tan^{-1}\left[\frac{\sqrt{1-x^{2}}}{x}\right]
\displaystyle \therefore \cos^{-1}x=\tan^{-1}\left[\frac{\sqrt{1-x^{2}}}{x}\right]
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 29. }\text{Solve for }x:\ \cos(2\sin^{-1}x)=\frac{1}{9},\ x>0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\sin^{-1}x.
\displaystyle \therefore \sin\theta=x
\displaystyle \cos2\theta=1-2\sin^2\theta
\displaystyle \therefore \cos(2\sin^{-1}x)=1-2x^2
\displaystyle \therefore 1-2x^2=\frac{1}{9}
\displaystyle \therefore 2x^2=\frac{8}{9}
\displaystyle \therefore x^2=\frac{4}{9}
\displaystyle \therefore x=\frac{2}{3}\quad(\because x>0)
\\

\displaystyle \textbf{Question 30. }\text{Write }\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right),\ x<\pi\text{ in the simplest form.}
\displaystyle \text{Answer:}
\displaystyle \frac{\cos x-\sin x}{\cos x+\sin x}=\frac{1-\tan x}{1+\tan x}
\displaystyle =\tan\left(\frac{\pi}{4}-x\right)
\displaystyle \therefore \tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)
\displaystyle =\tan^{-1}\left[\tan\left(\frac{\pi}{4}-x\right)\right]
\displaystyle =\frac{\pi}{4}-x
\\

\displaystyle \textbf{Question 31. }\text{Show that }\sin^{-1}\left(\frac{8}{17}\right)+\sin^{-1}\left(\frac{3}{5}\right)=\cos^{-1}\left(\frac{36}{85}\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let }\alpha=\sin^{-1}\left(\frac{8}{17}\right),\quad \beta=\sin^{-1}\left(\frac{3}{5}\right).
\displaystyle \therefore \sin\alpha=\frac{8}{17},\quad \cos\alpha=\frac{15}{17}
\displaystyle \text{and }\sin\beta=\frac{3}{5},\quad \cos\beta=\frac{4}{5}
\displaystyle \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta
\displaystyle =\frac{15}{17}\cdot\frac{4}{5}-\frac{8}{17}\cdot\frac{3}{5}
\displaystyle =\frac{60-24}{85}=\frac{36}{85}
\displaystyle \therefore \alpha+\beta=\cos^{-1}\left(\frac{36}{85}\right)
\displaystyle \therefore \sin^{-1}\left(\frac{8}{17}\right)+\sin^{-1}\left(\frac{3}{5}\right)=\cos^{-1}\left(\frac{36}{85}\right)
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 32. }\text{If }\sin^{-1}(1-x)-2\sin^{-1}x=\frac{\pi}{2},\text{ then find the value of }x.
\displaystyle \text{Answer:}
\displaystyle \sin^{-1}(1-x)-2\sin^{-1}x=\frac{\pi}{2}
\displaystyle \therefore \sin^{-1}(1-x)=\frac{\pi}{2}+2\sin^{-1}x
\displaystyle \text{Since }0\leq x\leq1,\text{ we have }\sin^{-1}x\geq0
\displaystyle \therefore \frac{\pi}{2}+2\sin^{-1}x\geq\frac{\pi}{2}
\displaystyle \text{But }\sin^{-1}(1-x)\leq\frac{\pi}{2}
\displaystyle \therefore \sin^{-1}x=0
\displaystyle \therefore x=0
\\

\displaystyle \textbf{Question 33. }\text{Prove that }\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\cos^{-1}x.
\displaystyle \therefore x=\cos\theta
\displaystyle \sqrt{1+x}=\sqrt{1+\cos\theta}=\sqrt{2}\cos\frac{\theta}{2}
\displaystyle \sqrt{1-x}=\sqrt{1-\cos\theta}=\sqrt{2}\sin\frac{\theta}{2}
\displaystyle \therefore \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}
\displaystyle =\frac{\sqrt2\cos\frac{\theta}{2}-\sqrt2\sin\frac{\theta}{2}}{\sqrt2\cos\frac{\theta}{2}+\sqrt2\sin\frac{\theta}{2}}
\displaystyle =\frac{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}
\displaystyle =\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)
\displaystyle \therefore \tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{\theta}{2}
\displaystyle =\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 34. }\text{Prove that }\tan^{-1}\left[\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right]=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\cos^{-1}x^{2}.
\displaystyle \therefore \cos\theta=x^{2}
\displaystyle \sqrt{1+x^{2}}=\sqrt{1+\cos\theta}=\sqrt2\cos\frac{\theta}{2}
\displaystyle \sqrt{1-x^{2}}=\sqrt{1-\cos\theta}=\sqrt2\sin\frac{\theta}{2}
\displaystyle \therefore \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}
\displaystyle =\frac{\sqrt2\cos\frac{\theta}{2}+\sqrt2\sin\frac{\theta}{2}}{\sqrt2\cos\frac{\theta}{2}-\sqrt2\sin\frac{\theta}{2}}
\displaystyle =\frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}
\displaystyle =\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)
\displaystyle \therefore \tan^{-1}\left[\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right]=\frac{\pi}{4}+\frac{\theta}{2}
\displaystyle =\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^{2}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 35. }\text{If }\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi,\text{ then prove that } \\ x+y+z=xyz.
\displaystyle \text{Answer:}
\displaystyle \tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi
\displaystyle \therefore \tan(\tan^{-1}x+\tan^{-1}y+\tan^{-1}z)=\tan\pi
\displaystyle \therefore \frac{x+y+z-xyz}{1-xy-yz-zx}=0
\displaystyle \therefore x+y+z-xyz=0
\displaystyle \therefore x+y+z=xyz
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 36. }\text{Prove that }\tan^{-1}\sqrt{x}=\frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right),\ x\in[0,1].
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\tan^{-1}\sqrt{x}.
\displaystyle \therefore \tan\theta=\sqrt{x}
\displaystyle \therefore \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}
\displaystyle =\frac{1-x}{1+x}
\displaystyle \therefore 2\theta=\cos^{-1}\left(\frac{1-x}{1+x}\right)
\displaystyle \therefore \theta=\frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)
\displaystyle \therefore \tan^{-1}\sqrt{x}=\frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 37. }\text{Prove that }\frac{9\pi}{8}-\frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right)=\frac{9}{4}\sin^{-1}\left(\frac{2\sqrt2}{3}\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\sin^{-1}\left(\frac{1}{3}\right).
\displaystyle \therefore \sin\theta=\frac{1}{3}
\displaystyle \therefore \cos\theta=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt2}{3}
\displaystyle \therefore \sin^{-1}\left(\frac{2\sqrt2}{3}\right)=\sin^{-1}(\cos\theta)
\displaystyle =\sin^{-1}\left[\sin\left(\frac{\pi}{2}-\theta\right)\right]
\displaystyle =\frac{\pi}{2}-\theta
\displaystyle =\frac{\pi}{2}-\sin^{-1}\left(\frac{1}{3}\right)
\displaystyle \therefore \frac{9}{4}\sin^{-1}\left(\frac{2\sqrt2}{3}\right)=\frac{9}{4}\left[\frac{\pi}{2}-\sin^{-1}\left(\frac{1}{3}\right)\right]
\displaystyle =\frac{9\pi}{8}-\frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right)
\displaystyle \text{Hence proved.}
\\

 

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.