\displaystyle \textbf{Question 1:}
\displaystyle \text{(i) }\cot\left(\sin^{-1}\frac{3}{4}+\sec^{-1}\frac{4}{3}\right)
\displaystyle \text{(ii) }\sin\left(\tan^{-1}x+\tan^{-1}\frac{1}{x}\right),\ x<0
\displaystyle \text{(iii) }\sin\left(\tan^{-1}x+\tan^{-1}\frac{1}{x}\right),\ x>0
\displaystyle \text{(iv) }\cot\left(\tan^{-1}a+\cot^{-1}a\right)\qquad\text{[CBSE 2012]}
\displaystyle \text{(v) }\cos\left(\sec^{-1}x+\mathrm{cosec}^{-1}x\right),\ |x|\geq1
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\cot\left(\sin^{-1}\frac{3}{4}+\sec^{-1}\frac{4}{3}\right)
\displaystyle =\cot\left(\sin^{-1}\frac{3}{4}+\cos^{-1}\frac{3}{4}\right)\qquad\left[\because\sec^{-1}x=\cos^{-1}\left(\frac1x\right)\right]
\displaystyle =\cot\left(\frac{\pi}{2}\right)=0

\displaystyle \text{(ii) }\sin\left(\tan^{-1}x+\tan^{-1}\frac{1}{x}\right)
\displaystyle =\sin\left(\tan^{-1}x-\pi+\cot^{-1}x\right)\qquad\left[\because x<0\right]
\displaystyle =\sin\left(\tan^{-1}x+\cot^{-1}x-\pi\right)
\displaystyle =\sin\left(\frac{\pi}{2}-\pi\right)\qquad\left[\because\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\right]
\displaystyle =\sin\left(-\frac{\pi}{2}\right)=-1

\displaystyle \text{(iii) }\sin\left(\tan^{-1}x+\tan^{-1}\frac{1}{x}\right)
\displaystyle =\sin\left(\tan^{-1}x+\cot^{-1}x\right)\qquad\left[\because x>0\right]
\displaystyle =\sin\left(\frac{\pi}{2}\right)\qquad\left[\because\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\right]
\displaystyle =1

\displaystyle \text{(iv) }\cot\left(\tan^{-1}a+\cot^{-1}a\right)
\displaystyle =\cot\left(\frac{\pi}{2}\right)\qquad\left[\because\tan^{-1}a+\cot^{-1}a=\frac{\pi}{2}\right]
\displaystyle =0

\displaystyle \text{(v) }\cos\left(\sec^{-1}x+\mathrm{cosec}^{-1}x\right)
\displaystyle =\cos\left(\frac{\pi}{2}\right)\qquad\left[\because\sec^{-1}x+\mathrm{cosec}^{-1}x=\frac{\pi}{2}\right]
\displaystyle =0
\displaystyle \\

\displaystyle \textbf{Question 2. If }\cos^{-1}x+\cos^{-1}y=\frac{\pi}{4},\text{ find the value of }\sin^{-1}x+\sin^{-1}y.
\displaystyle \text{Answer:}
\displaystyle \cos^{-1}x+\cos^{-1}y=\frac{\pi}{4}
\displaystyle \Rightarrow\left(\frac{\pi}{2}-\sin^{-1}x\right)+\left(\frac{\pi}{2}-\sin^{-1}y\right)=\frac{\pi}{4}\qquad\left[\because\ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\right]
\displaystyle \Rightarrow\pi-\left(\sin^{-1}x+\sin^{-1}y\right)=\frac{\pi}{4}
\displaystyle \Rightarrow\sin^{-1}x+\sin^{-1}y=\frac{3\pi}{4}
\displaystyle \\

\displaystyle \textbf{Question 3. If }\sin^{-1}x+\sin^{-1}y=\frac{\pi}{3},\text{ and }\cos^{-1}x-\cos^{-1}y=\frac{\pi}{6},\text{ find the values of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \cos^{-1}x-\cos^{-1}y=\frac{\pi}{6}
\displaystyle \Rightarrow\left(\frac{\pi}{2}-\sin^{-1}x\right)-\left(\frac{\pi}{2}-\sin^{-1}y\right)=\frac{\pi}{6}\qquad\left[\because\ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\right]
\displaystyle \Rightarrow-\left(\sin^{-1}x-\sin^{-1}y\right)=\frac{\pi}{6}
\displaystyle \Rightarrow\sin^{-1}x-\sin^{-1}y=-\frac{\pi}{6}
\displaystyle \text{Now solve }\sin^{-1}x+\sin^{-1}y=\frac{\pi}{3}\text{ and }\sin^{-1}x-\sin^{-1}y=-\frac{\pi}{6}
\displaystyle \Rightarrow2\sin^{-1}x=\frac{\pi}{6}
\displaystyle \Rightarrow\sin^{-1}x=\frac{\pi}{12}
\displaystyle \Rightarrow x=\sin\frac{\pi}{12}=\frac{\sqrt3-1}{2\sqrt2}
\displaystyle \text{Also,}
\displaystyle \sin^{-1}y=\frac{\pi}{3}-\sin^{-1}x
\displaystyle \Rightarrow\sin^{-1}y=\frac{\pi}{3}-\frac{\pi}{12}=\frac{\pi}{4}
\displaystyle \Rightarrow y=\sin\frac{\pi}{4}=\frac{1}{\sqrt2}
\displaystyle \\

\displaystyle \textbf{Question 4. If }\cot\left(\cos^{-1}\frac35+\sin^{-1}x\right)=0,\text{ find the value of }x.
\displaystyle \text{Answer:}
\displaystyle \cot\left(\cos^{-1}\frac35+\sin^{-1}x\right)=0
\displaystyle \Rightarrow \cos^{-1}\frac35+\sin^{-1}x=\frac{\pi}{2}\qquad\left[\because\ \cot^{-1}0=\frac{\pi}{2}\right]
\displaystyle \Rightarrow \cos^{-1}\frac35=\frac{\pi}{2}-\sin^{-1}x
\displaystyle \Rightarrow \cos^{-1}\frac35=\cos^{-1}x\qquad\left[\because\ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\right]
\displaystyle \Rightarrow x=\frac35
\displaystyle \\

\displaystyle \textbf{Question 5. If }\left(\sin^{-1}x\right)^2+\left(\cos^{-1}x\right)^2=\frac{17\pi^2}{36},\text{ find }x.
\displaystyle \text{Answer:}
\displaystyle \left(\sin^{-1}x\right)^2+\left(\cos^{-1}x\right)^2=\frac{17\pi^2}{36}
\displaystyle \Rightarrow\left(\sin^{-1}x\right)^2+\left(\frac{\pi}{2}-\sin^{-1}x\right)^2=\frac{17\pi^2}{36}
\displaystyle \text{Let }\sin^{-1}x=y.
\displaystyle \therefore y^2+\left(\frac{\pi}{2}-y\right)^2=\frac{17\pi^2}{36}
\displaystyle \Rightarrow y^2+\frac{\pi^2}{4}+y^2-\pi y=\frac{17\pi^2}{36}
\displaystyle \Rightarrow 2y^2-\pi y=\frac{2\pi^2}{9}
\displaystyle \Rightarrow 18y^2-9\pi y-2\pi^2=0
\displaystyle \Rightarrow 6y(3y-2\pi)+\pi(3y-2\pi)=0
\displaystyle \Rightarrow(3y-2\pi)(6y+\pi)=0
\displaystyle \Rightarrow y=\frac{2\pi}{3}\text{ or }y=-\frac{\pi}{6}
\displaystyle \text{Since }y=\sin^{-1}x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\text{ we reject }y=\frac{2\pi}{3}.
\displaystyle \therefore y=-\frac{\pi}{6}
\displaystyle \therefore x=\sin y=\sin\left(-\frac{\pi}{6}\right)=-\frac12
\displaystyle \\

\displaystyle \textbf{Question 6. }\text{Solve: }\sin\left(\sin^{-1}\frac15+\cos^{-1}x\right)=1\qquad\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \sin\left(\sin^{-1}\frac15+\cos^{-1}x\right)=1
\displaystyle \text{Since }\sin^{-1}\frac15\in\left(0,\frac{\pi}{2}\right)\text{ and }\cos^{-1}x\in[0,\pi],
\displaystyle \sin^{-1}\frac15+\cos^{-1}x\in\left(0,\frac{3\pi}{2}\right).
\displaystyle \text{Therefore, }\sin\left(\sin^{-1}\frac15+\cos^{-1}x\right)=1
\displaystyle \Rightarrow\sin^{-1}\frac15+\cos^{-1}x=\frac{\pi}{2}
\displaystyle \Rightarrow\sin^{-1}\frac15=\frac{\pi}{2}-\cos^{-1}x
\displaystyle \Rightarrow\sin^{-1}\frac15=\sin^{-1}x\qquad\left[\because\ \sin^{-1}x=\frac{\pi}{2}-\cos^{-1}x\right]
\displaystyle \Rightarrow x=\frac15
\displaystyle \\

\displaystyle \textbf{Question 7. }\text{Solve: }\sin^{-1}x=\frac{\pi}{6}+\cos^{-1}x.
\displaystyle \text{Answer:}
\displaystyle \sin^{-1}x=\frac{\pi}{6}+\cos^{-1}x
\displaystyle \Rightarrow\sin^{-1}x=\frac{\pi}{6}+\frac{\pi}{2}-\sin^{-1}x\qquad\left[\because\ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\right]
\displaystyle \Rightarrow2\sin^{-1}x=\frac{2\pi}{3}
\displaystyle \Rightarrow\sin^{-1}x=\frac{\pi}{3}
\displaystyle \Rightarrow x=\sin\frac{\pi}{3}=\frac{\sqrt3}{2}
\displaystyle \\

\displaystyle \textbf{Question 8. }\text{Solve: }4\sin^{-1}x=\pi-\cos^{-1}x.
\displaystyle \text{Answer:}
\displaystyle 4\sin^{-1}x=\pi-\cos^{-1}x
\displaystyle \Rightarrow4\sin^{-1}x=\pi-\left(\frac{\pi}{2}-\sin^{-1}x\right)\qquad\left[\because\ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\right]
\displaystyle \Rightarrow4\sin^{-1}x=\frac{\pi}{2}+\sin^{-1}x
\displaystyle \Rightarrow3\sin^{-1}x=\frac{\pi}{2}
\displaystyle \Rightarrow\sin^{-1}x=\frac{\pi}{6}
\displaystyle \Rightarrow x=\sin\frac{\pi}{6}=\frac12
\displaystyle \\

\displaystyle \textbf{Question 9. }\text{Solve: }\tan^{-1}x+2\cot^{-1}x=\frac{2\pi}{3}.
\displaystyle \text{Answer:}
\displaystyle \tan^{-1}x+2\cot^{-1}x=\frac{2\pi}{3}
\displaystyle \Rightarrow\tan^{-1}x+2\left(\frac{\pi}{2}-\tan^{-1}x\right)=\frac{2\pi}{3}\qquad\left[\because\ \cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x\right]
\displaystyle \Rightarrow\tan^{-1}x+\pi-2\tan^{-1}x=\frac{2\pi}{3}
\displaystyle \Rightarrow\pi-\tan^{-1}x=\frac{2\pi}{3}
\displaystyle \Rightarrow\tan^{-1}x=\frac{\pi}{3}
\displaystyle \Rightarrow x=\tan\frac{\pi}{3}=\sqrt3
\displaystyle \\

\displaystyle \textbf{Question 10. Solve: }5\tan^{-1}x+3\cot^{-1}x=2\pi.
\displaystyle \text{Answer:}
\displaystyle 5\tan^{-1}x+3\cot^{-1}x=2\pi
\displaystyle \Rightarrow5\tan^{-1}x+3\left(\frac{\pi}{2}-\tan^{-1}x\right)=2\pi\qquad\left[\because\ \cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x\right]
\displaystyle \Rightarrow5\tan^{-1}x+\frac{3\pi}{2}-3\tan^{-1}x=2\pi
\displaystyle \Rightarrow2\tan^{-1}x=\frac{\pi}{2}
\displaystyle \Rightarrow\tan^{-1}x=\frac{\pi}{4}
\displaystyle \Rightarrow x=\tan\frac{\pi}{4}=1
\displaystyle \\

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