\displaystyle \textbf{Question 1: }\text{Evaluate each of the following:}
\displaystyle \text{(i) }\sin^{-1}\left(\sin\frac{\pi}{6}\right)\qquad \text{(ii) }\sin^{-1}\left(\sin\frac{7\pi}{6}\right)
\displaystyle \text{(iii) }\sin^{-1}\left(\sin\frac{5\pi}{6}\right)\qquad \text{(iv) }\sin^{-1}\left(\sin\frac{13\pi}{7}\right)
\displaystyle \text{(vi) }\sin^{-1}\left(\sin\frac{17\pi}{8}\right)\qquad \text{(vii) }\sin^{-1}\left(\sin\left(-\frac{17\pi}{8}\right)\right)
\displaystyle \text{(viii) }\sin^{-1}(\sin3)\qquad \text{(ix) }\sin^{-1}(\sin4)
\displaystyle \text{(x) }\sin^{-1}(\sin12)\qquad \text{(xi) }\sin^{-1}(\sin2)
\displaystyle \text{Answer:}
\displaystyle \text{We know that }\sin^{-1}(\sin\theta)=\theta\text{ for }\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].

\displaystyle \text{(i) }\sin^{-1}\left(\sin\frac{\pi}{6}\right)=\frac{\pi}{6}

\displaystyle \text{(ii) }\sin^{-1}\left(\sin\frac{7\pi}{6}\right)
\displaystyle =\sin^{-1}\left(\sin\left(\pi+\frac{\pi}{6}\right)\right)
\displaystyle =\sin^{-1}\left(\sin\left(-\frac{\pi}{6}\right)\right)=-\frac{\pi}{6}

\displaystyle \text{(iii) }\sin^{-1}\left(\sin\frac{5\pi}{6}\right)
\displaystyle =\sin^{-1}\left(\sin\left(\pi-\frac{\pi}{6}\right)\right)
\displaystyle =\sin^{-1}\left(\sin\frac{\pi}{6}\right)=\frac{\pi}{6}

\displaystyle \text{(iv) }\sin^{-1}\left(\sin\frac{13\pi}{7}\right)
\displaystyle =\sin^{-1}\left(\sin\left(2\pi-\frac{\pi}{7}\right)\right)
\displaystyle =\sin^{-1}\left(\sin\left(-\frac{\pi}{7}\right)\right)=-\frac{\pi}{7}

\displaystyle \text{(vi) }\sin^{-1}\left(\sin\frac{17\pi}{8}\right)
\displaystyle =\sin^{-1}\left(\sin\left(2\pi+\frac{\pi}{8}\right)\right)
\displaystyle =\sin^{-1}\left(\sin\frac{\pi}{8}\right)=\frac{\pi}{8}

\displaystyle \text{(vii) }\sin^{-1}\left(\sin\left(-\frac{17\pi}{8}\right)\right)
\displaystyle =\sin^{-1}\left(\sin\left(-2\pi-\frac{\pi}{8}\right)\right)
\displaystyle =\sin^{-1}\left(\sin\left(-\frac{\pi}{8}\right)\right)=-\frac{\pi}{8}

\displaystyle \text{(viii) }\sin^{-1}(\sin3)
\displaystyle =\sin^{-1}\left(\sin(\pi-3)\right)=\pi-3

\displaystyle \text{(ix) }\sin^{-1}(\sin4)
\displaystyle =\sin^{-1}\left(\sin(\pi-4)\right)=\pi-4

\displaystyle \text{(x) }\sin^{-1}(\sin12)
\displaystyle =\sin^{-1}\left(\sin(12-4\pi)\right)
\displaystyle =12-4\pi

\displaystyle \text{(xi) }\sin^{-1}(\sin2)
\displaystyle =\sin^{-1}\left(\sin(\pi-2)\right)=\pi-2
\displaystyle \\

\displaystyle \textbf{Question 2: Evaluate each of the following:}
\displaystyle \text{(i) } \cos^{-1}\left(\cos\frac{-\pi}{4}\right)\qquad \text{(ii) } \cos^{-1}\left(\cos\frac{5\pi}{4}\right)
\displaystyle \text{(iii) } \cos^{-1}\left(\cos\frac{4\pi}{3}\right)\qquad \text{(iv) } \cos^{-1}\left(\cos\frac{13\pi}{6}\right)
\displaystyle \text{(v) } \cos^{-1}(\cos3)\qquad \text{(vi) } \cos^{-1}(\cos4)
\displaystyle \text{(vii) } \cos^{-1}(\cos5)\qquad \text{(viii) } \cos^{-1}(\cos12)
\displaystyle \text{Answer:}
\displaystyle \textbf{Note: }\cos^{-1}(\cos\theta)=\theta\text{ for all }\theta\in[0,\pi].

\displaystyle \text{(i) }\cos^{-1}\left(\cos\frac{-\pi}{4}\right)=\cos^{-1}\left(\cos\frac{\pi}{4}\right)=\frac{\pi}{4}

\displaystyle \text{(ii) }\cos^{-1}\left(\cos\frac{5\pi}{4}\right)=\cos^{-1}\left(\cos\left(2\pi-\frac{3\pi}{4}\right)\right)=\cos^{-1}\left(\cos\frac{3\pi}{4}\right)=\frac{3\pi}{4}

\displaystyle \text{(iii) }\cos^{-1}\left(\cos\frac{4\pi}{3}\right)=\cos^{-1}\left(\cos\left(2\pi-\frac{2\pi}{3}\right)\right)=\cos^{-1}\left(\cos\frac{2\pi}{3}\right)=\frac{2\pi}{3}

\displaystyle \text{(iv) }\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\cos^{-1}\left(\cos\left(2\pi+\frac{\pi}{6}\right)\right)=\cos^{-1}\left(\cos\frac{\pi}{6}\right)=\frac{\pi}{6}

\displaystyle \text{(v) }\cos^{-1}(\cos3)=3

\displaystyle \text{(vi) }\cos^{-1}(\cos4)=\cos^{-1}(\cos(2\pi-4))=2\pi-4

\displaystyle \text{(vii) }\cos^{-1}(\cos5)=\cos^{-1}(\cos(2\pi-5))=2\pi-5

\displaystyle \text{(viii) }\cos^{-1}(\cos12)=\cos^{-1}(\cos(4\pi-12))=4\pi-12
\displaystyle \\

\displaystyle \textbf{Question 3: Evaluate each of the following:}
\displaystyle \text{(i) } \tan^{-1}\left(\tan\frac{\pi}{3}\right)\qquad \text{(ii) } \tan^{-1}\left(\tan\frac{6\pi}{7}\right)
\displaystyle \text{(iii) } \tan^{-1}\left(\tan\frac{7\pi}{6}\right)\qquad \text{(iv) } \tan^{-1}\left(\tan\frac{9\pi}{4}\right)
\displaystyle \text{(v) } \tan^{-1}(\tan1)\qquad \text{(vi) } \tan^{-1}(\tan2)
\displaystyle \text{(vii) } \tan^{-1}(\tan4)\qquad \text{(viii) } \tan^{-1}(\tan12)
\displaystyle \text{Answer:}
\displaystyle \textbf{Note: }\tan^{-1}(\tan\theta)=\theta\text{ for all }\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).

\displaystyle \text{(i) }\tan^{-1}\left(\tan\frac{\pi}{3}\right)=\frac{\pi}{3}

\displaystyle \text{(ii) }\tan^{-1}\left(\tan\frac{6\pi}{7}\right)=\tan^{-1}\left(\tan\left(\pi-\frac{\pi}{7}\right)\right)=\tan^{-1}\left(\tan\left(-\frac{\pi}{7}\right)\right)=-\frac{\pi}{7}

\displaystyle \text{(iii) }\tan^{-1}\left(\tan\frac{7\pi}{6}\right)=\tan^{-1}\left(\tan\left(\pi+\frac{\pi}{6}\right)\right)=\tan^{-1}\left(\tan\frac{\pi}{6}\right)=\frac{\pi}{6}

\displaystyle \text{(iv) }\tan^{-1}\left(\tan\frac{9\pi}{4}\right)=\tan^{-1}\left(\tan\left(2\pi+\frac{\pi}{4}\right)\right)=\tan^{-1}\left(\tan\frac{\pi}{4}\right)=\frac{\pi}{4}

\displaystyle \text{(v) }\tan^{-1}(\tan1)=1

\displaystyle \text{(vi) }\tan^{-1}(\tan2)=\tan^{-1}\big(\tan(2-\pi)\big)=2-\pi

\displaystyle \text{(vii) }\tan^{-1}(\tan4)=\tan^{-1}\big(\tan(4-\pi)\big)=4-\pi

\displaystyle \text{(viii) }\tan^{-1}(\tan12)=\tan^{-1}\big(\tan(12-4\pi)\big)=12-4\pi
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Evaluate each of the following:}
\displaystyle \text{(i) }\sec^{-1}\left(\sec\frac{\pi}{3}\right)\qquad \text{(ii) }\sec^{-1}\left(\sec\frac{2\pi}{3}\right)
\displaystyle \text{(iii) }\sec^{-1}\left(\sec\frac{5\pi}{4}\right)\qquad \text{(iv) }\sec^{-1}\left(\sec\frac{7\pi}{3}\right)
\displaystyle \text{(v) }\sec^{-1}\left(\sec\frac{9\pi}{5}\right)\qquad \text{(vi) }\sec^{-1}\left(\sec\left(-\frac{7\pi}{3}\right)\right)
\displaystyle \text{(vii) }\sec^{-1}\left(\sec\frac{13\pi}{4}\right)\qquad \text{(viii) }\sec^{-1}\left(\sec\frac{25\pi}{6}\right)
\displaystyle \text{Answer:}
\displaystyle \textbf{Note: }\sec^{-1}(\sec\theta)=\theta\text{ for all }\theta\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right].

\displaystyle \text{(i) }\sec^{-1}\left(\sec\frac{\pi}{3}\right)=\frac{\pi}{3}

\displaystyle \text{(ii) }\sec^{-1}\left(\sec\frac{2\pi}{3}\right)=\frac{2\pi}{3}

\displaystyle \text{(iii) }\sec^{-1}\left(\sec\frac{5\pi}{4}\right)=\sec^{-1}\left(\sec\left(2\pi-\frac{3\pi}{4}\right)\right)=\sec^{-1}\left(\sec\frac{3\pi}{4}\right)=\frac{3\pi}{4}

\displaystyle \text{(iv) }\sec^{-1}\left(\sec\frac{7\pi}{3}\right)=\sec^{-1}\left(\sec\left(2\pi+\frac{\pi}{3}\right)\right)=\sec^{-1}\left(\sec\frac{\pi}{3}\right)=\frac{\pi}{3}

\displaystyle \text{(v) }\sec^{-1}\left(\sec\frac{9\pi}{5}\right)=\sec^{-1}\left(\sec\left(2\pi-\frac{\pi}{5}\right)\right)=\sec^{-1}\left(\sec\frac{\pi}{5}\right)=\frac{\pi}{5}

\displaystyle \text{(vi) }\sec^{-1}\left(\sec\left(-\frac{7\pi}{3}\right)\right)=\sec^{-1}\left(\sec\frac{7\pi}{3}\right)=\sec^{-1}\left(\sec\left(2\pi+\frac{\pi}{3}\right)\right)=\sec^{-1}\left(\sec\frac{\pi}{3}\right)=\frac{\pi}{3}

\displaystyle \text{(vii) }\sec^{-1}\left(\sec\frac{13\pi}{4}\right)=\sec^{-1}\left(\sec\left(4\pi-\frac{3\pi}{4}\right)\right)=\sec^{-1}\left(\sec\frac{3\pi}{4}\right)=\frac{3\pi}{4}

\displaystyle \text{(viii) }\sec^{-1}\left(\sec\frac{25\pi}{6}\right)=\sec^{-1}\left(\sec\left(4\pi+\frac{\pi}{6}\right)\right)=\sec^{-1}\left(\sec\frac{\pi}{6}\right)=\frac{\pi}{6}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Evaluate each of the following:}
\displaystyle \text{(i) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{\pi}{4}\right)\qquad \text{(ii) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{3\pi}{4}\right)
\displaystyle \text{(iii) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{6\pi}{5}\right)\qquad \text{(iv) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{11\pi}{6}\right)
\displaystyle \text{(v) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{13\pi}{6}\right)\qquad \text{(vi) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{9\pi}{4}\right)\right)
\displaystyle \text{Answer:}
\displaystyle \textbf{Note: }\mathrm{cosec}^{-1}(\mathrm{cosec}\,\theta)=\theta\text{ for all }\theta\in\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right].

\displaystyle \text{(i) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{\pi}{4}\right)=\frac{\pi}{4}

\displaystyle \text{(ii) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{3\pi}{4}\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(\pi-\frac{\pi}{4}\right)\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{\pi}{4}\right)=\frac{\pi}{4}

\displaystyle \text{(iii) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{6\pi}{5}\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(\pi+\frac{\pi}{5}\right)\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{5}\right)\right)=-\frac{\pi}{5}

\displaystyle \text{(iv) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{11\pi}{6}\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(2\pi-\frac{\pi}{6}\right)\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{6}\right)\right)=-\frac{\pi}{6}

\displaystyle \text{(v) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{13\pi}{6}\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(2\pi+\frac{\pi}{6}\right)\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{\pi}{6}\right)=\frac{\pi}{6}

\displaystyle \text{(vi) }\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{9\pi}{4}\right)\right)=\mathrm{cosec}^{-1}\left(-\mathrm{cosec}\left(2\pi+\frac{\pi}{4}\right)\right)=\mathrm{cosec}^{-1}\left(-\mathrm{cosec}\frac{\pi}{4}\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{4}\right)\right)=-\frac{\pi}{4}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Evaluate each of the following:}
\displaystyle \text{(i) }\cot^{-1}\left(\cot\frac{\pi}{3}\right)\qquad \text{(ii) }\cot^{-1}\left(\cot\frac{4\pi}{3}\right)
\displaystyle \text{(iii) }\cot^{-1}\left(\cot\frac{9\pi}{4}\right)\qquad \text{(iv) }\cot^{-1}\left(\cot\frac{19\pi}{6}\right)
\displaystyle \text{(v) }\cot^{-1}\left(\cot\left(-\frac{8\pi}{3}\right)\right)\qquad \text{(vi) }\cot^{-1}\left(\cot\frac{21\pi}{4}\right)
\displaystyle \text{Answer:}
\displaystyle \textbf{Note: }\cot^{-1}(\cot\theta)=\theta\text{ for all }\theta\in(0,\pi).

\displaystyle \text{(i) }\cot^{-1}\left(\cot\frac{\pi}{3}\right)=\frac{\pi}{3}

\displaystyle \text{(ii) }\cot^{-1}\left(\cot\frac{4\pi}{3}\right)=\cot^{-1}\left(\cot\left(\pi+\frac{\pi}{3}\right)\right)=\cot^{-1}\left(\cot\frac{\pi}{3}\right)=\frac{\pi}{3}

\displaystyle \text{(iii) }\cot^{-1}\left(\cot\frac{9\pi}{4}\right)=\cot^{-1}\left(\cot\left(2\pi+\frac{\pi}{4}\right)\right)=\cot^{-1}\left(\cot\frac{\pi}{4}\right)=\frac{\pi}{4}

\displaystyle \text{(iv) }\cot^{-1}\left(\cot\frac{19\pi}{6}\right)=\cot^{-1}\left(\cot\left(3\pi+\frac{\pi}{6}\right)\right)=\cot^{-1}\left(\cot\frac{\pi}{6}\right)=\frac{\pi}{6}

\displaystyle \text{(v) }\cot^{-1}\left(\cot\left(-\frac{8\pi}{3}\right)\right)=\cot^{-1}\left(-\cot\frac{8\pi}{3}\right)=\cot^{-1}\left(-\cot\left(3\pi-\frac{\pi}{3}\right)\right)=\cot^{-1}\left(\cot\frac{\pi}{3}\right)=\frac{\pi}{3}

\displaystyle \text{(vi) }\cot^{-1}\left(\cot\frac{21\pi}{4}\right)=\cot^{-1}\left(\cot\left(5\pi+\frac{\pi}{4}\right)\right)=\cot^{-1}\left(\cot\frac{\pi}{4}\right)=\frac{\pi}{4}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Write each of the following in the simplest form:}
\displaystyle \text{(i) }\cot^{-1}\left(\frac{a}{\sqrt{x^2-a^2}}\right),\ |x|>a,\ a>0
\displaystyle \text{(ii) }\tan^{-1}\left(x+\sqrt{1+x^2}\right),\ x\in R
\displaystyle \text{(iii) }\tan^{-1}\left(\sqrt{1+x^2}-x\right),\ x\in R
\displaystyle \text{(iv) }\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right),\ x\ne0
\displaystyle \text{(v) }\tan^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right),\ x\ne0
\displaystyle \text{(vi) }\tan^{-1}\sqrt{\frac{a-x}{a+x}},\ -a<x<a,\ a>0
\displaystyle \text{(vii) }\tan^{-1}\left(\frac{x}{a+\sqrt{a^2-x^2}}\right),\ -a<x<a,\ a>0
\displaystyle \text{(viii) }\sin^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt2}\right),\ -\frac12<x<\frac1{\sqrt2}
\displaystyle \text{(ix) }\sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}2\right),\ 0<x<1
\displaystyle \text{(x) }\sin\left(2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right),\ -1<x\leq1
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }\theta=\cot^{-1}\left(\frac{a}{\sqrt{x^2-a^2}}\right).
\displaystyle \text{Then }\cot\theta=\frac{a}{\sqrt{x^2-a^2}}\text{ and }\tan\theta=\frac{\sqrt{x^2-a^2}}a.
\displaystyle \therefore \sec\theta=\sqrt{1+\tan^2\theta}=\sqrt{1+\frac{x^2-a^2}{a^2}}=\frac{|x|}{a}
\displaystyle \therefore \cot^{-1}\left(\frac{a}{\sqrt{x^2-a^2}}\right)=\sec^{-1}\left(\frac{|x|}{a}\right)
\displaystyle \\

\displaystyle \text{(ii) Let }\theta=\tan^{-1}x,\text{ so that }-\frac{\pi}{2}<\theta<\frac{\pi}{2}.
\displaystyle x+\sqrt{1+x^2}=\tan\theta+\sec\theta=\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)
\displaystyle \therefore \tan^{-1}\left(x+\sqrt{1+x^2}\right)=\frac{\pi}{4}+\frac{\theta}{2}
\displaystyle =\frac{\pi}{4}+\frac12\tan^{-1}x
\displaystyle \\

\displaystyle \text{(iii) Let }\theta=\tan^{-1}x,\text{ so that }-\frac{\pi}{2}<\theta<\frac{\pi}{2}.
\displaystyle \sqrt{1+x^2}-x=\sec\theta-\tan\theta=\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)
\displaystyle \therefore \tan^{-1}\left(\sqrt{1+x^2}-x\right)=\frac{\pi}{4}-\frac{\theta}{2}
\displaystyle =\frac{\pi}{4}-\frac12\tan^{-1}x
\displaystyle \\

\displaystyle \text{(iv) Let }\theta=\tan^{-1}x,\text{ so that }-\frac{\pi}{2}<\theta<\frac{\pi}{2}.
\displaystyle \frac{\sqrt{1+x^2}-1}{x}=\frac{\sec\theta-1}{\tan\theta}
\displaystyle =\frac{1-\cos\theta}{\sin\theta}=\tan\frac{\theta}{2}
\displaystyle \therefore \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{\theta}{2}=\frac12\tan^{-1}x
\displaystyle \\

\displaystyle \text{(v) Let }\theta=\tan^{-1}x,\text{ so that }-\frac{\pi}{2}<\theta<\frac{\pi}{2}.
\displaystyle \frac{\sqrt{1+x^2}+1}{x}=\frac{\sec\theta+1}{\tan\theta}=\cot\frac{\theta}{2}
\displaystyle \therefore \tan^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right)
\displaystyle =\begin{cases}\displaystyle \frac{\pi}{2}-\frac12\tan^{-1}x,&x>0,\\[6pt]\displaystyle -\frac{\pi}{2}-\frac12\tan^{-1}x,&x<0.\end{cases}
\displaystyle \\

\displaystyle \text{(vi) Let }x=a\cos\theta,\text{ where }0<\theta<\pi.
\displaystyle \tan^{-1}\sqrt{\frac{a-x}{a+x}}=\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}
\displaystyle =\tan^{-1}\sqrt{\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}}=\tan^{-1}\left(\tan\frac{\theta}{2}\right)
\displaystyle =\frac{\theta}{2}=\frac12\cos^{-1}\left(\frac{x}{a}\right)
\displaystyle \\

\displaystyle \text{(vii) Let }x=a\sin\theta,\text{ where }-\frac{\pi}{2}<\theta<\frac{\pi}{2}.
\displaystyle \tan^{-1}\left(\frac{x}{a+\sqrt{a^2-x^2}}\right)=\tan^{-1}\left(\frac{\sin\theta}{1+\cos\theta}\right)
\displaystyle =\tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{2}
\displaystyle =\frac12\sin^{-1}\left(\frac{x}{a}\right)
\displaystyle \\

\displaystyle \text{(viii) Let }x=\sin\theta.
\displaystyle \text{Since }-\frac12<x<\frac1{\sqrt2},\text{ we have }-\frac{\pi}{6}<\theta<\frac{\pi}{4}.
\displaystyle \sin^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt2}\right)=\sin^{-1}\left(\frac{\sin\theta+\cos\theta}{\sqrt2}\right)
\displaystyle =\sin^{-1}\left(\sin\left(\theta+\frac{\pi}{4}\right)\right)=\theta+\frac{\pi}{4}
\displaystyle =\sin^{-1}x+\frac{\pi}{4}
\displaystyle \\

\displaystyle \text{(ix) Let }x=\cos\theta,\text{ where }0<\theta<\frac{\pi}{2}.
\displaystyle \sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}2\right)
\displaystyle =\sin^{-1}\left(\frac{\sqrt2\cos\frac{\theta}{2}+\sqrt2\sin\frac{\theta}{2}}2\right)
\displaystyle =\sin^{-1}\left(\sin\left(\frac{\theta}{2}+\frac{\pi}{4}\right)\right)
\displaystyle =\frac{\theta}{2}+\frac{\pi}{4}=\frac12\cos^{-1}x+\frac{\pi}{4}
\displaystyle \\

\displaystyle \text{(x) Let }x=\cos\theta,\text{ where }0\leq\theta<\pi.
\displaystyle \sin\left(2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right)
\displaystyle =\sin\left(2\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\right)
\displaystyle =\sin\left(2\tan^{-1}\left(\tan\frac{\theta}{2}\right)\right)=\sin\theta
\displaystyle =\sqrt{1-\cos^2\theta}=\sqrt{1-x^2}
\displaystyle \\

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