Question

Evaluate each of the following:

(i) $\sin \left({\sin }^{-1}\frac{7}{25}\right)$
(ii) $\sin \left({\cos }^{-1}\frac{5}{13}\right)$
(iii) $\sin \left({\tan }^{-1}\frac{24}{7}\right)$
(iv) $\sin \left({\sec }^{-1}\frac{17}{8}\right)$
(v) $\mathrm{cosec}\left({\cos }^{-1}\frac{3}{5}\right)$
(vi) $\sec \left({\sin }^{-1}\frac{12}{13}\right)$
(vii) $\tan \left({\cos }^{-1}\frac{8}{17}\right)$
(viii) $\cot \left({\cos }^{-1}\frac{3}{5}\right)$
(ix) $\cos \left({\tan }^{-1}\frac{24}{7}\right)$

Answer 1

(i) $\sin \left({\sin }^{-1}\frac{7}{25}\right)=\frac{7}{25}$
(ii)

$$\begin{gathered} \sin \left({\cos }^{-1}\frac{5}{13}\right)=\sin \left[{\sin }^{-1}\sqrt{1-{\left(\frac{5}{13}\right)}^2}\right]\left[\because {\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}\right] \\ =\sin \left[{\sin }^{-1}\left(\sqrt{1-\frac{25}{169}}\right)\right] \\ =\sin \left[{\sin }^{-1}\left(\sqrt{\frac{144}{169}}\right)\right] \\ =\sin \left[{\sin }^{-1}\frac{12}{13}\right] \\ =\frac{12}{13} \end{gathered}$$
(iii)
$$\begin{gathered} \sin \left({\tan }^{-1}\frac{24}{7}\right)=\sin \left({\sin }^{-1}\frac{{\displaystyle \frac{24}{7}}}{\sqrt{1+{\left({\displaystyle \frac{24}{7}}\right)}^2}}\right)\left[\because {\tan }^{-1}x=\frac{x}{\sqrt{1+x^2}}\right] \\ =\sin \left({\sin }^{-1}\frac{{\displaystyle \frac{24}{7}}}{\sqrt{1+{\displaystyle \frac{576}{49}}}}\right) \\ =\sin \left({\sin }^{-1}\frac{{\displaystyle \frac{24}{7}}}{\sqrt{{\displaystyle \frac{625}{49}}}}\right) \\ =\sin \left({\sin }^{-1}\frac{{\displaystyle \frac{24}{7}}}{{\displaystyle \frac{25}{7}}}\right) \\ =\frac{24}{25} \end{gathered}$$
(iv)
$$\begin{gathered} \sin \left({\sec }^{-1}\frac{17}{8}\right)=\sin \left({\cos }^{-1}\frac{8}{17}\right) \\ =\sin \left[{\sin }^{-1}\sqrt{1-{\left(\frac{8}{17}\right)}^2}\right]\left[\because {\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}\right] \\ =\sin \left[{\sin }^{-1}\left(\sqrt{1-\frac{64}{289}}\right)\right] \\ =\sin \left[{\sin }^{-1}\left(\sqrt{\frac{225}{289}}\right)\right] \\ =\sin \left[{\sin }^{-1}\frac{15}{17}\right] \\ =\frac{15}{17} \end{gathered}$$
(v)
$$\begin{gathered} \mathrm{cosec}\left({\cos }^{-1}\frac{3}{5}\right)=\mathrm{cosec}\left[{\sin }^{-1}\sqrt{1-{\left(\frac{3}{5}\right)}^2}\right]\left[\because {\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}\right] \\ =\mathrm{cosec}\left[{\sin }^{-1}\left(\sqrt{1-\frac{9}{25}}\right)\right] \\ =\mathrm{cosec}\left[{\sin }^{-1}\left(\sqrt{\frac{16}{25}}\right)\right] \\ =\mathrm{cosec}\left[{\sin }^{-1}\frac{4}{5}\right] \\ =\mathrm{cosec}\left[{\mathrm{cosec}}^{-1}\frac{5}{4}\right] \\ =\frac{5}{4} \end{gathered}$$
(vi)
$$\begin{gathered} \sec \left({\sin }^{-1}\frac{12}{13}\right)=\sec \left[{\cos }^{-1}\sqrt{1-{\left(\frac{12}{13}\right)}^2}\right]\left[\because {\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}\right] \\ =\sec \left[{\cos }^{-1}\left(\sqrt{1-\frac{144}{169}}\right)\right] \\ =\sec \left[{\cos }^{-1}\left(\sqrt{\frac{25}{169}}\right)\right] \\ =\sec \left[{\cos }^{-1}\frac{5}{13}\right] \\ =\sec \left[{\sec }^{-1}\frac{13}{5}\right] \\ =\frac{13}{5} \end{gathered}$$
(vii)
$$\begin{gathered} \tan \left({\cos }^{-1}\frac{8}{17}\right)=\tan \left\{{\tan }^{-1}\frac{\sqrt{1-{\left({\displaystyle \frac{8}{17}}\right)}^2}}{{\displaystyle \frac{8}{17}}}\right\}\left[\because {\cos }^{-1}x={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right] \\ =\tan \left({\tan }^{-1}\frac{{\displaystyle \frac{15}{17}}}{{\displaystyle \frac{8}{17}}}\right) \\ =\frac{15}{8} \end{gathered}$$
(viii)
$$\begin{gathered} \cot \left({\cos }^{-1}\frac{3}{5}\right)=\cot \left\{{\tan }^{-1}\frac{\sqrt{1-{\left({\displaystyle \frac{3}{5}}\right)}^2}}{{\displaystyle \frac{3}{5}}}\right\}\left[\because {\cos }^{-1}x={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right] \\ =\cot \left({\tan }^{-1}\frac{{\displaystyle \frac{4}{5}}}{{\displaystyle \frac{3}{5}}}\right) \\ =\cot \left({\cot }^{-1}\frac{{\displaystyle 3}}{{\displaystyle 4}}\right) \\ =\frac{{\displaystyle 3}}{{\displaystyle 4}} \end{gathered}$$
(ix)
$$\begin{gathered} \cos \left({\tan }^{-1}\frac{24}{7}\right)=\cos \left[{\cos }^{-1}\frac{1}{\sqrt{1+{\left({\displaystyle \frac{24}{7}}\right)}^2}}\right]\left[\because {\tan }^{-1}x={\cos }^{-1}\frac{1}{\sqrt{1+x^2}}\right] \\ =\cos \left[{\cos }^{-1}\frac{1}{\sqrt{1+{\displaystyle \frac{576}{49}}}}\right] \\ =\cos \left[{\cos }^{-1}\frac{1}{{\displaystyle \frac{25}{7}}}\right] \\ =\cos \left[{\cos }^{-1}\frac{7}{{\displaystyle 25}}\right] \\ =\frac{7}{{\displaystyle 25}} \end{gathered}$$