Question

Evaluate the following:
(i) $\cos \left({\sin }^{-1}\frac{3}{5}\right)$
(ii) $\sin \left({\cos }^{-1}\frac{4}{5}\right)$
(iii) $\cos \left({\sin }^{-1}-\frac{3}{5}\right)$
(iv) $\tan \left({\cos }^{-1}\frac{8}{17}\right)$
(v) $\mathrm{cosec}\left\{{\cos }^{-1}\left(-\frac{12}{13}\right)\right\}$
(vi) $\tan \left\{2{\tan }^{-1}\frac{1}{5}-\frac{\mathrm{\pi }}{4}\right\}$
(vii) $\tan \frac{1}{2}\left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)$
(viii) $\sin \left(\frac{1}{2}{\cos }^{-1}\frac{4}{5}\right)$
(ix) $\cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}\right)$
(x) sin (tan⁻¹ x + cot⁻¹ x)

Answer 1

(i)
$$\begin{gathered} \cos \left({\sin }^{-1}\frac{3}{5}\right)=\cos \left\{{\cos }^{-1}\sqrt{1-{\left(\frac{3}{5}\right)}^2}\right\}\left[\because {\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}\right] \\ =\cos \left({\cos }^{-1}\frac{4}{5}\right) \\ =\frac{4}{5} \end{gathered}$$

(ii)
$$\begin{gathered} \sin \left({\cos }^{-1}\frac{4}{5}\right)=\sin \left\{{\sin }^{-1}\sqrt{1-{\left(\frac{4}{5}\right)}^2}\right\}\left[\because {\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}\right] \\ =\sin \left({\sin }^{-1}\frac{3}{5}\right) \\ =\frac{3}{5} \end{gathered}$$


(iii)
$$\begin{gathered} \cos \left({\sin }^{-1}-\frac{3}{5}\right)=\cos \left\{{\cos }^{-1}\sqrt{1-{\left(-\frac{3}{5}\right)}^2}\right\}\left[\because {\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}\right] \\ =\cos \left({\cos }^{-1}\frac{4}{5}\right) \\ =\frac{4}{5} \end{gathered}$$


(iv)
$$\begin{gathered} \tan \left({\cos }^{-1}\frac{8}{17}\right)=\tan \left\{{\tan }^{-1}\frac{\sqrt{1-{\left({\displaystyle \frac{8}{17}}\right)}^2}}{{\displaystyle \frac{8}{17}}}\right\}\left[\because {\cos }^{-1}x={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right] \\ =\tan \left({\tan }^{-1}\frac{{\displaystyle \frac{15}{17}}}{{\displaystyle \frac{8}{17}}}\right) \\ =\frac{15}{8} \end{gathered}$$

(v)

$$\begin{gathered} \mathrm{cosec}\left\{{\cos }^{-1}\left(-\frac{12}{13}\right)\right\}=\mathrm{cosec}\left\{{\mathrm{cosec}}^{-1}\frac{1}{\sqrt{1-{\left(-\frac{12}{13}\right)}^2}}\right\}\left[\because {\cos }^{-1}x={\mathrm{cosec}}^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)\right] \\ =\mathrm{cosec}\left({\mathrm{cosec}}^{-1}\frac{1}{{\displaystyle \frac{5}{13}}}\right) \\ =\frac{13}{5} \end{gathered}$$
(vi)

$$\begin{gathered} \tan \left(2{\tan }^{-1}\frac{1}{5}-\frac{\mathrm{\pi }}{4}\right)=\tan \left(2{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}1\right) \\ =\tan \left[{\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{1}{5}}}{1-{\left({\displaystyle \frac{1}{5}}\right)}^2}\right\}-{\tan }^{-1}1\right]\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ =\tan \left[{\tan }^{-1}\left\{\frac{{\displaystyle \frac{2}{5}}}{{\displaystyle \frac{24}{25}}}\right\}-{\tan }^{-1}1\right] \\ =\tan \left[{\tan }^{-1}\frac{5}{12}+{\tan }^{-1}1\right] \\ =\tan \left[{\tan }^{-1}\left(\frac{{\displaystyle \frac{5}{12}-}1}{1+{\displaystyle \frac{5}{12}}}\right)\right]\left[\because {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}-y}{1+{\displaystyle xy}}\right)\right] \\ =\tan \left[{\tan }^{-1}\left(\frac{{\displaystyle \frac{-7}{12}}}{{\displaystyle \frac{17}{12}}}\right)\right] \\ =\tan \left[{\tan }^{-1}\frac{-7}{17}\right] \\ =\frac{-7}{17} \end{gathered}$$
(vii)

$$\begin{gathered} \mathrm{Let},{\cos }^{-1}\frac{\sqrt{5}}{3}=\theta \\ \Rightarrow \cos \theta =\frac{\sqrt{5}}{3} \\ \Rightarrow 2{\cos }^2\frac{\theta }{2}-1=\frac{\sqrt{5}}{3} \\ \Rightarrow {\cos }^2\frac{\theta }{2}=\frac{3+\sqrt{5}}{6} \\ \Rightarrow \cos \frac{\theta }{2}=\sqrt{\frac{3+\sqrt{5}}{6}} \\ \Rightarrow \frac{\theta }{2}={\cos }^{-1}\left(\sqrt{\frac{3+\sqrt{5}}{6}}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{1-{\left(\sqrt{\frac{3+\sqrt{5}}{6}}\right)}^2}}{\sqrt{\frac{3+\sqrt{5}}{6}}}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{1-\frac{3+\sqrt{5}}{6}}}{\sqrt{\frac{3+\sqrt{5}}{6}}}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{\frac{3-\sqrt{5}}{6}}}{\sqrt{\frac{3+\sqrt{5}}{6}}}\right) \\ ={\tan }^{-1}\left(\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}\right) \\ ={\tan }^{-1}\left(\sqrt{\frac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}\right) \\ ={\tan }^{-1}\left(\sqrt{\frac{{\left(3-\sqrt{5}\right)}^2}{9-5}}\right) \\ ={\tan }^{-1}\left(\frac{3-\sqrt{5}}{2}\right) \\ \mathrm{i}.\mathrm{e}.,\frac{1}{2}\left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)={\tan }^{-1}\left(\frac{3-\sqrt{5}}{2}\right) \\ \Rightarrow \tan \frac{1}{2}\left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)=\tan \left[{\tan }^{-1}\left(\frac{3-\sqrt{5}}{2}\right)\right] \\ \therefore \tan \frac{1}{2}\left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)=\frac{3-\sqrt{5}}{2} \end{gathered}$$


(viii)
$$\begin{gathered} \sin \left(\frac{1}{2}{\cos }^{-1}\frac{4}{5}\right)=\sin \left\{\frac{1}{2}\times 2{\sin }^{-1}\pm \sqrt{\frac{1-\frac{4}{5}}{2}}\right\}\left[\because {\cos }^{-1}x=2{\sin }^{-1}\pm \sqrt{\frac{1-x}{2}}\right] \\ =\sin \left({\sin }^{-1}\pm \frac{1}{\sqrt{10}}\right) \\ =\pm \frac{1}{\sqrt{10}} \end{gathered}$$


(ix)
$$\begin{gathered} \cos \left({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}\right)=\cos \left\{{\sin }^{-1}\left(\frac{3}{5}\sqrt{1-{\left(\frac{5}{13}\right)}^2}+\frac{5}{13}\sqrt{1-{\left(\frac{3}{5}\right)}^2}\right)\right\}\left[\because {\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)\right] \\ =\cos \left\{{\sin }^{-1}\left(\frac{3}{5}\times \frac{12}{13}+\frac{5}{13}\times \frac{4}{5}\right)\right\} \\ =\cos \left\{{\sin }^{-1}\left(\frac{36}{65}+\frac{4}{13}\right)\right\} \\ =\cos \left\{{\sin }^{-1}\left(\frac{56}{65}\right)\right\} \\ =\cos \left\{{\cos }^{-1}\sqrt{1-{\left(\frac{56}{65}\right)}^2}\right\}\left[\because {\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}\right] \\ =\cos \left\{{\cos }^{-1}\frac{33}{65}\right\} \\ =\frac{33}{65} \end{gathered}$$

(x)
sin (tan⁻¹ x + cot⁻¹ x)$=\sin \frac{\mathrm{\pi }}{2}\left[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}\right]$
= 1