Evaluate each of the following:

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) $\operatorname{cosec}\left(\cos ^{-1} \frac{3}{5}\right)$

(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) $\cos \left(\tan ^{-1} \frac{24}{7}\right)$

Solution:

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

(ii)

$\sin \left(\cos ^{-1} \frac{5}{13}\right)=\sin \left[\sin ^{-1} \sqrt{1-\left(\frac{5}{13}\right)^{2}}\right] \quad\left[\because \cos ^{-1} x=\sin ^{-1} \sqrt{1-x^{2}}\right]$

$=\sin \left[\sin ^{-1}\left(\sqrt{1-\frac{25}{169}}\right)\right]$

$=\sin \left[\sin ^{-1}\left(\sqrt{\frac{144}{169}}\right)\right]$

$=\sin \left[\sin ^{-1} \frac{12}{13}\right]$

$=\frac{12}{13}$

$\sin \left(\tan ^{-1} \frac{24}{7}\right)=\sin \left(\sin ^{-1} \frac{\frac{24}{7}}{\sqrt{1+\left(\frac{24}{7}\right)^{2}}}\right) \quad\left[\because \tan ^{-1} x=\frac{x}{\sqrt{1+x^{2}}}\right]$

$=\sin \left(\sin ^{-1} \frac{\frac{24}{7}}{\sqrt{1+\frac{576}{49}}}\right)$

$=\sin \left(\sin ^{-1} \frac{\frac{24}{7}}{\sqrt{\frac{625}{49}}}\right)$

$=\sin \left(\sin ^{-1} \frac{\frac{24}{7}}{\frac{24}{7}}\right)$

$=\frac{24}{25}$

(iv)

$\sin \left(\sec ^{-1} \frac{17}{8}\right)=\sin \left(\cos ^{-1} \frac{8}{17}\right)$

$=\sin \left[\sin ^{-1} \sqrt{1-\left(\frac{8}{17}\right)^{2}}\right] \quad\left[\because \cos ^{-1} x=\sin ^{-1} \sqrt{1-x^{2}}\right]$

$=\sin \left[\sin ^{-1}\left(\sqrt{1-\frac{64}{289}}\right)\right]$

$=\sin \left[\sin ^{-1}\left(\sqrt{\frac{225}{289}}\right)\right]$

$=\sin \left[\sin ^{-1} \frac{15}{17}\right]$

$=\frac{15}{17}$

(v)

$\operatorname{cosec}\left(\cos ^{-1} \frac{3}{5}\right)=\operatorname{cosec}\left[\sin ^{-1} \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right] \quad\left[\because \cos ^{-1} x=\sin ^{-1} \sqrt{1-x^{2}}\right]$

$=\operatorname{cosec}\left[\sin ^{-1}\left(\sqrt{1-\frac{9}{25}}\right)\right]$

$=\operatorname{cosec}\left[\sin ^{-1}\left(\sqrt{\frac{16}{25}}\right)\right]$

$=\operatorname{cosec}\left[\sin ^{-1} \frac{4}{5}\right]$

$=\operatorname{cosec}\left[\operatorname{cosec}^{-1} \frac{5}{4}\right]$

$=\frac{5}{4}$

(vi)

$\sec \left(\sin ^{-1} \frac{12}{13}\right)=\sec \left[\cos ^{-1} \sqrt{1-\left(\frac{12}{13}\right)^{2}}\right] \quad\left[\because \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}\right]$

$=\sec \left[\cos ^{-1}\left(\sqrt{1-\frac{144}{169}}\right)\right]$

$=\sec \left[\cos ^{-1}\left(\sqrt{\frac{25}{169}}\right)\right]$

$=\sec \left[\cos ^{-1} \frac{5}{13}\right]$

$=\sec \left[\sec ^{-1} \frac{13}{5}\right]$

$=\frac{13}{5}$

(vii)

$\tan \left(\cos ^{-1} \frac{8}{17}\right)=\tan \left\{\tan ^{-1} \frac{\sqrt{1-\left(\frac{8}{17}\right)^{2}}}{\frac{8}{17}}\right\} \quad\left[\because \cos ^{-1} x=\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)\right]$

$=\tan \left(\tan ^{-1} \frac{\frac{15}{17}}{\frac{8}{17}}\right)$

$=\frac{15}{8}$

(viii)

$\cot \left(\cos ^{-1} \frac{3}{5}\right)=\cot \left\{\tan ^{-1} \frac{\sqrt{1-\left(\frac{3}{5}\right)^{2}}}{\frac{3}{5}}\right\} \quad\left[\because \cos ^{-1} x=\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)\right]$

$=\cot \left(\tan ^{-1} \frac{\frac{4}{5}}{\frac{3}{5}}\right)$

$=\cot \left(\cot ^{-1} \frac{3}{4}\right)$

$=\frac{3}{4}$

(ix)

$\cos \left(\tan ^{-1} \frac{24}{7}\right)=\cos \left[\cos ^{-1} \frac{1}{\sqrt{1+\left(\frac{24}{7}\right)^{2}}}\right] \quad\left[\because \tan ^{-1} x=\cos ^{-1} \frac{1}{\sqrt{1+x^{2}}}\right]$

$=\cos \left[\cos ^{-1} \frac{1}{\sqrt{1+\frac{576}{49}}}\right]$

$=\cos \left[\cos ^{-1} \frac{1}{\frac{25}{7}}\right]$

$=\cos \left[\cos ^{-1} \frac{7}{25}\right]$

$=\frac{7}{25}$

 

 

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now