Solve the following

​$$\sin\lparen cos^{-1}\tfrac{3}{5}\rparen+\cos\lparen tan^{-1}\tfrac{5}{12}\rparen$$​​

Correct option is C

Given:

​$$\sin \left(\cos^{-1} \frac{3}{5} \right) + \cos \left(\tan^{-1} \frac{5}{12} \right)$$​​

Concept Used:

​$$\sin (\cos^{-1} x) = \sqrt{1 - x^2}$$​​

​$$\cos (\tan^{-1} x) = \frac{1}{\sqrt{1 + x^2}}$$​​

Solution:

Since,    $$cos \theta = \frac{3}{5}​, \sin^2 \theta + \cos^2 \theta = 1$$​:

​$$\sin \theta = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$​​

Thus,

​$$\sin \left(\cos^{-1} \frac{3}{5} \right) = \frac{4}{5}$$​​

Since,  $$\tan \theta = \frac{5}{12}​, \sec^2 \theta = 1 + \tan^2 \theta;$$​​

​$$\cos \theta = \frac{1}{\sqrt{1 + \left(\frac{5}{12}\right)^2}} = \frac{1}{\sqrt{1 + \frac{25}{144}}} = \frac{1}{\sqrt{\frac{169}{144}}} = \frac{12}{13}$$​​

Thus,

​$$\cos \left(\tan^{-1} \frac{5}{12} \right) = \frac{12}{13}$$

Using these values in equation,

$$\sin \left(\cos^{-1} \frac{3}{5} \right) + \cos \left(\tan^{-1} \frac{5}{12} \right)$$​​​

=$$\frac{4}{5} + \frac{12}{13} = \frac{4 \times 13 + 12 \times 5}{5 \times 13} = \frac{52 + 60}{65} = \frac{112}{65}$$

Option (c) is right.