Question

$co{t}^{-1}\left(\frac{3}{4}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)=?$

• A. ${\sin }^{-1}\frac{63}{65}$
• B. ${\sin }^{-1}\frac{12}{13}$
• C. ${\sin }^{-1}\frac{65}{68}$
• D. ${\sin }^{-1}\frac{5}{12}$

Answer 1

The correct option is A

${\sin }^{-1}\frac{63}{65}$

Explanation for the correct option:

Step 1. Let $co{t}^{-1}\frac{3}{4}=\theta$ ……(1)

$\Rightarrow$ $cot\theta =\frac{3}{4}$

As we know, $\sin \theta =\frac{1}{\sqrt{1+co{t}^2\theta }}$

$=\frac{1}{\sqrt{1+\left({\displaystyle \frac{9}{16}}\right)}}$

$=\frac{1}{\sqrt{{\displaystyle \frac{25}{16}}}}$

$=\frac{4}{5}$

$\Rightarrow$ $\theta ={\sin }^{-1}\left(\frac{4}{5}\right)$ ……(2)

Given, $co{t}^{-1}\left(\frac{3}{4}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)$

Step 2. From equation (1) and (2):

$={\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)$

Step 3. Use the identity $\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{+}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{y}\mathbf{}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{\{}\mathbf{x}\sqrt{\left(1-y^2\right)}\mathbf{+}\mathbf{y}\sqrt{\left(1-x^2\right)}\mathbf{\}}$

$={\sin }^{-1}\left[\frac{4}{5}\times \sqrt{1–{\left(\frac{5}{13}\right)}^2}+\frac{5}{13}\times \sqrt{1–{\left(\frac{4}{5}\right)}^2}\right]$

$={\sin }^{-1}\left[\frac{4}{5}\times \sqrt{1–\left(\frac{25}{169}\right)}+\frac{5}{13}\times \sqrt{1–\left(\frac{16}{25}\right)}\right]$

$={\sin }^{-1}\left(\frac{4}{5}\times \frac{12}{13}+\frac{5}{13}\times \frac{3}{5}\right)$

$={\sin }^{-1}\left[\frac{48+15}{65}\right]$

$=\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{63}}{\mathbf{65}}$

Hence, Option ‘A’ is Correct.