Question

$\cos ec\left[2co{t}^{–1}\left(5\right)+{\cos }^{–1}\left(\frac{4}{5}\right)\right]$ is equal to

• A. $\frac{75}{56}$
• B. $\frac{65}{56}$
• C. $\frac{56}{33}$
• D. $\frac{65}{33}$

Answer 1

The correct option is B

$\frac{65}{56}$

Explanation for the correct option:

Step 1. Find the value of given equation:

$\cos ec\left[2co{t}^{–1}\left(5\right)+{\cos }^{–1}\left(\frac{4}{5}\right)\right]$

$=\cos ec\left[2{\tan }^{–1}\left(\frac{1}{5}\right)+{\cos }^{–1}\left(\frac{4}{5}\right)\right]$

$=\cos ec\left[{\tan }^{-1}\left(\frac{2\left({\displaystyle \frac{1}{5}}\right)}{1-{\left({\displaystyle \frac{1}{5}}\right)}^2}\right)+{\cos }^{–1}\left(\frac{4}{5}\right)\right]$ ; $\left[\mathbf{\because }\mathbf{2}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{2\theta }{1-{\theta }^2}\right)\right]$

$=\cos ec\left[{\tan }^{–1}\left(\frac{5}{12}\right)+{\cos }^{–1}\left(\frac{4}{5}\right)\right]$ …….(1)

Step 2. Let ${\tan }^{-1}\left(\frac{5}{12}\right)=\theta$

$\Rightarrow \tan \theta =\frac{5}{12},\sin \theta =\frac{5}{13},\cos \theta =\frac{12}{13}$

and ${\cos }^{-1}\left(\frac{4}{5}\right)=\varphi$

$\Rightarrow \cos \varphi =\frac{4}{5},\sin \varphi =\frac{3}{5}$

Step 3. Put the values of ${\tan }^{-1}\left(\frac{5}{12}\right),{\cos }^{-1}\left(\frac{4}{5}\right)$ in equation (1):

$=\cos ec\left(\theta +\varphi \right)$

$=\frac{1}{\sin \theta \cos \varphi +\cos \theta \sin \varphi }$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\mathbf{(}\mathit{A}\mathbf{+}\mathit{B}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{A}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{B}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{A}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{B}}\right]$

$=\frac{1}{{\displaystyle \frac{5}{13}}\times {\displaystyle \frac{4}{5}}+{\displaystyle \frac{12}{13}}\times {\displaystyle \frac{3}{5}}}$

$=\frac{1}{{\displaystyle \frac{20}{65}}+{\displaystyle \frac{36}{65}}}$

$=\frac{65}{20+36}$

$=\frac{\mathbf{65}}{\mathbf{56}}$

Hence, Option ‘B’ is Correct.