Question

The value of $cot\left[\cos e{c}^{-1}\left(\frac{5}{3}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\right]$ is

• A. $\frac{3}{17}$
• B. $\frac{4}{17}$
• C. $\frac{5}{17}$
• D. $\frac{6}{17}$

Answer 1

The correct option is D

$\frac{6}{17}$

Explanation for the correct option:

Find the required value.

Given: $cot\left[\cos e{c}^{-1}\left(\frac{5}{3}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\right]....\left(i\right)$

Let $\cos e{c}^{-1}\left(\frac{5}{3}\right)=A$

$\Rightarrow \cos ecA=\frac{5}{3}$

By using the identity, $\cos e{c}^{2}A–co{t}^{2}A=1$

$$\begin{gathered} \Rightarrow {\left(\frac{5}{3}\right)}^2-co{t}^{2}A=1 \\ \Rightarrow {\left(\frac{5}{3}\right)}^2-1=co{t}^{2}A \\ \Rightarrow \frac{25}{9}-1=co{t}^{2}A \\ \Rightarrow \frac{16}{9}=co{t}^{2}A \\ \Rightarrow cotA=\frac{4}{3} \end{gathered}$$

Now let ${\tan }^{-1}\left(\frac{2}{3}\right)=B$

$$\begin{gathered} \Rightarrow \tan B=\frac{2}{3} \\ \therefore cotB=\frac{3}{2}\left[\because cotB=\frac{1}{\tan B}\right] \end{gathered}$$

From $\left(i\right)$,

$$\begin{gathered} cot\left[\cos e{c}^{-1}\left(\frac{5}{3}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\right]=cot(A+B) \\ =\frac{(cotAcotB–1)}{(cotA+cotB)} \\ =\frac{\left({\displaystyle \frac{4}{3}}\right)\left({\displaystyle \frac{3}{2}}\right)-1}{\left(\frac{4}{3}\right)+\left(\frac{3}{2}\right)} \\ =\frac{1}{\frac{17}{6}} \\ =\frac{6}{17} \end{gathered}$$

Hence, option (D), $\frac{6}{17}$ is the correct answer.