Question

The value of$co{t}^{-1}\left(9\right)+\cos e{c}^{-1}\left(\frac{\sqrt{41}}{4}\right)$ is given by

• A. $0$
• B. ${\tan }^{-1}x$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{4}$

Explanation for the correct option:

Find the required value.

Given: $co{t}^{-1}\left(9\right)+\cos e{c}^{-1}\left(\frac{\sqrt{41}}{4}\right)$

Let $\cos e{c}^{-1}\frac{\sqrt{41}}{4}=A$

$$\begin{gathered} \Rightarrow \cos ecA=\frac{\sqrt{41}}{4} \\ \therefore \sin A=\frac{4}{\sqrt{41}}\left[\because \cos ecA=\frac{1}{\sin A}\right] \end{gathered}$$

By using the identity

$$\begin{gathered} {\sin }^{2}A+{\cos }^{2}A=1 \\ \Rightarrow {\left(\frac{4}{\sqrt{41}}\right)}^2+{\cos }^{2}A=1 \\ \Rightarrow {\cos }^{2}A=1-{\left(\frac{4}{\sqrt{41}}\right)}^2 \\ \Rightarrow {\cos }^{2}A=1-\frac{16}{41} \\ \Rightarrow {\cos }^{2}A=\frac{25}{41} \\ \Rightarrow \cos A=\frac{5}{\sqrt{41}} \end{gathered}$$,

$$\begin{gathered} \therefore \tan A=\frac{4}{5}\left[\tan A=\frac{\sin A}{\cos A}\right] \\ \Rightarrow A={\tan }^{-1}\frac{4}{5} \end{gathered}$$

Now,

$co{t}^{-1}\left(9\right)+\cos e{c}^{-1}\left(\frac{\surd 41}{4}\right)={\tan }^{-1}\left(\frac{1}{9}\right)+{\tan }^{-1}\left(\frac{4}{5}\right)\left[\because co{t}^{-1}x=t{\mathrm{an}}^{-1}\left(\frac{1}{x}\right)\right]$

$$\begin{gathered} =ta{n}^{-1}\frac{\left({\displaystyle \frac{1}{9}}+{\displaystyle \frac{4}{5}}\right)}{\left(1-\left(\frac{1}{9}\right)\left(\frac{4}{5}\right)\right)}\left[\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\frac{\left(x+y\right)}{1-xy}\right] \\ =ta{n}^{-1}\frac{\left({\displaystyle \frac{5+36}{45}}\right)}{\left(\frac{45-4}{45}\right)} \\ =ta{n}^{-1}\left(\frac{41}{41}\right) \\ ={\tan }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

Hence, option (C) is the correct answer.