Question

$co{t}^{-1}\left(3\right)+\cos e{c}^{-1}(\surd 5)=$

• A. $\frac{\mathrm{\pi }}{3}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{6}$
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is B

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

Explanation for the correct option:

Step 1. Suppose $\mathit{c}\mathit{o}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{-}\mathbf{1}}\sqrt{\mathbf{5}}\mathbf{}\mathbf{=}\mathbf{}\mathit{\theta }\mathbf{}$

$\Rightarrow$ $\cos ec\theta =\sqrt{5}$

Given, $co{t}^{-1}\left(3\right)+\cos e{c}^{-1}(\surd 5)$ …..(1)

$\because co{t}^2\theta =cose{c}^2\theta -1$

$=5-1$

$=4$

$\therefore$ $cot\theta =2$

$\Rightarrow$$\theta =co{t}^{-1}2$

$\therefore \cos e{c}^{-1}\sqrt{5}=\theta$

$\Rightarrow$ $\cos e{c}^{-1}\sqrt{5}=co{t}^{-1}2$

Step 2. Put the value of $\mathit{c}\mathit{o}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{-}\mathbf{1}}\sqrt{\mathbf{5}}$ in Equation (1):

$\Rightarrow$$co{t}^{-1}3+co{t}^{-1}2$

$\Rightarrow {\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{2}$

$\Rightarrow {\tan }^{-1}\left[\frac{{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{1}{3}}.{\displaystyle \frac{1}{2}}}\right]$ $\left[\mathbf{\because }\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathbf{+}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{B}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{A+B}{1-A.B}\right)\right]$

$\Rightarrow {\tan }^{-1}\left(\frac{{\displaystyle \frac{5}{6}}}{{\displaystyle \frac{5}{6}}}\right)$

$\Rightarrow {\tan }^{-1}\left(1\right)$

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

Hence, Option ‘B’ is Correct.