Question

$\tan \left[\frac{\mathrm{\pi }}{4}+\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\mathrm{\pi }}{4}-\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{a}{b}\right)\right]$ is equal to

• A. $\frac{2a}{b}$
• B. $\frac{2b}{a}$
• C. $\frac{a}{b}$
• D. $\frac{b}{a}$

Answer 1

The correct option is B

$\frac{2b}{a}$

Explanation for correct answer option:

Finding the value of the given Equation.

We have the given Equation as

$\tan \left[\frac{\mathrm{\pi }}{4}+\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\mathrm{\pi }}{4}-\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{a}{b}\right)\right]$

Let, $\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)=x$

$\therefore \frac{a}{b}=\cos 2x$

Then, substituting the assumed values, we have

$$\begin{gathered} \text{LHS}=\tan \left[\frac{\mathrm{\pi }}{4}+x\right]+\tan \left[\frac{\mathrm{\pi }}{4}-x\right] \\ =\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan x}{1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\times \tan x}+\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\tan x}{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\times \tan x}\mathbf{\left[}\mathbf{\because }\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{a}\mathbf{)}\mathbf{=}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{a}}{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}\mathbf{\times }\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{a}}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}\mathbf{)}\mathbf{=}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{a}}{\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}\mathbf{\times }\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{a}}\mathbf{\right]} \\ =\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}\mathbf{}\mathbf{[}\mathbf{taking}\mathbf{}\mathbf{tan}\mathbf{}\frac{\mathbf{\pi }}{4}\mathbf{}\mathbf{common}\mathbf{]} \\ =\frac{{\displaystyle \left(1+\tan x\right)}\left(1+\tan x\right)+\left(1-\tan x\right)\left(1-\tan x\right)}{\left(1-\tan x\right)\left(1+\tan x\right)} \\ =\frac{1-{\tan }^{2}x+2\tan x+1+{\tan }^{2}x+2\tan x}{1-{\tan }^{2}x} \\ =\frac{2\left(1+{\tan }^{2}x\right)}{1-{\tan }^{2}x} \end{gathered}$$

$$\begin{gathered} =\frac{2se{c}^{2}x}{1+1-se{c}^{2}x}\left[\because 1+{\tan }^{2}x={\sec }^{2}x\right] \\ =\frac{2se{c}^{2}x}{2-se{c}^{2}x} \end{gathered}$$

$$\begin{gathered} =\frac{{\displaystyle \frac{2}{{\cos }^{2}x}}}{2-{\displaystyle \frac{1}{{\cos }^{2}x}}}\left[\because {\sec }^{2}x=\frac{1}{{\cos }^{2}x}\right] \\ =\frac{2}{2{\cos }^{2}x-1} \\ =\frac{2}{\cos 2x}\mathbf{}\left[\cos 2x=2{\cos }^{2}x-1\right] \\ =\frac{2}{{\displaystyle \frac{a}{b}}}\left[\text{Where,}\cos 2x=\frac{a}{b}\right] \\ =\frac{2b}{a} \end{gathered}$$

thus, $\tan \left[\frac{\mathrm{\pi }}{4}+\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\mathrm{\pi }}{4}-\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{a}{b}\right)\right]=\frac{2b}{a}$

Therefore, the correct answer is option B.