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(s)

Find the value of the given Equation.

Consider the given Equation as,

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

Let,

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

Then,

$$\begin{gathered} y=\tan \left[\frac{\mathrm{\pi }}{4}+x\right]+\tan \left[\frac{\mathrm{\pi }}{4}-x\right] \\ y=\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan x}{1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan x}+\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\tan x}{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan x} \\ y=\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}\left[\tan \frac{\mathrm{\pi }}{4}=1\right] \\ y=\frac{\left(1+\tan x\right)\left(1+\tan x\right)+\left(1-\tan x\right)\left(1-\tan x\right)}{\left(1-\tan x\right)\left(1+\tan x\right)} \\ y=\frac{1+{\tan }^{2}x+2\tan x+1+{\tan }^{2}x-2\tan x}{1-{\tan }^{2}x} \\ y=\frac{2\left(1+{\tan }^{2}x\right)}{1-{\tan }^{2}x} \end{gathered}$$

We know that

$$\begin{gathered} \Rightarrow se{c}^{2}x-{\tan }^{2}x=1 \\ \Rightarrow se{c}^{2}x=1+{\tan }^{2}x \end{gathered}$$

Then,

$$\begin{gathered} y=\frac{2se{c}^{2}x}{1+1-se{c}^{2}x} \\ y=\frac{2se{c}^{2}x}{2-se{c}^{2}x} \\ y=\frac{{\displaystyle \frac{2}{{\cos }^{2}x}}}{2-{\displaystyle \frac{1}{{\cos }^{2}x}}}\left[se{c}^{2}x=\frac{1}{{\cos }^{2}x}\right] \\ y=\frac{2}{2{\cos }^{2}x-1} \\ y=\frac{2}{\cos 2x}\left[2{\cos }^{2}x-1=\cos 2x\right]......\left(2\right) \end{gathered}$$

From the Equation (1) we can write it as,

$$\begin{gathered} x=\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right) \\ 2x={\cos }^{-1}\left(\frac{a}{b}\right) \\ \cos 2x=\left(\frac{a}{b}\right) \end{gathered}$$

Substitute $\cos 2x=\left(\frac{a}{b}\right)$ in the above Equation (2)

$$\begin{gathered} y=\frac{2}{{\displaystyle \frac{a}{b}}} \\ y=\frac{2b}{a} \end{gathered}$$

Since,

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

Therefore, the correct answer is Option B.