Question

If $a>b>0$, then the value of ${\tan }^{-1}\left(\frac{a}{b}\right)+{\tan }^{-1}\left(\frac{a+b}{a-b}\right)$

• A. both $a$ and $b$
• B. $b$ but not $a$
• C. $a$ but not $b$
• D. neither $a$ nor $b$

Answer 1

The correct option is D

neither $a$ nor $b$

Explanation for the correct option:

Find the value ${\tan }^{-1}\left(\frac{a}{b}\right)+{\tan }^{-1}\left(\frac{a+b}{a-b}\right)$ depends on:

${\tan }^{-1}\left(\frac{a}{b}\right)+{\tan }^{-1}\left(\frac{a+b}{a-b}\right)$$={\tan }^{-1}\left[\frac{\left({\displaystyle \frac{a}{b}}\right)+\left({\displaystyle \frac{a+b}{a–b}}\right)}{1–\left({\displaystyle \frac{a}{b}}\right)\left({\displaystyle \frac{a+b}{a–b}}\right)}\right]$ $\left[\mathbf{\because }\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{\varphi }\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{\theta +\varphi }{1-\theta \varphi }\right)\right]$

$={\tan }^{-1}\left(\frac{a^2–ab+ab+b^2}{ab–b^2–a^2–ab}\right)$

$={\tan }^{-1}\left(\frac{a^2+b^2}{-a^2-b^2}\right)$

$={\tan }^{-1}(-1)$

$\therefore$The value neither depends neither on $a$ nor on $b$.

Hence, option ‘D’ is Correct.