Question

Evaluate ${\tan }^{-1}\left(\frac{3\mathrm{x}-{\mathrm{x}}^3}{1-3{\mathrm{x}}^2}\right)-{\tan }^{-1}\left(\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\right)$:

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

Answer 1

The correct option is C

${\tan }^{-1}\mathrm{x}$

Explanation for correct option

Evaluating the given trigonometric expression:

Given, the expression is: ${\tan }^{-1}\left(\frac{3\mathrm{x}-{\mathrm{x}}^3}{1-3{\mathrm{x}}^2}\right)-{\tan }^{-1}\left(\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\right)$

We know that:

${\tan }^{-1}\mathrm{a}-{\tan }^{-1}\mathrm{b}={\tan }^{-1}\left(\frac{\mathrm{a}-\mathrm{b}}{1+\mathrm{ab}}\right)...\left(\mathrm{i}\right)$

Here,

$\mathrm{a}=\left(\frac{3\mathrm{x}-{\mathrm{x}}^3}{1-3{\mathrm{x}}^2}\right)$ and $\mathrm{b}=\left(\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\right)$

Now, applying $a$ and $b$ in equation $\left(\mathrm{i}\right)$:

${\tan }^{-1}\left(\frac{3\mathrm{x}-{\mathrm{x}}^3}{1-3{\mathrm{x}}^2}\right)-{\tan }^{-1}\left(\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\right)={\tan }^{-1}\left[\frac{\frac{3\mathrm{x}-{\mathrm{x}}^3}{1-3{\mathrm{x}}^2}-\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}}{1+\frac{3\mathrm{x}-{\mathrm{x}}^3}{1-3{\mathrm{x}}^2}\times \frac{2x}{1-{\mathrm{x}}^2}}\right]$

Cross-multiplying further,

$={\tan }^{-1}\left[\frac{\left(3\mathrm{x}-{\mathrm{x}}^3\right)\left(1-{\mathrm{x}}^2\right)-2\mathrm{x}\left(1-3{\mathrm{x}}^2\right)}{\left(1-3{\mathrm{x}}^2\right)\left(1-{\mathrm{x}}^2\right)+2x\left(3\mathrm{x}-{\mathrm{x}}^3\right)}\right]$

$={\tan }^{-1}\left[\frac{3\mathrm{x}-3{\mathrm{x}}^3-{\mathrm{x}}^3+{\mathrm{x}}^5-2\mathrm{x}+6{\mathrm{x}}^3}{1-{\mathrm{x}}^2-3{\mathrm{x}}^2+3{\mathrm{x}}^4+6{\mathrm{x}}^2-2{\mathrm{x}}^4}\right]={\tan }^{-1}\left[\frac{{\mathrm{x}}^5+2{\mathrm{x}}^3+\mathrm{x}}{{\mathrm{x}}^4+2{\mathrm{x}}^2+1}\right]$

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

$={\tan }^{-1}\mathrm{x}$

Therefore, the value is ${\tan }^{-1}\mathrm{x}$.

Hence, the correct answer is Option (C).