Question

If $Ta{n}^{-1}\left(\frac{3a^{2}X-X^3}{a^3-3a{X}^2}\right)=KTa{n}^{-1}\left(\frac{x}{a}\right)$ then k =

• A. 2
• B. 3
• C. -2
• D. 4

Answer 1

The correct option is B 3

${\text{tan}}^{-1}(\frac{3a^{2}x-x^3}{a^3-3a{x}^2})={\text{tan}}^{-1}(\frac{3\frac{x}{a}-\frac{x^3}{a^3}}{1-3\frac{x^2}{a^2}})$

Put $\frac{x}{a}=\tan \theta ⟹\theta ={\text{Tan}}^{-1}\frac{x}{a}$

As we know that $\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

$⟹3\theta ={\text{tan}}^{-1}(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta })$

$⟹{\text{tan}}^{-1}(\frac{3\frac{x}{a}-(\frac{x}{a}{)}^3}{1-3(\frac{x}{a}{)}^2})=3{\text{tan}}^{-1}\frac{x}{a}$

$⟹K{\text{tan}}^{-1}\frac{x}{a}=3{\text{tan}}^{-1}\frac{x}{a}$

$K=3$