Question

The differential coefficient of $\tan^{- 1}\frac{2x}{1 - x^{2}}$ w.r.t. $\sin^{- 1}\frac{2x}{1 + x^{2}}$is

• A. 1
• B. –1
• C. 0
• D. None of these

Answer 1

Answer: (A)

1

Answer 2

Let $y_{1} = \tan^{- 1}\frac{2x}{1 - x^{2}}$ and $y_{2} = \sin^{- 1}\frac{2x}{1 + x^{2}}$

Putting x = tanθ

∴ $y_{1} = \tan^{- 1}{\tan 2}\theta = 2\theta = 2\tan^{- 1}x$ and

$y_{2} = \sin^{- 1}{\sin 2}\theta = 2\tan^{- 1}x$

Again $\frac{dy_{1}}{dx} = \frac{d}{dx}\lbrack 2\tan^{- 1}x\rbrack = \frac{2}{1 + x^{2}}$ ........(i)

and $\frac{dy_{2}}{dx} = \frac{d}{dx}\lbrack 2\tan^{- 1}x\rbrack = \frac{2}{1 + x^{2}}$ ........(ii)

Hence $\frac{dy_{1}}{dy_{2}} = 1$