Question

Obtain differential co-efficient of $Tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})$ w.r.t $Cos^{-1}\frac{\sqrt{1+\sqrt{1+x^2}}}{2\sqrt{1+x^2}}$ ($x>0, x \in R$)

Answer 1

1

Answer 2

Let $y = Tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})$ and $z = Cos^{-1}\frac{\sqrt{1+\sqrt{1+x^2}}}{2\sqrt{1+x^2}}$. We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.

For $y$: Let $x = \tan \theta$. Then $\frac{\sqrt{1+x^2}-1}{x} = \frac{\sec \theta - 1}{\tan \theta} = \tan(\theta/2)$. So, $y = Tan^{-1}(\tan(\theta/2)) = \theta/2$. Since $x = \tan \theta$, $\theta = Tan^{-1}x$. Thus, $y = \frac{1}{2} Tan^{-1}x$.

For $z$: Let $\sqrt{1+x^2} = \sec \alpha$. Then $\frac{\sqrt{1+\sec \alpha}}{2\sec \alpha} = \sqrt{\frac{1+\cos \alpha}{2}} = \cos(\alpha/2)$. So, $z = Cos^{-1}(\cos(\alpha/2)) = \alpha/2$. Since $\sqrt{1+x^2} = \sec \alpha$, we have $x = \tan \alpha$. Thus $\alpha = Tan^{-1}x$. Thus, $z = \frac{1}{2} Tan^{-1}x$.

Since $y=z$, $\frac{dy}{dx} = \frac{dz}{dx}$. Therefore, $\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = 1$.