Question

If $y^2=100{\tan }^{-1}x+45{\sec }^{-1}x+100{\cot }^{-1}x+45{\text{cosec}}^{-1}x$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{x^2-1}{x^2+1}$
• B. $\frac{x^2+1}{x^2-1}$
• C. $1$
• D. $0$
• E. $\frac{1}{x\sqrt{x^2-1}}$

Answer 1

The correct option is B $\frac{x^2+1}{x^2-1}$

Given, $y^2=100{\tan }^{-1}x+45{\sec }^{-1}x+100{\cot }^{-1}x+45{\text{cosec}}^{-1}x$
$=100{\tan }^{-1}x+100{\cot }^{-1}x+45{\sec }^{-1}x+45{\text{cosec}}^{-1}x$
$=100({\tan }^{-1}x+{\cot }^{-1}x)+45({\sec }^{-1}x+{\text{cosec}}^{-1}x)$
$=100\times \frac{\pi }{2}+45\times \frac{\pi }{2}$
On differentiating both sides with respect to $x$, we get
$2y{y}^{{}^{\prime }}=0$
$\Rightarrow y^{{}^{\prime }}=0$ $[\because 2\ne 0,y\ne 0]$