Question

If $y=\sqrt{\frac{secx-1}{secx+1}}$ then $\frac{dy}{dx}$ =

• A. $\frac{1}{2}se{c}^2\frac{x}{2}$
• B. $se{c}^2\frac{x}{2}$
• C. $\frac{1}{2}tan\frac{x}{2}$
• D. $tan\frac{x}{2}$

Answer 1

The correct option is A $\frac{1}{2}se{c}^2\frac{x}{2}$

Given,

$y=\sqrt{\frac{\sec x-1}{\sec x+1}}$

$=\sqrt{\frac{\sec x-1}{\sec x+1}}\times \sqrt{\frac{\sec x-1}{\sec x-1}}$

$=\sqrt{\frac{(secx-1{)}^2}{(\sec x-1)(\sec x+1)}}$

$=\sqrt{\frac{(secx-1{)}^2}{{\sec }^{2}x-1}}$

$=\sqrt{\frac{(secx-1{)}^2}{{\tan }^{2}x}}$

$=\frac{secx-1}{\tan x}$

$=\frac{secx}{\tan x}-\frac{1}{\tan x}$

$=\csc x-\cot x$

$\therefore y=\csc x-\cot x$

Now,

$\frac{dy}{dx}=\frac{d}{dx}(\csc x-\cot x)$

$=-\cot \left(x\right)\csc \left(x\right)-(-{\csc }^2\left(x\right))$

$=-\cot \left(x\right)\csc \left(x\right)+{\csc }^2\left(x\right)$

$={\left(\frac{1}{\sin \left(x\right)}\right)}^2-\frac{1}{\sin \left(x\right)}\cdot \frac{\cos \left(x\right)}{\sin \left(x\right)}$

$=\frac{1}{{\sin }^2\left(x\right)}-\frac{\cos \left(x\right)}{{\sin }^2\left(x\right)}$

$=\frac{1-\cos \left(x\right)}{{\sin }^2\left(x\right)}$

$=\frac{1-\cos \left(x\right)}{1-{\cos }^2\left(x\right)}$

$=-\frac{1-\cos \left(x\right)}{(\cos \left(x\right)+1)(\cos \left(x\right)-1)}$

$=\frac{1}{\cos \left(x\right)+1}$

$=\frac{1}{2{\cos }^2\left(\frac{x}{2}\right)-1+1}$

$=\frac{1}{2}se{c}^2\frac{x}{2}$