#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 75 maths textbook solution

Answer: $\frac{2}{3}$

Hint: you must know the rules of solving derivatives of trigonometric function

Given: $f(x)=\sqrt{\frac{\sec x-1}{\sec x+1}}$

Find : $f^{\prime}\left(\frac{\pi}{3}\right)$

Solution:

$f(x)=\sqrt{\frac{\sec x-1}{\sec x+1}}$

$=\sqrt{\frac{1-\cos x}{1+\cos x}} \quad\left[\therefore \sec x=\frac{1}{\cos x}\right]$

Now rationalize

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

$f(x)=\frac{1-\cos x}{\sin x}$

$=\frac{1}{\sin x}-\frac{\cos x}{\sin x}$

$f(x)=\operatorname{cosec} x-\cot x$

Differentiate with respect to x,

\begin{aligned} &f^{\prime}(x)=-\operatorname{cosec} x \cot x-\left(-\operatorname{cosec}^{2} x\right)\\ \\ &f^{\prime}\left(\frac{\pi}{3}\right)=-\operatorname{cosec}\left(\frac{\pi}{3}\right) \cot \left(\frac{\pi}{3}\right)+\operatorname{cosec}^{2}\left(\frac{\pi}{3}\right) \end{aligned}

\begin{aligned} &=\frac{-2}{\sqrt{3}} \times \frac{1}{\sqrt{3}}+\left(\frac{2}{\sqrt{3}}\right)^{2} \\\\ &=\frac{-2}{3}+\frac{4}{3} \\\\ &\Rightarrow \frac{-2+4}{3} \Rightarrow \frac{2}{3} \end{aligned}