frac{d}{dx}sqrt{sec^2 x + text{cosec}^2 x} = | vedclass

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Question

(B) Let $y = \sqrt{\sec^2 x + 	ext{cosec}^2 x}$.First,simplify the expression inside the square root:$y = \sqrt{\frac{1}{\cos^2 x} + \frac{1}{\sin^2

Vedclass · Accepted Answer

$-4 	ext{cosec } 2x \cdot \cot 2x$

Vedclass · Answer

(B) Let $y = \sqrt{\sec^2 x + 	ext{cosec}^2 x}$.First,simplify the expression inside the square root:$y = \sqrt{\frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}} = \sqrt{\frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x}} = \sqrt{\frac{1}{\sin^2 x \cos^2 x}}$Using the identity $\sin 2x = 2 \sin x \cos x$,we have $\sin x \cos x = \frac{\sin 2x}{2}$.Thus,$y = \sqrt{\frac{1}{(\frac{\sin 2x}{2})^2}} = \sqrt{\frac{4}{\sin^2 2x}} = \frac{2}{\sin 2x} = 2 	ext{cosec } 2x$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(2 	ext{cosec } 2x) = 2 \cdot (-	ext{cosec } 2x \cot 2x) \cdot \frac{d}{dx}(2x)$$\frac{dy}{dx} = 2 \cdot (-	ext{cosec } 2x \cot 2x) \cdot 2 = -4 	ext{cosec } 2x \cot 2x$.

$\frac{d}{dx}\sqrt{\sec^2 x + \text{cosec}^2 x} = $

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