Question

$\small{ \tt {If \: \: \: y \: = \sqrt{\dfrac{ \sec x - 1 }{\sec x + 1}}} }$
$\small{\tt {Prove \: \: \: that \: \: \: \dfrac{ dy}{dx} = \cosec x( \cosec x - \cot x)}}$
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Answer 1

Given,

$\longrightarrow \sf y = \sqrt{\dfrac{\sec x -1 }{\sec x + 1}}$

As we know that,

• sec x = 1/cos x

$\longrightarrow \sf y = \sqrt{\dfrac{\frac{1}{\cos x } -1 }{\frac{1}{\cos x } + 1}}$

$\longrightarrow \sf y = \sqrt{\dfrac{\dfrac{1-\cos x}{\cos x}}{\dfrac{1+\cos x }{\cos x } }}$

$\longrightarrow \sf y = \sqrt{\dfrac{1-\cos x}{1+\cos x }}$

$\longrightarrow \sf y = \sqrt{\dfrac{1-\cos x}{1+\cos x }\times \dfrac{1-\cos x }{1 - \cos x }}$

$\longrightarrow \sf y = \sqrt{\dfrac{(1-\cos x)^2}{(1-\cos ^2x) }}$

Apply formula, 1 - cos²x = sin² x

$\longrightarrow \sf y = \sqrt{\dfrac{(1-\cos x)^2}{sin^2x}}$

$\longrightarrow \sf y = \dfrac{(1-\cos x)}{sinx}$

$\longrightarrow \sf y = \dfrac{1}{\sin x}-\dfrac{\cos x }{\sin x}$

$\longrightarrow \sf y = cosec \:x -cot \: x$

Now,

$\sf\leadsto \dfrac{dy}{dx} = \dfrac{d}{dx}(cosec \:x - cot\: x)$

$\sf\leadsto \dfrac{dy}{dx} = \dfrac{d}{dx}(cosec \:x) - \dfrac{d}{dx}(cot\: x)$

We know that, derivative of cosec x is - cot x.cosec x and derivative of cot x is - cosec²x.

$\sf\leadsto \dfrac{dy}{dx} =- \cot x. cosec\: x - (-cosec^2x)$

$\sf\leadsto \dfrac{dy}{dx} =- \cot x. cosec\: x +cosec^2x$

$\sf\leadsto \dfrac{dy}{dx} = cosec \:x(cosec \:x - \cot x)$

Hence we have proved.

Formula used :-

• ( cosec x )' = - cot x . cosec x
• ( cot x )' = - cosec²x

Here [ ' represents the derivative ]

Answer 2

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