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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 61 of 72
Marks: +1, -0
sec⁡x−1sec⁡x+1\frac{\sec x - 1}{\sec x + 1}
Solution:  
Let f (x) = sec⁡x−1sec⁡x+1\frac{\sec x - 1}{\sec x + 1}
⇒ f (x) = 1cos⁡x−11cos⁡x+1\frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1}
⇒ f (x) = 1−cos⁡x1+cos⁡x\frac{1 - \cos x}{1 + \cos x} ... (i)
Differentiating (i) with respect to x, we get
ddx\frac{d}{dx} (f (x)) =
(1+cos⁡x)(1−cos⁡x)′−(1−cos⁡x)(1+cos⁡x)′(1+cos⁡x)2\frac{(1 + \cos x)(1 - \cos x)' - (1 - \cos x)(1 + \cos x)'}{(1 + \cos x)^2}
=
(1+cos⁡x)sin⁡x−(1−cos⁡x)(−sin⁡x)(1+cos⁡x)2\frac{(1 + \cos x)\sin x - (1 - \cos x)(-\sin x)}{(1 + \cos x)^2}
=
sin⁡x+sin⁡xcos⁡x+sin⁡x−sin⁡xcos⁡x(1+cos⁡x)2\frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{(1 + \cos x)^2}
= 2sin⁡x(1+cos⁡x)2\frac{2\sin x}{(1 + \cos x)^2}
= 2⋅1csc⁡x(1+1sec⁡x)2\frac{2 \cdot \frac{1}{\csc x}}{\left(1 + \frac{1}{\sec x}\right)^2}
= 2csc⁡x(sec⁡x+1)2sec⁡2x\frac{\frac{2}{\csc x}}{\frac{(\sec x + 1)^2}{\sec^2 x}}
= 2csc⁡x⋅sec⁡2x(sec⁡x+1)2\frac{2}{\csc x} \cdot \frac{\sec^2 x}{(\sec x + 1)^2}
= 2sin⁡x⋅1cos⁡x⋅sec⁡x(sec⁡x+1)2\frac{2 \sin x \cdot \frac{1}{\cos x} \cdot \sec x}{(\sec x + 1)^2}
= 2tan⁡x⋅sec⁡x(sec⁡x+1)2\frac{2 \tan x \cdot \sec x}{(\sec x + 1)^2}
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