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Question 71 of 96

Q.The value of ∫0π/2tan⁡x−cot⁡x1+tan⁡xcot⁡x dx\displaystyle\int_{0}^{\pi/2} \dfrac{\tan x - \cot x}{1 + \tan x \cot x} \, dx is :

(a) π4\dfrac{\pi}{4}
(b) π\pi
(c) π2\dfrac{\pi}{2}
(d) 00
Puducherry TnboardTamil Nadu HSC (DGE) Board 2019MCQ· 1mImportance★★★★★
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Using f(π/2−x)=−f(x)f(\pi/2-x)=-f(x), the integral ∫0π/2tan⁡x−cot⁡x1+tan⁡xcot⁡x dx\displaystyle\int_0^{\pi/2}\dfrac{\tan x-\cot x}{1+\tan x\cot x}\,dx equals 00.

  1. Since tan⁡x⋅cot⁡x=1\tan x\cdot\cot x=1 (wherever both are defined), the denominator 1+tan⁡xcot⁡x=1+1=21+\tan x\cot x=1+1=2 throughout the interval.
  2. So the integrand simplifies to f(x)=tan⁡x−cot⁡x2f(x)=\dfrac{\tan x-\cot x}{2}.
  3. Apply the substitution x→π2−xx\to \dfrac{\pi}{2}-x: tan⁡(π2−x)=cot⁡x\tan\left(\dfrac{\pi}{2}-x\right)=\cot x and cot⁡(π2−x)=tan⁡x\cot\left(\dfrac{\pi}{2}-x\right)=\tan x.
  4. So f(π2−x)=cot⁡x−tan⁡x2=−f(x)f\left(\dfrac{\pi}{2}-x\right)=\dfrac{\cot x-\tan x}{2}=-f(x). …

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