Q67 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics

The expression $\dfrac{1 - \tan^2\left(\dfrac{\pi}{4} - A\right)}{1 + \tan^2\left(\dfrac{\pi}{4} - A\right)}$ equals

  • A
    $\cos 2A$
  • B
    $\cos A$
  • Correct answer
    C
    $\sin 2A$
  • D
    $\sin A$

Solution

Using the identity $\cos 2\theta = \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta}$

Let $\theta = \dfrac{\pi}{4} - A$

The given expression becomes $\cos\left(2\left(\dfrac{\pi}{4} - A\right)\right)$

$\cos\left(\dfrac{\pi}{2} - 2A\right)$

Using $\cos\left(\dfrac{\pi}{2} - x\right) = \sin x$, we get $\sin 2A$

Answer: $\sin 2A$

Hence, the correct answer is option: 'C'.

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