Q84 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
The expression $\dfrac{\tan\left(x - \dfrac{\pi}{2}\right)\cos\left(\dfrac{3\pi}{2} + x\right) - \sin^3\left(\dfrac{7\pi}{2} - x\right)}{\cos\left(x - \dfrac{\pi}{2}\right)\tan\left(\dfrac{3\pi}{2} + x\right)}$ simplifies to:
- Correct answerA$\sin^2 x$
- B$1 + \cos^2 x$
- C$\cos^2 x - \sin^2 x$
- D$-(1 + \cos^2 x)$
Solution
We simplify each term in the given expression using trigonometric identities:
$\tan\left(x - \dfrac{\pi}{2}\right) = -\tan\left(\dfrac{\pi}{2} - x\right) = -\cot x$
$\cos\left(\dfrac{3\pi}{2} + x\right) = \sin x$
$\sin\left(\dfrac{7\pi}{2} - x\right) = \sin\left(3\pi + \dfrac{\pi}{2} - x\right) = -\sin\left(\dfrac{\pi}{2} - x\right) = -\cos x$
$\cos\left(x - \dfrac{\pi}{2}\right) = \cos\left(\dfrac{\pi}{2} - x\right) = \sin x$
$\tan\left(\dfrac{3\pi}{2} + x\right) = -\cot x$
Substituting these values into the given expression:
Numerator $= (-\cot x)(\sin x) - (-\cos x)^3 = -\cos x + \cos^3 x$
Denominator $= (\sin x)(-\cot x) = -\cos x$
The expression becomes:
$\dfrac{-\cos x + \cos^3 x}{-\cos x} = 1 - \cos^2 x = \sin^2 x$
Answer: $\sin^2 x$
$\tan\left(x - \dfrac{\pi}{2}\right) = -\tan\left(\dfrac{\pi}{2} - x\right) = -\cot x$
$\cos\left(\dfrac{3\pi}{2} + x\right) = \sin x$
$\sin\left(\dfrac{7\pi}{2} - x\right) = \sin\left(3\pi + \dfrac{\pi}{2} - x\right) = -\sin\left(\dfrac{\pi}{2} - x\right) = -\cos x$
$\cos\left(x - \dfrac{\pi}{2}\right) = \cos\left(\dfrac{\pi}{2} - x\right) = \sin x$
$\tan\left(\dfrac{3\pi}{2} + x\right) = -\cot x$
Substituting these values into the given expression:
Numerator $= (-\cot x)(\sin x) - (-\cos x)^3 = -\cos x + \cos^3 x$
Denominator $= (\sin x)(-\cot x) = -\cos x$
The expression becomes:
$\dfrac{-\cos x + \cos^3 x}{-\cos x} = 1 - \cos^2 x = \sin^2 x$
Answer: $\sin^2 x$
Hence, the correct answer is option: 'A'.