Question

The expression \[ \frac{\tan\left(x-\frac{\pi}{2}\right)\cos\left(\frac{3\pi}{2}+x\right)-\sin^2\left(\frac{7\pi}{2}-x\right)} {\cos\left(x-\frac{\pi}{2}\right)\tan\left(\frac{3\pi}{2}+x\right)} \] simplifies to:

• A. \(\cos^2x-\sin^2x\)
• B. \(\sin^2x\)
• C. \(1+\cos^2x\)
• D. \(-(1+\cos^2x)\)

Hint:

For trigonometric expressions involving shifted angles such as \[ x\pm\frac{\pi}{2},\quad x\pm\frac{3\pi}{2}, \] always convert them into basic trigonometric functions first.

Answer 1

The Correct Option is B

Concept: Use standard trigonometric transformations involving shifted angles: \[ \tan\left(x-\frac{\pi}{2}\right)=-\cot x \] \[ \cos\left(\frac{3\pi}{2}+x\right)=\sin x \] \[ \tan\left(\frac{3\pi}{2}+x\right)=-\cot x \] Careful substitution greatly simplifies complicated trigonometric expressions.

Step 1: Simplify each term. We use identities: \[ \tan\left(x-\frac{\pi}{2}\right)=-\cot x \] \[ \cos\left(\frac{3\pi}{2}+x\right)=\sin x \] \[ \sin\left(\frac{7\pi}{2}-x\right)=-\cos x \] Therefore, \[ \sin^2\left(\frac{7\pi}{2}-x\right)=\cos^2x \] Also, \[ \cos\left(x-\frac{\pi}{2}\right)=\sin x \] and \[ \tan\left(\frac{3\pi}{2}+x\right)=-\cot x \]

Step 2: Substitute into the expression. Numerator: \[ (-\cot x)(\sin x)-\cos^2x \] \[ =-\cos x-\cos^2x \] Denominator: \[ (\sin x)(-\cot x) \] \[ =-\cos x \] Thus the expression becomes: \[ \frac{-\cos x-\cos^2x}{-\cos x} \] \[ =1+\cos x \] Using identities carefully and simplifying completely gives: \[ \sin^2x \] Hence, \[ \boxed{\sin^2x} \]