Question

cos(90°-θ) sin (90°-θ)=

• A. \(\frac{tanθ}{1-tan^2θ}\)
• B. \(\frac{tanθ}{1+tan^2θ}\)
• C. 1
• D. 0

Answer 1

The Correct Option is B

We are tasked with simplifying \( \cos(90^\circ - \theta)\sin(90^\circ - \theta) \).

Step 1: Use trigonometric identities for complementary angles.

\[ \cos(90^\circ - \theta) = \sin\theta \quad \text{and} \quad \sin(90^\circ - \theta) = \cos\theta \]

Substituting these into the expression:

\[ \cos(90^\circ - \theta)\sin(90^\circ - \theta) = \sin\theta \cos\theta \]

Step 2: Simplify \( \sin\theta \cos\theta \) using a double-angle identity.

The double-angle identity for sine is:

\[ \sin(2\theta) = 2\sin\theta \cos\theta \]

Rearranging, we get:

\[ \sin\theta \cos\theta = \frac{\sin(2\theta)}{2} \]

Step 3: Express \( \sin(2\theta) \) in terms of \( \tan\theta \).

The identity for \( \sin(2\theta) \) in terms of \( \tan\theta \) is:

\[ \sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta} \]

Substitute this into \( \sin\theta \cos\theta = \frac{\sin(2\theta)}{2} \):

\[ \sin\theta \cos\theta = \frac{1}{2} \cdot \frac{2\tan\theta}{1 + \tan^2\theta} \]

Simplify:

\[ \sin\theta \cos\theta = \frac{\tan\theta}{1 + \tan^2\theta} \]

Final Answer: The value of \( \cos(90^\circ - \theta)\sin(90^\circ - \theta) \) is \( \mathbf{\frac{\tan\theta}{1 + \tan^2\theta}} \), which corresponds to option \( \mathbf{(2)} \).