Question

If msinθ=nsin(θ+2α), then tan(θ+α) is equal to:

• A. \( \dfrac{m+n}{m-n}\tan\alpha \)
• B. \( \dfrac{m+n}{m-n}\tan\theta \)
• C. \( \dfrac{m+n}{m-n}\cot\alpha \)
• D. (m+n)/(m-n)cotθ

Hint:

Trigonometric equations often simplify after converting to tan.

Answer 1

The Correct Option is A

Step 1: Expand: sin(θ+2α)=sinθ\cos2α+cosθ\sin2α.
Step 2: Substitute into the equation and rearrange: (m-n\cos2α)sinθ=n\sin2αcosθ.
Step 3: Divide by cosθ: tanθ=(n\sin2α)/(m-n\cos2α).
Step 4: Using identities, obtain: tan(θ+α)=(m+n)/(m-n)tanα.