Question

If \(\alpha\) and \(\beta\) are acute angles and \[ \cos\alpha(1+\tan\alpha\tan\beta)=1, \] then \[ \sin\left(\frac{\alpha-2\beta}{3}\right)= \]

• A. \(\frac12\)
• B. \(\frac{\sqrt3}{2}\)
• C. \(\frac35\)
• D. \(0\)

Hint:

Whenever an equation reduces to \(\cos A=\cos B\), use angle restrictions before writing the general solution.

Answer 1

The Correct Option is D

Step 1: Simplify the given equation.
\[ \cos\alpha \left( 1+\frac{\sin\alpha\sin\beta} {\cos\alpha\cos\beta} \right)=1 \] \[ \cos\alpha+ \frac{\sin\alpha\sin\beta}{\cos\beta} =1. \] Multiplying by \(\cos\beta\), \[ \cos\alpha\cos\beta+\sin\alpha\sin\beta = \cos\beta. \] \[ \cos(\alpha-\beta) = \cos\beta. \]

Step 2: Use acute angle condition.
Since \(\alpha,\beta\) are acute, \[ \alpha-\beta=\beta. \] Hence \[ \alpha=2\beta. \]

Step 3: Evaluate required expression.
\[ \sin \left( \frac{\alpha-2\beta}{3} \right) = \sin0 = 0. \] \[ \boxed{0} \]