Question

For \( \alpha, \beta \in \left( 0, \frac{\pi}{2} \right) \), let \( 3 \sin(\alpha + \beta) = 2 \sin(\alpha - \beta) \) and a real number \( k \) be such that \( \tan \alpha = k \tan \beta \). Then the value of \( k \) is equal to:

• A. \( -\frac{2}{3} \)
• B. –5
• C. \( \frac{2}{3} \)
• D. 5

Answer 1

The Correct Option is B

To solve this problem, we have the trigonometric identities involving angles \( \alpha \) and \( \beta \) in the interval \( (0, \frac{\pi}{2}) \).

We are given two equations:

1. \( 3 \sin(\alpha + \beta) = 2 \sin(\alpha - \beta) \)
2. \( \tan \alpha = k \tan \beta \)

Let's start by expanding the trigonometric identities:

• \(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\)
• \(\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\)

Using these, let's rewrite the first equation:

\(3(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2(\sin \alpha \cos \beta - \cos \alpha \sin \beta)\)

Expanding and simplifying:

\(3 \sin \alpha \cos \beta + 3 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta - 2 \cos \alpha \sin \beta\)

Rearranging the terms:

\(3 \sin \alpha \cos \beta - 2 \sin \alpha \cos \beta = -3 \cos \alpha \sin \beta - 2 \cos \alpha \sin \beta\)

This leads to:

\(\sin \alpha \cos \beta = -5 \cos \alpha \sin \beta\)

Divide both sides by \(\cos \alpha \cos \beta\):

\(\tan \alpha = -5 \tan \beta\)

Comparing this with the given second equation \(\tan \alpha = k \tan \beta\), we find the value of \(k\):

\(k = -5\)

Hence, the value of \(k\) is –5, which matches the correct option.

Answer 2

Given that

\(3 \sin(\alpha + \beta) = 2 \sin(\alpha - \beta)\)

and a real number \(k\) such that \(\tan \alpha = k \tan \beta\).

Expanding \(\sin(\alpha + \beta)\) and \(\sin(\alpha - \beta)\) using trigonometric identities:

\(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta,\)

\(\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta.\)

Substituting these into the given equation:

\(3(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2(\sin \alpha \cos \beta - \cos \alpha \sin \beta).\)

Expanding both sides:

\(3 \sin \alpha \cos \beta + 3 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta - 2 \cos \alpha \sin \beta.\)

Collecting terms involving \(\sin \alpha \cos \beta\) and \(\cos \alpha \sin \beta\), we get:

\(3 \sin \alpha \cos \beta - 2 \sin \alpha \cos \beta = -2 \cos \alpha \sin \beta - 3 \cos \alpha \sin \beta.\)

Simplifying this:

\(\sin \alpha \cos \beta = -5 \cos \alpha \sin \beta.\)

Dividing both sides by \(\cos \alpha \cos \beta\) (assuming \(\cos \alpha \cos \beta \neq 0\)), we get:

\(\tan \alpha = -5 \tan \beta.\)

Therefore, the value of \(k\) is:

\(k = -5.\)