If 3 tan θ = 2, what is the value of (3sinθ -cosθ)/(3sinθ +cosθ) ?

3 tan θ = 2

tan θ = \(\frac{2}{3}\)

{ we know,  tan A = \(\frac{P}{B}\) }

By using pythagoras theorem ,

P² + B² = H²

2² + 3² = H²

H = √13

Now,

\(\frac{3 sinθ - cosθ }{3 sinθ + cosθ}\)

= \(\frac{3× 2/√13 - 3/√13 }{3 × 2/√13 + 3/√13 }\)

= \(\frac{3 }{9 }\)

= \(\frac{1}{3}\)