If $\sqrt{13} sin \theta = 2, $ then the value of $\frac{3tanθ+\sqrt{13}sinθ}{\sqrt{13}cosθ-3 tan θ}$ is :

4

\(\sqrt {13 }\) sinθ = 2

sinθ = \(\frac{2}{√13}\)

{ sinθ = \(\frac{P}{H}\) }

By using pythagoras theorem,

P² + B² = H²

4 + B² = 13

B = 3

Now,

\(\frac{3tanθ + √13sinθ }{√13cosθ - 3tanθ}\)

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

= \(\frac{4}{1}\)

= 4