If $\sqrt{13} sin \theta = 2, $ then the value of $\frac{3tanθ+\sqrt{13}sinθ}{\sqrt{13}cosθ-3 tan θ}$ is :
Answer & explanation
Correct answer: option 2
\(\sqrt {13 }\) 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