If $\frac{sinθ+cosθ}{sinθ-cosθ}= 3$ and θ is an acute angle, then the value of $\frac{3sinθ+4cosθ}{8cosθ-3sinθ}$ is :

5

Given :-

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

Divide LHS by cosθ

\(\frac{tanθ  + 1}{tanθ  - 1}\) = 3

tanθ  + 1 = 3tanθ  - 3

tanθ = 2

So , P = 2 & B = 1

P² + B² = H²

2² + 1² = H²

H = √5

Now,

 \(\frac{3sinθ  + 4cosθ}{8cosθ  - 3sinθ}\)

=  \(\frac{3× 2/√5  + 4× 1/√5}{8× 1/√5  - 3× 2/√5}\)

= \(\frac{6  + 4}{8  - 6)

= 5