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