If $\frac{secθ+tanθ}{secθ-tanθ}= 5$ and θ is an acute angle, then the value of $\frac{3cos^2θ+1}{3cos^2θ-1}$ is:
Answer & explanation
Correct answer: option 4
Given :-
\(\frac{secθ + tanθ }{secθ - tanθ }\) = 5
Now,
5 = \(\frac{secθ + tanθ }{secθ - tanθ }\)
5 = \(\frac{1 + sinθ }{1 - sinθ }\)
5 - 5sinθ = 1 + sinθ
sinθ = \(\frac{2 }{3}\)
Now,
sinθ = \(\frac{P }{H}\)
P² + B² = H²
2² + B² = 3²
B = √5
cosθ = \(\frac{√5 }{3}\)
Now, \(\frac{3 cos²θ + 1 }{3 cos²θ - 1}\)
= \(\frac{3 × 5/9 + 1 }{3 × 5/9 - 1}\)
= \(\frac{ 5 + 3 }{5 - 3}\)
= 4