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:

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