If $5 \cos ^2 \theta+1=3 \sin ^2 \theta, 0^{\circ}<\theta<90^{\circ}$, then what is the value of $\frac{\tan \theta+\sec \theta}{\cot \theta+{cosec} \theta}$.
Answer & explanation
Correct answer: option 4
5 cos²θ + 1 = 3 sin²θ
{ sin²θ + cos²θ = 1 }
5(1 - sin²θ) + 1 = 3 sin²θ
5 - 5sin²θ) + 1 = 3 sin²θ
8sin²θ = 6
sinθ = \(\frac{√3}{2}\)
{ we know, sin60º = \(\frac{√3}{2}\) }
So, θ = 60º
Now,
\(\frac{ tanθ + secθ }{cotθ + cosecθ}\)
= \(\frac{ tan 60º + sec 60º }{cot 60º + cosec 60º}\)
= \(\frac{ √3 + 2 }{1/√3 + 2/√3}\)
= \(\frac{ 3 + 2√3 }{3}\)