If $7 \cos ^2 \theta+3 \sin ^2 \theta=6,0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\cot ^2 2 \theta+\sec ^2 2 \theta}{\tan ^2 2 \theta-\sin ^2 2 \theta}$ is:
Answer & explanation
Correct answer: option 1
7 cos²θ + 3 sin²θ = 6
4cos²θ + 3cos²θ + 3 sin²θ = 6
{ sin²θ + cos²θ = 1 }
4cos²θ = 3
cosθ= \(\frac{ √3}{2}\)
{ we know, cos30º= \(\frac{ √3}{2}\) }
So, θ = 30º
Now,
\(\frac{ cot²2θ + sec²2θ}{tan²2θ- sin²2θ}\)
= \(\frac{ cot²60º + sec²60º}{tan²60º- sin²60º}\)
= \(\frac{ 1/3 + 4}{3- 3/4}\)
= \(\frac{ 13/3}{9/4}\)
= \(\frac{ 52}{27}\)