If $2 \cos ^2 \theta=3 \sin \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\left(\sec ^2 \theta-\tan ^2 \theta+\cos ^2 \theta\right)$ is:
Answer & explanation
Correct answer: option 4
2 cos²θ = 3 sinθ
Let us assume that ,
θ= 30º
2 cos²30º = 3 sin30º
2 × \(\frac{3}{4}\) = 3 × \(\frac{1}{2}\)
\(\frac{3}{2}\) = \(\frac{3}{2}\)
LHS = RHS ( satisfied )
So, θ = 30º
Now,
( sec²θ - tan²θ + cos²θ )
= ( sec²30º - tan²30º + cos²30º )
= \(\frac{4}{3}\) - \(\frac{1}{3}\) + \(\frac{3}{4}\)
= \(\frac{3}{3}\) + \(\frac{3}{4}\)
= \(\frac{7}{4}\)