If $\sqrt{3} cos \theta = sin \theta $, then the value of $\frac{4sin^2\theta -5cos\theta}{3cos\theta + 1}$ is :
Answer & explanation
Correct answer: option 2
Given:-
\(\sqrt { 3}\) cosθ = sinθ
tanθ = \(\sqrt { 3}\)
{ tan60º = \(\sqrt { 3}\) }
Now,
\(\frac{4sin²θ - 5cosθ }{3cosθ + 1 }\)
= \(\frac{4sin²60º - 5cos60º }{3cos60º + 1 }\)
= \(\frac{4 × 3/4 - 5 × 1/2 }{3× 1/2 + 1 }\)
= \(\frac{1 }{5 }\)