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