If $4 - 2 \sin^2 \theta - 5 \cos \theta = 0, 0^\circ < \theta < 90^\circ$, then the value of $\cos \theta - \tan \theta$ is:
Answer & explanation
Correct answer: option 4
We are given that ,
4 - 2 sin²θ - 5 cosθ= 0
{ using sin²θ + cos²θ = 1 }
4 - 2 ( 1 - cos²θ ) - 5 cosθ= 0
2cos²θ - 5 cosθ + 2 = 0
2cos²θ - 4 cosθ - cosθ + 2 = 0
2 cosθ ( cosθ - 2 ) - 1 ( cosθ - 2 ) = 0
Either ( 2cosθ - 1 ) = 0 Or ( cosθ - 2 ) = 0
( cosθ - 2 ) = 0 is not possible.
So, ( 2cosθ - 1 ) = 0
cosθ = \(\frac{1}{2}\)
{ we know , cos60º = \(\frac{1}{2}\) }
So, θ = 60º
Now,
cosθ - tanθ
= cos60º - tan60º
= \(\frac{1}{2}\) - \(\sqrt { 3}\)
= \(\frac{1 - 2√3}{2}\)