Question

4-2sin^2-5cos =0,sin+tan

Answer 1

Answer:

3√3/2.

Step-by-step explanation:

4-2sin²θ - 5cosθ =0

=> 4 - 2(1 - Cos²θ) - 5Cosθ  = 0

=> 2Cos²θ  - 5Cosθ  + 2 = 0

=> 2Cos²θ - 4Cosθ - Cosθ + 2 = 0

=> 2Cosθ(Cosθ - 2) - 1(Cosθ - 2) = 0

=> (2Cosθ - 1)(Cosθ - 2) = 0

Since Cosθ cannot be greater than 1, Cosθ ≠2)

=> Cosθ = 1/2

=> θ = π/3.

Now Sinθ + Tanθ

= Sinπ/3 + Tanπ/3  ( Substitute θ = π/3 or 60°)

= √3/2 + √3  (∵ Sin 60° = √3/2 and Tan60° = √3)

= 3√3/2.