- 
					 If 5 sin2θ + 4cos2θ = 9 and 0 < θ < π then tanθ is equal to 2 2 
- 
                        - 1
- 0
- –1
-  1 4 
 
Correct Option: A
| 5 sin2θ + 4cos2θ = | |
| 2 | 
⇒ 10 sin2θ + 8 cos2θ = 9
On dividing by cos2θ
| + | = | = 9 sec2θ | |||
| cos2θ | cos2θ | cos2θ | 
⇒ 10 tan2θ + 8 = 9 (1 + tan2θ)
⇒ 10 tan2θ + 8 = 9 + 9 tan2θ
⇒ 10 tan2θ – 9 tan2θ = 9 – 8
⇒ tan2θ = 1 ⇒ tan θ = ± 1
| ∵ 0 < θ < | , ∴ tanθ = 1 | |
| 2 | 
 
	