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If 5 sin2θ + 4cos2θ = 9 and 0 < θ < π then tanθ is equal to 2 2
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- 1
- 0
- –1
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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 |