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If tanθ + secθ = 3, θ being acute, the value of 5 sinθ is :
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5 2 -
√3 5 -
5 √3 - 4
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Correct Option: D
tanθ + secθ = 3 ......(i)
∵ sec² θ – tan² θ = 1
⇒ (secθ – tanθ) (secθ + tanθ) = 1
| ⇒ secθ – tanθ = | .... (ii) | 3 |
On adding equations (i) and (ii),
| 2 secθ = 3 + | = | 3 | 3 |
On subtracting equation (ii) from (i),
| 2tanθ = 3 - | 3 |
| = | = | 3 | 3 |
| ∴ sin θ = | sec θ |
| = | × | = | 3 | 10 | 5 |
| ∴ 5 sinθ = 5 × | = 4 | 5 |
