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If sec θ + tan θ = &radic3 (0° ≤ θ ≤ 90°), then tan 3θ is
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- undefined
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1 √3 -
1 √2
- 3
Correct Option: A
sec θ + tan θ = √3.....(i)
∵ sec² θ - tan² θ = 1
⇒ (secθ - tanθ )(secθ + tanθ ) = 1
| ⇒ secθ - tanθ = | ....(ii) | √3 |
By subtracting (ii) from (i)
secθ + tanθ - secθ + tanθ
| = √3 - | √3 |
| ⇒ 2tan θ = | √3 |
| = tan θ = | = tan 30° | √3 |
⇒ θ = 30°
∴ tan3 θ =tan 90° = undefined
