If secθ + tanθ = √3 (0°≤ θ ≤ 90°), then tan 3θ is
secθ + tanθ = √3
$$sec 30^o = \frac{2}{\sqrt{3}}$$
$$tan30^o = \frac{1}{\sqrt{3}}$$
$$ sec 30^o + tan 30^o = \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \sqrt{3} $$
$$\theta = 30^o$$
$$tan 3\theta = tan3(30) = tan 90 = infinity$$
Option C is the correct answer
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