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Miscellaneous Exercise 2 · Q58

Q.If θ = 60°, then 1+tan⁡2θ2tan⁡θ\dfrac{1+\tan^2\theta}{2\tan\theta} is equal to: A) 23\dfrac{2}{\sqrt3} B) 32\dfrac{\sqrt3}{2} C) 13\dfrac{1}{\sqrt3} D) 3\sqrt3

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Step 1. 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta, so the expression is sec⁡2θ2tan⁡θ\dfrac{\sec^2\theta}{2\tan\theta}.

Step 2. At θ=60°\theta=60°: tan⁡60°=3\tan60°=\sqrt3, sec⁡260°=1+3=4\sec^260°=1+3=4. …

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