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Question

If y=2sin2θ+tanθ, then dydθ will be

A
4sinθcosθ+secθtanθ
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B
2sin2θ+sec2θ
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C
4sinθ+sec2θ
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D
2cos2θ+sec2θ
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Solution

The correct option is B 2sin2θ+sec2θ
y=2sin2θ+tanθdydθ=2×2sinθcosθ+sec2θ =2sin2θ+sec2θ

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