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Question

If 7 sin2θ+3cos2θ=4, then sec θ+cosec θ is equal to


A

232

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B

23+2

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C

23

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D

None of these

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Solution

The correct option is B

23+2


7sin2θ+3cos2θ=47sin2θ+3(1sin2θ)=4[cos2θ+sin2θ=1]7sin2θ+33sin2θ=44sin2θ=1sin2θ=14sinθ=12θ=30

We have,
sec 30=23 and cosec 30=2sec 30+cosec 30=23+2


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