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Question

If 7 sin2θ+3cos2θ=4, then secθ+cosecθ is equal to


A

232

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B

23+63

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C

23

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D

3

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Solution

The correct option is B

23+63


7sin2θ+3cos2θ=47sin2θ+3(1sin2θ)=47sin2θ+33sin2θ=44sin2θ=1sin2θ=14sinθ=12θ=30sec30+cosec30=(23)+2

= 2+233
=2(1+3)3
Multiplying numerator and denominator by 3 , we get
= 23+63


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