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Questions  

The equation tan2x+cot2x=4acosec22x has a real solution if 

a
0
b
1/2≤a≤1
c
1/4≤a≤1/2
d
−1≤a≤1

detailed solution

Correct option is B

a=sin2⁡x+cos2⁡x2−2sin2⁡xcos2⁡x⇒sin2⁡2x=−2(a−1)⇒1−cos⁡4x=−4(a−1)⇒cos⁡4x−4a−3so−1≤4a−3≤1⇒1/2≤a≤1

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