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Question

If (cos2x+1cos2x)(1+tan22y)(3+sin3z)=4,then

A
x may be a multiple of π
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B
x can not be an even multiple of π
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C
z can be a multiple of π
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D
y can be a multiple of π/2
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Solution

The correct option is D y can be a multiple of π/2

minvalueof(cos2x+(1cos2x))=2minvalueof1+tan22y=1andminvalueof3+sin3z=31=2Hencexmaybeamultipleofπbecauseminimumvaluewilloccuronlywhencos2x=(1cos2x)=1ycanbemultipleof(π2)suchthattan(2y)=0


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