1
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Question

# The value of limx→π2tan2x(√2sin2x+3sinx+4−√sin2x+6sinx+2) is equal to:

A
110
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B
111
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C
112
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D
18
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Solution

## The correct option is A 112Let L=limx→π2tan2x(√2sin2x+3sinx+4−√sin2x+6sinx+2)Rationalizing, we getL=limx→π2sin2x−3sinx+2cot2x(√2sin2x+3sinx+4+√sin2x+6sinx+2)Substituting limit in denominator ;L=limx→π216sin2x−3sinx+2cot2xLimit is in the form 0/0 ;applying L'Hopitals rule we get ;L=limx→π2162sinxcosx−3cosx−2cotxcosec2xCancelling cosx and rewriting the limit ;L=limx→π2162sinx−3−2(sin3x)Substituting x=π2 in limit, we get ;L=112

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