Q.

The value of the limit limθ→0tanπcos2θsin2πsin2θ is equal to:

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a

−12

b

14

c

−14

d

0

answer is A.

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Detailed Solution

limθ→0tanπcos2θsin2πsin2θ=limθ→0tanπcos2θsin2π−2πcos2θ=limθ→0tanπcos2θsin2π−2πcos2θ=−limθ→0tanπcos2θsin2πcos2θ=−limθ→0tanπcos2θ2sinπcos2θcosπcos2θ=−12limθ→0sec2πcos2θ=−12sec20=−12
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