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Question: Answered & Verified by Expert
The value of limxπtanπcos2xsin22x is equal to
MathematicsLimitsJEE Main
Options:
  • A 1
  • B π
  • C -π4
  • D π2
Solution:
2390 Upvotes Verified Answer
The correct answer is: -π4
Let, x=π+t
limt0tanπcos2π+tsin22π+2t=limt0tanπcos2tsin22t
=limt0tanπ-πsin2tsin22t=limt0-tanπsin2t4πsin2tcos2t·π=-π4

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