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Question Number 132079 by liberty last updated on 11/Feb/21
IfL=limx→π/4tan3x−tanxcos(x+π4)thenL4=?
Answered by EDWIN88 last updated on 11/Feb/21
L=limx→π/4tanx(tanx+1)(tanx−1)cos(x+π4)L=1×2×limx→π/4(sinx−cosxcosx).(1cos(x+π4))L=−1×4×limx→π/412cosx−12sinxcos(x+π4)L=−4×limx→π/4cos(x+π4)cos(x+π4)=−4thenL4=−1
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