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Question Number 82176 by oyemi kemewari last updated on 19/Feb/20
Commented by jagoll last updated on 19/Feb/20
(1b)32−2135=27+5−227×5=27−5=33−5
(2b)α×(α+3)=28α2+3α−28=0(α+7)(α−4)=0(i)k=2α+3=−11(ii)k=2α+3=11
Commented by john santu last updated on 19/Feb/20
cosx1+sinx+1+sinxcosx=cos2x+1+2sinx+sin2xcosx(1+sinx)=?2(1+sinx)cosx(1+sinx)=2cosx=2secx
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