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Question Number 54480 by 951172235v last updated on 04/Feb/19
showthattanα+tan(α+2Λ−5)+tan(α+4Λ−5)+tan(α+6Λ−5)+tan(α+8Λ−5)=5tan5α
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19
tanα+tan(α+72o)+tan(α+144o)+tan(α+216o)+tan(α+288o)tanα=atan72o=cot18o=5+25=btan1440=tan(180−36)=−tan36o=−5−25=−ctan(216)=tan(180+36)=tan36=5−25=ctan(288)=tan(3×90+18)=−cot18=−5+25=−b(5+25)(5−25)=5solettanα=a5+25=b5−25=cbc=5tanα+tanα+tan721−tanαtan72+tanα+tan1441−tanαtan144+tanα+tan2161−tanαtan216+tanα+tan2881−tanαtan288a+a+b1−ab+a−c1+ac+a+c1−ac+a−b1+aba+(a+b)(1+ab)+(a−b)(1−ab)1−a2b2+(a+c)(1+ac)+(a−c)(1−ac)1−a2c2a+a+b+a2b+ab2+a−b−a2b+ab21−a2b2+a+c+a2c+ac2+a−c−a2c+ac21−a2c2a+2a+2ab21−a2b2+2a+2ac21−a2c2a+2a(1+ab)(1+ab)(1−ab)+2a(1+ac)(1+ac)(1−ac)a+2a1−ab+2a1−aca+2a(1−ac+1−ab1−ab−ac+a2bc)a+2a(2−a(b+c)1−a(b+c)+a2bc)a−a2(b+c)+a3bc+4a−2a2(b+c)1−ab−ac+a2bc5a−3a2(b+c)+a3bc1−ab−ac+a2bcwait...
Answered by 951172235v last updated on 05/Feb/19
tan(α+8Λ−5)=tan(2Λ−−2Λ−5+α)=tan(α−2Λ−5)tan(α+2Λ−5)+tan(α+8Λ−5)=x+y1−xy+x−y1+xywheretanα=xtan2Λ−5=y=2x(1+y2)1−x2y2=2x(5+1)21−(5+25)x2similarlytan(α+4Λ−5)+tan(α+6Λ−5)=2x(5−1)21−(5−25)x2L.H.S=x+2x[(5−1)2{1−(5+25)x2+(5+1)2{1−(5−25)x2}1−10x2+5x4=x+2x(12−(5×12−8×5)x21−10x2+5x4)=5(5x−10x3+x5)1−10x2+5x4=5tan5αans.
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