if A+B+C=180° prove that cotAcotB+cotBcotC+cotAcotA=1
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Hey!!!
A + B + C = 180°
A + B = 180°-C
Cot(A+B ) = cot( 180°- C)
As we know that
cot( A + B ) = cotA ×cotB -1/ cotA + cotB
CotA ×cotB - 1/ CotA + CotB = -cotC
= CotA × CotB - 1 = -CotA×CotC - CotB×CotC
=> CotA×cotB +CotB×Cot+CotA×CotC = 1
prooved
Hope it helps !!!
#Rajukumar111
A + B + C = 180°
A + B = 180°-C
Cot(A+B ) = cot( 180°- C)
As we know that
cot( A + B ) = cotA ×cotB -1/ cotA + cotB
CotA ×cotB - 1/ CotA + CotB = -cotC
= CotA × CotB - 1 = -CotA×CotC - CotB×CotC
=> CotA×cotB +CotB×Cot+CotA×CotC = 1
prooved
Hope it helps !!!
#Rajukumar111
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