If tanA+tanB=a, cotA+cotB=b, prove that : cot(A+B) = (b-a)/ab.?

Answer 1

Remember that:

tan(x) = sin(x)/cos(x)
cot(x) = cos(x)/sin(x)

Also recall that:

cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)
sin(A+B) = cos(A)*sin(B) + sin(A)*cos(B)

So:

cot(A+B) = cos(A+B)/sin(A+B) = cos(A)*cos(B)/(cos(A)*sin(B) + sin(A)*cos(B)) - sin(A)*sin(B)/(cos(A)*sin(B) + sin(A)*cos(B))

and:

tan(A) + tan(B) = sin(A)/cos(A) + sin(B)/cos(B) = ( sin(A)*cos(B) + sin(B)*cos(A) )/( cos(A)*cos(B) ) = a

and:

cot(A) + cot(B) = cos(A)/sin(A) + cos(B)/sin(B) = ( cos(A)*sin(B) + cos(B)*sin(A) )/( sin(A)*sin(B) ) = b

However, you can see that the reciprocals of these two expressions fit nicely into the cot(A+B) expression above, so:

cot(A+B) = cos(A+B)/sin(A+B) = 1/a - 1/b = (b-a)/ab

That is what was to be shown.

Answer 2

cot(A+B) = (1-tanA.tanB)/(tanA+tanB)
(standard identity, tan(A+B) reciprocal)

= 1/(tanA+tanB)-
(tanA.tanB)/(tanA+tanB)

=1/a -1/((1/tanB)+1/(tanA))
{a given as tanA+tanB}
{dividing tanA.tanB/(tanA+tanB) by tanA.tanB}
=1/a -1/(cotB+cotA){1/tanx=cosx}
=1/a-1/b {given cosB+cosA=b}

=(b-a)/ab as required

i hope that this helps

Answer 3

cot(A+B) = ( cot A cot B - 1 ) / ( cot A + cot B )
= ( cot A cot B - 1 ) / b.

Now show cot A cot B - 1 = (b-a)/a = b/a - 1. That is, show that

cot A cot B = b/a.

b/a = (cot A + cot B) / (tan A + tan B)
= (cot A + cot B) / (1/cot A + 1/cot B)
= (cot A + cot B) / [ ( cot A + cot B ) / (cot A cot B) ]
= cot A cot B.

Answer 4

(b-a)/ab
=cotA+cotB-tanA-tanB/(tanA+tanB)(cotA+cotB)
=1/tanA+1/tanB-tanA-tanB/
(1+tanA/tanB+tanB/tanA+1)
tanA+tanB+-tan^2AtanB-tanAtan^2B
=2tanAtanB+tan^2A+tan^2B
tanA+tanB+tanAtanB(tanA+tanB)
=(tanA+tanB)^2
(tanA+tan)(1+tanAtanB)
=(tanA+tanB)^2
(1+tanAtanB)/(tanA+tanB)
=1/tan(A+B)=cot(A+B)
hence proved