how do u solve tanAtanB=tanA+tanB/cotA+cotB?

Answer 1

tan(A)tan(B) = [tan(A) + tan(B)] / [cot(A) + cot(B)]

Choose the more complex side, which would be the right hand side.

RHS = [tan(A) + tan(B)] / [cot(A) + cot(B)]

Convert everything to sines and cosines.

RHS = [ sin(A)/cos(A) + sin(B)/cos(B) ] / [ cos(A)/sin(A) + cos(B)/sin(B) ]

To eliminate all fractions, multiply everything by sin(A)cos(A)sin(B)cos(B).

RHS = [ sin^2(A)sin(B)cos(B) + sin(A)cos(A)sin^2(B) ] /
[ cos^2(A)sin(B)cos(B) + sin(A)cos(A)cos^2(B) ]

Factor the numerator and denominator.

RHS = [ sin(A)sin(B) (sin(A)cos(B) + cos(A)sin(B)) ] /
[ cos(A)cos(B) (cos(A)sin(B) + sin(A)cos(B)) ]

Note that sin(A)cos(B) + cos(A)sin(B) is the same as
cos(A)sin(B) + sin(A)cos(B); after some manipulation, we can get them to be the same, because multiplication and addition are commutative, which means we can swap and switch. This means they cancel each other out, leaving us with

RHS = [sin(A)sin(B)] / [cos(A)cos(B)]

Splitting this up into two fractions,

RHS = [sin(A)/cos(A)] [sin(B)/cos(B)]

And, by definition

RHS = tan(A)tan(b) = LHS

Answer 2

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