prove that tan(A-B)= (1-cot A tanB)/ cot A + tan B?

Answer 1

RHS = N / D where:-
N = 1 - (cos A / sin A)(sin B / cos B)
N = (sin A cos B - cos A sin B) / (sin A cos B)
N = sin (A - B) / (sin A cos B)

D = cot A + tan B
D = cos A / sin A + sin B / cos B
D = (cos A cos B + sin A sin B) / (sin A cos B)
D = cos (A - B) / (sin A cos B)

N / D = sin (A - B) cos (A - B)
N / D = tan (A - B)

Thus RHS = LHS as required

Answer 2

Left hand side of the equation is:
tan(a-b) = [tan(a) - tan(b)]/[1 + tan(a) tan(b)] --- (1)

The right hand side should be (1-cot A tanB)/(cot A + tan B) and not (1-cot A tanB)/ cot A + tan B.

Right hand side of the equation is:
(1-cot A tanB)/(cot A + tan B)
(1-tanB/tanA)/(1/tanA + tanB)
(1-tanB/tanA)/[(1 + tanA tanB)/tanA]
tanA[(tanA - tanB)/tanA]/(1 + tanA tanB)
(tanA - tanB)/(1 + tanA tanB) --- (2)

Left hand side = right hand side --- refer (1) and (2). Hence proved.

Answer 3

tan(A-B)
= [tan A - tan B]/[1+tan A tan B]
= [(tan A - tan B)(cot A)]/[(1+tan A tan B)(cot A)]
= (1-cot A tanB)/(cot A + tan B)
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Ideas: Multiply by cot A / cot A

Answer 4

tan(A-B)= tanA-tanB/(1+tanAtanB)QED
=1-cotAtanB/(cotA+tanB), where we have divided the numerator and denominator by tanA and used i/tanA=cotA.