How to prove cot (A + B) = (cot A × cot B - 1) / (cot A + cot B)?

We will make use of trigonometric identities to prove this result.

Answer: Cot(A+B) =(cotAcotB-1) / (cotA+cotB) can be proved using trigonometric conversions from a cot function into cosine and sine function.

Explanation:

Given below is the step-by-step proof of the above-mentioned problem.

Consider LHS, cot (A + B) = cos (A + B) / sin (A + B) ---- (1)

using the sine and cosine addition formula  
cos(A+B) = cosAcosB-sinA sinB and
sin(A+B) = sinAcosB +cosAsinB

substituting in equation (1) we have

cot (A + B) =  (cos A × cos B − sin A × sin B) / (sin A × cos B + cos A × sin B)

Now divide the terms on both numerator and denominator by sin A × sin B. we get

[  (cos A × cos B − sin A × sin B) / (sin A × cos B + cos A × sin B) ]  ×  [sin A × sin B / sin A × sin B]

=   [cos A × cos B/ sin A × sin B  -  sin A × sin B / sin A × sin B] / [sin A × cos B/ [sin A × sin B + cos A × sin B / sin A × sin B  ]

 = [cosA/sinA × cosB/sinB - 1 ] / [cos B / sin B + cos  A/sin A]  [∵cosθ/sinθ = cotθ]

 = (cot A cot B - 1) /  (cot B + cot A) [∵cosθ/sinθ = cotθ]

= RHS

Hence Proved that cot (A + B) = (cot A × cot B - 1) /  (cot A + cot B).