Its easy. Just remember tan x =1/cot x

If we substitute tan A = 1/cot A in the left hand side, we get

(1-tan A)/(1+ tan A) = (1-1/cot A)/(1+1/cot A) = [(cot A-1)/cot A] /[(cot A + 1)/cot A]

cancelling out cot A from both numerator and denominator, we...

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Its easy. Just remember tan x =1/cot x

If we substitute tan A = 1/cot A in the left hand side, we get

(1-tan A)/(1+ tan A) = (1-1/cot A)/(1+1/cot A) = [(cot A-1)/cot A] /[(cot A + 1)/cot A]

cancelling out cot A from both numerator and denominator, we get

cot A-1 / cot A + 1 = right hand side. Hence proved.

Start with the left hand side, and show that it is equivalent to the right hand side:

`(1-tanA)/(1+tanA) `

`=(1-(sinA)/(cosA))/(1+(sinA)/(cosA)) `

`=(1-(sinA)/(cosA))/(1+(sinA)/(cosA))*((cosA)/(sinA))/((cosA)/(sinA)) `

`=((cosA)/(sinA)-1)/((cosA)/(sinA)+1) `

`=(cotA-1)/(cotA+1) ` as required.

** Recognizing that the cot is the reciprocal of the tan, you can shorten this by multiplying the left side by `(cotA)/(cotA) ` to get the desired result immediately.**

**Further Reading**