please solve tanA/1-cotA+cotA/1-tanA=1+secAcosecAeverything in this question in detial . every step in detail 

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You need to check if the expression `tanA/(1-cotA) + cotA/(1-tanA) = 1+secAcscA`  is an identity, hence, you need to substitute, to the left side,`sin A/cos A`  for `tan A`  and `cos A/sin A`  for `cot A`  such that:

`(sin A/cos A)/(1-cos A/sin A) + (cos A/sin A)/(1-sin A/cos A) = 1+secAcscA `

You need to bring the terms `1-cos A/sin A`  and `1-sin A/cos A`  to a common denominator such that:

`1-cos A/sin A = (sin A - cos A)/sin A `

`1-sin A/cos A = -(sin A - cos A)/cos A`

`(sin A/cos A)/(1-cos A/sin A) = (sin^2 A)/(cosA(sin A - cos A))`

`(cos A/sin A)/(1-sin A/cos A) = -(cos^2 A)/(sin A(sin A - cos A))`

Hence, you need to check if the left side is equal to the right side such that:

`(sin^2 A)/(cosA(sin A - cos A)) - (cos^2 A)/(sin A(sin A - cos A)) = 1+secAcscA `

You need to bring the terms `(sin^2 A)/(cosA(sin A - cos A)) and (cos^2 A)/(sin A(sin A - cos A))`  to a common denominator such that:

`(sin^3 A - cos^3 A)/(sin A*cos A(sin A - cos A)) = 1+secAcscA` 

You need to remember that `sec A = 1/cos A`  and `cscA = 1/sin A`  such that:

`(sin^3 A - cos^3 A)/(sin A*cos A(sin A - cos A)) = 1+ 1/(sin A*cos A)`

`(sin^3 A - cos^3 A)/(sin A*cos A(sin A - cos A)) = (sin A*cos A + 1)/(sin A*cos A)`

`(sin^3 A - cos^3 A)/(sin A - cos A) = (sin A*cos A + 1)`

You need to substitute the difference of cubes by the special product such that:

`sin^3 A - cos^3 A = (sin A - cos A)(sin^2A + sinA*cos A + cos^2A)`

You should remember the fundamental formula of trigonometry such that:

`sin^2A + cos^2A = 1`

`sin^3 A - cos^3 A = (sin A - cos A)(1+ sinA*cos A)`

`(sin^3 A - cos^3 A)/(sin A - cos A) = (sin A*cos A + 1)`

`((sin A - cos A)(1+ sinA*cos A))/(sin A - cos A) = (sin A*cos A + 1)`

Reducing by `(sin A - cos A)`  yields:

`1+ sinA*cos A = sin A*cos A + 1`

Hence, evaluating both sides yields that they are identically, hence, the expression `tanA/(1-cotA) + cotA/(1-tanA) = 1+secAcscA`  is an identity.

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