The given identity is,
`(tan(A))/(sec(A)-1) - (sin(A))/(1+cos(A)) = 2cot(A)`
I will start from LHS:
LHS =
`= (tan(A))/(sec(A)-1) - (sin(A))/(1+cos(A))`
I can rewrite sec(A) and tan(A) as,
`= ((sin(A))/(cos(A)))/((1/(cos(A)))-1) - (sin(A))/(1+cos(A))`
Simplifying,
`= (sin(A))/(1-cos(A)) - (sin(A))/(1+cos(A))`
Taking both terms into a common denominator,
`= (sin(A)(1+cos(A))-(sin(A)(1-cos(A))))/((1-cos(A))(1+cos(A)))`
=This gives,
`= (sin(A)+sin(A)cos(A)-sin(A)+sin(A)cos(A))/(1-cos^2(A))`
`=...
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The given identity is,
`(tan(A))/(sec(A)-1) - (sin(A))/(1+cos(A)) = 2cot(A)`
I will start from LHS:
LHS =
`= (tan(A))/(sec(A)-1) - (sin(A))/(1+cos(A))`
I can rewrite sec(A) and tan(A) as,
`= ((sin(A))/(cos(A)))/((1/(cos(A)))-1) - (sin(A))/(1+cos(A))`
Simplifying,
`= (sin(A))/(1-cos(A)) - (sin(A))/(1+cos(A))`
Taking both terms into a common denominator,
`= (sin(A)(1+cos(A))-(sin(A)(1-cos(A))))/((1-cos(A))(1+cos(A)))`
=This gives,
`= (sin(A)+sin(A)cos(A)-sin(A)+sin(A)cos(A))/(1-cos^2(A))`
`= (2sin(A)cos(A))/(sin^2(A))`
(Because,` sin^2(A)+cos^2(A) =1` )
LHS =
`= (2cos(A))/(sin(A))`
`= 2cot(A)`
Therefore, LHS = RHS, and
`(tan(A))/(sec(A)-1) - (sin(A))/(1+cos(A)) = 2cot(A)` Proved.