# prove that tan A/sec A-1 - sin A/1+cos A = 2cot Athis is a question from trignometry.

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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. **

In a right triangle with a as hypotenuse and b and c as other legs,

Let sinA = c/a,

Then cosA = b/a, secA = a/b, tanA = c/b, cotA = b/c

tanA/(secA-1) - sinA/(1+cosA)

=(c/b)/(b/a-1) - (c/a)/(1+b/a)

=c/(a-b)-c/(a+b)

=(c)(2b)/{(a-b)(a+b)}

=(2bc)/(a^2-b^2)

=(2bc)/(c^2) (by pythagoras theorem, a^2=b^2+c^2)

=(2b)/c

=2(b/c)

= 2cotA

Therefore, tanA/(secA-1) - sinA/(1+cosA) = 2cotA