prove that tan A/sec A-1 - sin A/1+cos A = 2cot Athis is a question from trignometry.
2 Answers
thilina-g | College Teacher | (Level 1) Educator
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.
leedygood7 | High School Teacher | (Level 2) eNoter
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