# Trigonometry

posted by
**Dirk**
.

I need to prove that the following is true. Thanks.

csc^2(A/2)=2secA/secA-1

Right Side=(2/cosA)/(1/cosA - 1)

= (2/cosA)/[(1-cosA)/cosA]

=2/cosA x (cosA)/(1-cosA)

=2/(1-cosA)

now recall cos 2X = cos^2 X - sin^2 X and we could say

cos A = cos^2 A/2 - sin^2 A/2 and of course sin^2 A/2 + cos^2 A/2=1

so the above 2/(1 - cosA)

=2/[sin^2 A/2 + cos^2 A/2 - (cos^2 A/2 - sin^2 A/2)]

= 2/(2sin^2 A/2)

= csc^2 A/2

= Left Side

Ok, so just using a different form of the double/half-angle formula allowed you to get the sin and therein getting csc.

Thanks Reiny! For a detailed how to and detailed steps. I will be using this service in the future. Thanks again.