Question

How to prove this: sin^3x+cos^3x/2sin^2x-1=secx-sinx/tanx-1

Answer 1

Does the denominator of tanx-1 apply to secx also? Please state.

Answer 2

http://www.math.com/tables/trig/identities.htm

Answer 3

lazy^

Answer 4

yes. tai. secx-sinx's denominator is tanx-1

Answer 5

u lazy! @Tai

Answer 6

\[\sin ^{3}x+\cos ^{3}x / 2\sin ^{2}x-1 = secx-sinx/tanx-1\]

Answer 7

Simplify rhs to (1-sincos)/(sin-cos)
Numerator lhs is (sin+cos)(1-sincos)
Denominator lhs is (sin^2-cos^2) = ((sin + cos)(sin-cos)
QED

Answer 8

Thank you. I'll try to figure out how did that happen! :)