on the LS,
sin^3+cos^3 = (sin+cos)(sin^2 - sin*cos + cos^2) = (sin+cos)(1-sin*cos)
1-2cos^2 = sin^2-cos^2 = (sin+cos)(sin-cos)
divide, giving (1-sin*cos)/(sin-cos)
on the RS,
sec-sin = 1/cos - sin = (1-sin*cos)/cos
tan-1 = sin/cos - 1 = (sin-cos)/cos
divide, giving (1-sin*cos)/(sin-cos)
LS = RS
Verify this identity. How would you prove it
(sin^3(x)+cos^3(x))÷(1-2cos^2(x)) = (sec(x)-sin(x))÷(tan(x)-1)
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