Prove the following identity :
Sin α . Cos α .
Tan α = (1 – Cos
α) (1 + Cos
Sin α . Cos α . Tan α = (1 – Cos α) (1 + Cos α)
Left side = Sin β . Tan β + Cos β
= Sin β . Sin β / Cos β + Cos β
= Sin² β / Cos β + Cos² β / Cos β
= 1 / Cos β = Sec β = Right side proven
Let's work on the left side first. And remember that
the tangent is the same as sin/cos.
sin(a) cos(a) tan(a)
Substitute for the tangent:
[ sin(a) cos(a) ] [ sin(a)/cos(a) ]
Cancel the cos(a) from the top and bottom, and you're left with
[ sin(a) ] . . . . . [ sin(a) ] which is [ sin²(a) ] That's the left side.
Now, work on the right side:
[ 1 - cos(a) ] [ 1 + cos(a) ]
Multiply that all out, using FOIL:
[ 1 + cos(a) - cos(a) - cos²(a) ]
= [ 1 - cos²(a) ] That's the right side.
Do you remember that for any angle, sin²(b) + cos²(b) = 1 ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.
So, on the right side, you could write [ sin²(a) ] .
Now look back about 9 lines, and compare that to the result we got for the left side .
They look quite similar. In fact, they're identical. And so the identity is proven.