Urgent Proving Identities

**ImKo** · January 12th 2009, 02:45 AM

Prove that
tan A - (sec A sin^3A / 1 + cos A ) = sin A

Obtained from my Workings

(sin A / cos A) - (sin^3A / cos A + cos^2 A )(From here on I don't know what to do aside from subtracting it)

Prove that

sin^3 A / cos A - cos^3A = tan A
Obtained from my Workings: Can't think of Any

Thanks, hoping for a fast reply

Arnold,

**Chop Suey** · January 12th 2009, 02:59 AM

1. $\tan{A} - \frac{\sec{A}\sin^3{A}}{1+\cos{A}}$

This can be rewritten as:
$\tan{A} \left( 1 - \frac{\sin^2{A}}{1+\cos{A}}\right)$

Recall that $\sin^2{\theta} = 1-\cos^2{\theta} = (1-\cos{\theta})(1+\cos{\theta})<br />$
2. $\frac{\sin^3{A}}{\cos{A}-\cos^3{A}}$

Pull one factor of each sine and cosine out:
$\frac{\sin{A}}{\cos{A}} \cdot \frac{\sin^2{A}}{1-\cos^2{A}}$

Can you do it now?

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