The common error in proving trigonometric identities is to assume the equality of the left-hand side (LHS) and the right-hand side (RHS) of what is being proved. To prove an identity means to transform the LHS and RHS in the same form.

Prove the identity:

$\frac {(1- cos x)}{sin x} = \frac {sin x}{(1+cos x)}$

Solution sure to get 0/5 mark:

This is a ratio so we cross multiply. That is,

$(1 - cos x)(1+cos x) = (sin x)(sin x)$

$(1 - cos^2 x) = sin^2 x$

So,

$sin^2 x = sin^2 x$

Why you get 0 mark? When you cross-multiply, you assume that you have an equation. You assume that the left-hand side is equal to the right-hand side. Isn’t that’s what you are exactly asked to do – To show that they are equal and not assume that they are equal?

Solution sure to get 5/5 mark:

Left Hand Side

$\frac {(1- cos x)}{sin x} \frac {(1+cos x)}{(1+cos x)}$

$=\frac {(1-cos^2 x)}{sin x (1+cos x)}$

$=\frac{sin^2 x}{sin x (1+cos x)}$

Right Hand Side

$\frac {sin x}{(1+cos x)} \frac {sin x}{sin x}$

$=\frac{sin^2 x}{sin x (1+cos x)}$

The right hand side is identical to the left hand side hence the identity is proved.