Vivekanand Vellanki
2

One commonly used formula is \sin^2\theta+\cos^2\theta=1. You have to look for ways to apply this formula repeatedly.

In this case, notice that if you multiply the numerator and denominator of the LHS by \left(1+\sin\theta\right), you get to use the above formula.

Lets try that:

\frac{\cos\theta}{1-\sin\theta}=\frac{\cos\theta\left(1+\sin\theta\right)}{\left(1-\sin\theta\right)\left(1+\sin\theta\right)}=\frac{\cos\theta\left(1+\sin\theta\right)}{1-\sin^2\theta}=\frac{\cos\theta\left(1+\sin\theta\right)}{\cos^2\theta}=\frac{\left(1+\sin\theta\right)}{\cos\theta}

Also, notice that by multiplying by \left(1+\sin\theta\right), you are already incorporating the numerator in the right hand side.

Think of trigonometric identifies as successive transformations that make the LHS equal to the RHS. The easiest way to do that is by including terms in the RHS in the LHS by multiplying the numerator and denominator with those terms.

Mahesh Godavarti
0

Are you sure you got this right?

This is not true for \theta = 45 \degree .

\cos 45 \degree \div 1 - \sin 45 \degree = 0 \neq 1 + \sin 45 \degree \div \cos 45 \degree = 1 + 1 = 2 .

Ah, you mean to show \frac{\cos \theta}{1 - \sin \theta} = \frac{1 + \sin \theta}{\cos \theta} . Please edit your question and fix it to show the right expression (parentheses are very important).

You should have written the expression as follows:

\cos\theta\div(1-\sin\theta)=(1+\sin\theta)\div\cos\theta.

To prove or disprove such identities, some tiimes it is useful to perform same operations on both side and end up with an identify that is easily true.

In this particular case, we first multiply both sides by \cos \theta and then by 1 - \sin \theta to get

\cos^2 \theta = (1 - \sin \theta) (1 + \sin \theta) = 1 - \sin^2 \theta .

Which is true. Therefore, it has been proved.