Question

Sec tita - 1/sec tita +1 = sin square tita/ (1+cos tita) whole square

Answer 1

Proof is in the explanation.

Explanation:

I figure out what you were saying:

`(\sec(x)-1)/(\sec(x)+1)=(\sin^2(x))/((1+\cos(x))^2)`

I'm going to try to rewrite left hand side as right hand side.

I'm going to rewrite
`\sec(x)=(1)/(\cos(x))`:

`((1)/(\cos(x))-1)/((1)/(\cos(x))+1)`

Multiply both numerator and denominator by
`\cos(x)`:

`(1-\cos(x))/(1+\cos(x))`

Multiply both numerator and denominator by
`1+\cos(x)`:

`((1-\cos(x))(1+\cos(x)))/((1+\cos(x))^2)`

When multiplying conjugates you only have to do first time first and last times last:

`(1-\cos^2(x))/((1+\cos(x))^2)`

We can rewrite the numerator using the Pythagorean Identity:
`\cos^2(x)+\sin^2(x)=1`.

This gives us:

`(\sin^2(x))/((1+\cos(x))^2)`