Solving Trig Identities: tan^2x - sin^2x = sin^2x*tan^2x Explanation

[Nelo]

Homework Statement

tan^2x - sin^2x = sin^2x*tan^2x

Homework Equations

The Attempt at a Solution

This is how my teacher solved it.

Sin^2x sin^2x
______ - ______
Cos^2x 1

sin^2x - cos^2x*sin^2x
_________________________
cos^2x

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

=sin^2x *sin^2x
______________
cos^2x


thus giving the answer. I don't understand what happends on the third line. How can 1-cos^2x be bracketed if cos^2x*sin^2x are one term, even though they equal one, how can that equal 1-cos^2x?

 

[rock.freak667]

Because in sin²x - cos²xsin²x, sin²x is common in both terms, thus it can be factored as sin²x(1-cos²x).

If you expand the bracket you will get back sin²x - cos²xsin²x, sin²x.

 

[PeterO]

thus giving the answer. I don't understand what happends on the third line. How can 1-cos^2x be bracketed if cos^2x*sin^2x are one term, even though they equal one, how can that equal 1-cos^2x?

In the third line, what happened was factorisation.