Trig identities

[Nelo]

1. The problem statement, all variables and given/known data

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

2. Relevant equations





3. 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 dont 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]

[b]

thus giving the answer. I dont 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.