How do you simplify #((1-costheta)(1 + costheta))/((1-sintheta)(1+sintheta))#?

1 Answer
Feb 15, 2016

#tan^2theta#

Explanation:

#((1-cos theta)(1+cos theta))/((1-sin theta)(1+sin theta)#

We know that #(a+b)(a-b)=a^2-b^2#, so we can convert the expresssion in:

#(1-cos^2 theta)/(1-sin^2 theta#

Then we know that #cos^2 theta+sin^2 theta=1#:

#(sin^2 theta)/(cos^2 theta)=tan^2theta#