Question #1a94c

1 Answer
May 14, 2017

See below.

Explanation:

Prove: tan^2x/(tan^2x+1)=sin^2x

Let us start on the left-hand side (LHS) and work toward the right-hand side (RHS).

Consider Pythagorean's identity:
sin^2x+cos^2x=1

which we can divide by cos^2x to get:
sin^2x/cos^2x+cos^2x=1/cos^2x

Note that secx=1/cosx and tanx=sinx/cosx:
tan^2x+1=sec^2x, which is most applicable to this problem

We can substitute tan^2x+1=sec^2x into the LHS:
LHS=tan^2x/sec^2x

Note that secx=1/cosx and tanx=sinx/cosx:
LHS=tan^2x/(1/cos^2x)

=(sin^2x/cos^2x)/(1/cos^2x)

=sin^2x/cos^2x*cos^2x

=sin^2x=RHS

Therefore, tan^2x/(tan^2x+1)=sin^2x