Question

Question #1a94c

Answer 1

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`