Question

How do you verify `sin^2(-x)=tan^2x/(tan^2x+1)`?

Answer 1

`sin^2(-x)=sin^2x`

`(tan^2x)/(tan^2x+1)=sin^2x`

`rArr(tan^2x)/(tan^2x+1)=sin^2(-x)`.

Explanation:

Verifying the equality of the given trigonometric expressions is determined by evaluating each side separately then comparing their results.

These trigonometric identities are used
`color(red)(sin(-x)=-sinx)`

`color(red)(tanx=sinx/cosx)`

`color(red)(cos^2+sin^2x=1)`

`sin^2(-x)=(sin(-x))^2=(color(red)(-sinx))^2=sin^2x`
`color(blue)(sin^2(-x)=sin^2x" " EQ1)`

`(tan^2x)/(tan^2x+1)`

`=((color(red)(sinx/cosx))^2)/((color(red)(sinx/cosx))^2+1)`

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

`=(sin^2x/cos^2x)/(color(red)(sin^2x+cos^2x)/cos^2x)`

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

`=sin^2x/cos^2x*cos^2x/1`

`=sin^2x`

`color(blue)((tan^2x)/(tan^2x+1)=sin^2x)`

`color(blue)(sin^2(-x)=sin^2x " "EQ1)`

Therefore,

`(tan^2x)/(tan^2x+1)=sin^2(-x)`