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

Question

14232 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-27T21:03:04

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

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

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

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,

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

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