Question #75a89
2 Answers
Explanation:
"using the "color(blue)"trigonometric identities"using the trigonometric identities
•color(white)(x)1+tan^2x=sec^2x∙x1+tan2x=sec2x
•color(white)(x)tanx=sinx/cosx∙xtanx=sinxcosx
•color(white)(x)secx=1/cosx∙xsecx=1cosx
"consider the LHS"consider the LHS
rArr(tanx)/(1+tan^2x)⇒tanx1+tan2x
=(sinx/cosx)/(sec^2x)=sinxcosxsec2x
=(sinx/cosx)/(1/cos^2x)=sinxcosx1cos2x
=sinx/cancel(cosx) xxcancel(cos^2x)^(cosx)
=sinxcosx="RHS"rArr"proven"
See below.
Explanation:
We're trying to prove
Let's manipulate the left side since it's more complicated.
There is a trigonometric identity that states
Thus,
tan x / (1+tan^2 x)
= tan x / (sec^2 x)
= (sin x / cos x)/(1/cos^2x)
=sin x / cancel(cos x) * cancel(cos^2 x)^cos x /1
=sin x cos x