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Mathematics 16 Online
OpenStudy (anonymous):

trig identities : tan^2x-sin^2x=tan^2sin^2x

OpenStudy (anonymous):

\[\tan^2x-\sin^2x=\tan^2xsin^2x\] prove the identity.

ganeshie8 (ganeshie8):

take left hand side, rewrite tan as sin/cos

OpenStudy (anonymous):

let me show you the work i've done so far.

ganeshie8 (ganeshie8):

or a better method would be to leave the `tan ` as it is, and write sin as tan*cos

OpenStudy (anonymous):

\[\frac{ 1-sinx }{1+sinx }*\frac{ \cos^2x }{ \cos^2x }= \frac{ \cos^2x(1-sinx )}{\cos^2x(1+sinx }\]

ganeshie8 (ganeshie8):

from where u pulled up (1-sinx)/(1+sinx) ?

OpenStudy (anonymous):

\[(1-sinx)\cos^2x= -(sinx-1)\cos^2x=(-\cos^2x)(sinx-1)\]

OpenStudy (anonymous):

sorry we wrote the wrong problem one sec

OpenStudy (anonymous):

starting over lol

ganeshie8 (ganeshie8):

take ur time

OpenStudy (anonymous):

brain is fried, can u tell?

OpenStudy (shamim):

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

OpenStudy (shamim):

do the rest

OpenStudy (anonymous):

tan^2xsin^2x= sin^2x=1-cos^2x tan^2x(1-cos^2x)=tan^2xcos^2x =tan^2x(-(sinx/cosx)^2cos^2x

ganeshie8 (ganeshie8):

whats going on ? :/

OpenStudy (shamim):

sin^2x.sin^2x/cos^2x =tan^2x.sin^2x

OpenStudy (anonymous):

\[\tan^2xsin^2x=\tan^2x(1-\cos^2x)\]

ganeshie8 (ganeshie8):

oh you're doing it from right hand side is it ?

OpenStudy (anonymous):

\[\tan^2x-\tan^2xcos^2x=\tan^2x-(\frac{ \sin^2x }{ \cos^2x})\cos^2x\]

OpenStudy (anonymous):

yes

ganeshie8 (ganeshie8):

looks great !

OpenStudy (anonymous):

now here's my problem

OpenStudy (anonymous):

\[\tan^2x-\sin^2x=\tan^2x-(\frac{ \sin^2x }{ \cos^2x })\cos^2x\]

ganeshie8 (ganeshie8):

what problem ? you're done lol

OpenStudy (anonymous):

\[-\sin^2x= -(\frac{ \sin^2x }{ \cos^2x })\cos^2x\]

OpenStudy (anonymous):

i'm supposed to prove that the identities equal one another

OpenStudy (anonymous):

but i'm at a loss for how to make -(sin^2x/cos^2x)cos^2x into tan^2x

ganeshie8 (ganeshie8):

Right hand side : \(\tan^2x \sin^2x\) \(= \tan^2x (1-\cos^2x)\) \(= \tan^2x - \tan^2x \cos^2x\) \(= \tan^2x - \dfrac{\sin^2x }{\cos^2x} \cos^2x\) \(= \tan^2x - \sin^2x\) = Left hand side. QED.

OpenStudy (anonymous):

got it idk what iwas thinking

ganeshie8 (ganeshie8):

if you're stuck here : \[-\sin^2x= -(\frac{ \sin^2x }{ \cos^2x })\cos^2x\]

ganeshie8 (ganeshie8):

they're both are equal

OpenStudy (anonymous):

okay, thank you.

ganeshie8 (ganeshie8):

simply cancel out \(\cos^2x\) okay ?

OpenStudy (anonymous):

yes

ganeshie8 (ganeshie8):

agree^ looks ur brain is friend lol put it in freezer somtime :P

ganeshie8 (ganeshie8):

*fried

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