Mathematics
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OpenStudy (anonymous):
trig identities :
tan^2x-sin^2x=tan^2sin^2x
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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):
\[\frac{ 1-sinx }{1+sinx }*\frac{ \cos^2x }{ \cos^2x }= \frac{ \cos^2x(1-sinx )}{\cos^2x(1+sinx }\]
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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
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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 ? :/
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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
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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):
\[-\sin^2x= -(\frac{ \sin^2x }{ \cos^2x })\cos^2x\]
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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 :
\[-\sin^2x= -(\frac{ \sin^2x }{ \cos^2x })\cos^2x\]
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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
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ganeshie8 (ganeshie8):
*fried