verify the trigonometric equation by substituting identities to match the right hand side with the left hand side:
(sinx / 1-cosx) + (sinx / 1+cosx) = 2cscx
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OpenStudy (wampominater):
looks like this
\[\frac{ sinx }{ 1-cosx } + \frac{ sinx }{ 1+cosx }=2cscx\]
been stuck on this forever...
OpenStudy (maheshmeghwal9):
on left hand side
take L.C.M.
OpenStudy (butterflydreamer):
start with your left hand side :)
hint: You'll want put the two fractions over the same denominator .
OpenStudy (wampominater):
so, would i get them both under 1-cos^2x?
OpenStudy (wampominater):
over*
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OpenStudy (butterflydreamer):
correct:)
OpenStudy (maheshmeghwal9):
It Becomes
\[\frac{ sinx(1+cosx)+sinx(1-cosx) }{ \sin^2x }\]
OpenStudy (wampominater):
alright then do I distribute sin?
OpenStudy (butterflydreamer):
yyeeepp
OpenStudy (butterflydreamer):
distribute sinx :)
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OpenStudy (wampominater):
\[\frac{ 2sinx }{ \sin^2x }\]
like this?
OpenStudy (maheshmeghwal9):
yes
OpenStudy (butterflydreamer):
then from there, you can simplify the fraction and you'll get the answer (RHS) :)
OpenStudy (wampominater):
Ok so how do I simplify this, Sorry! this is where i have been stuck at.
OpenStudy (butterflydreamer):
oh you can divide the numerator and denominator by sinx
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