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starkeyfish
Group Title
Verify the identity:
(1sinx)/cosx = cosx/(1+sinx)
 one year ago
 one year ago
starkeyfish Group Title
Verify the identity: (1sinx)/cosx = cosx/(1+sinx)
 one year ago
 one year ago

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RolyPoly Group TitleBest ResponseYou've already chosen the best response.2
LS = (1sinx)/cosx = 1/cosx  sinx/cosx RS = \(\large\frac{cosx}{1+sinx}\) = \(\large\frac{cosx(1sinx)}{(1+sinx)(1sinx)}\) = \( \large\frac{cosx}{1sin^2x}  \frac{cosxsinx}{1sin^2x}\) =.... Note: \[1 = cos^2x + sin^2x \]
 one year ago

Jhannybean Group TitleBest ResponseYou've already chosen the best response.0
why is it so tiny!
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
Because its a spoiler
 one year ago

starkeyfish Group TitleBest ResponseYou've already chosen the best response.0
I'm very bad at basic math , so I'm finding all of this to be very foreign.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.2
Actually, simply working from the left is ok :D \[RS = \frac{cosx}{(1+sinx)}\]\[=\frac{cos}{1+sinx}\times \frac{1sinx}{1sinx}\]By the identity a^2  b^2 = (a+b)(ab), we get\[=\frac{cos(1sinx)}{1sin^2x}\]Then, \(sin^2x+cos^2x =1\) to express \(1sin^2x\) in terms of \(cos^2x\). Finally, cancel the common factor, and you'll get the left side!
 one year ago

starkeyfish Group TitleBest ResponseYou've already chosen the best response.0
so 1sin^2 = cos^2x?
 one year ago

starkeyfish Group TitleBest ResponseYou've already chosen the best response.0
okay, thank you!
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.2
You're welcome~~
 one year ago
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