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anonymous
 3 years ago
Prove:
(sinx + cosx) (Tan^2x+1/tanx) = 1/cosx + 1/sinx
Picture below for what it looks like properly
anonymous
 3 years ago
Prove: (sinx + cosx) (Tan^2x+1/tanx) = 1/cosx + 1/sinx Picture below for what it looks like properly

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0tell me if you need clarification i can drop LS and RS

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm seems i really can't figure out what you mean. maybe it's time to show the picture

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352211230059:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352211285143:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here's a hint that might help: \[\huge \tan^2 x + 1 = \sec^2 x\] and \[\huge \sec^2 x = \frac1{\cos^2 x}\] try solving it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I ended up getting (sinx+cosx) (1/cosxsinx) for my left side

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How do I prove identities that are like this:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01+tanx/1tanx = Tan (x+pi/4)

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1+\tan(x)}{1\tan(x)}=\tan(x+\frac{\pi}{4})\] I would write each side so that the function tan( ) only contains x and not both x & x+pi/4. You the sum identity for tan. If you don't know it write both sides in terms of sine and cosine. Use the sum identities for them to expand tan(x+pi/4)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sum identity for tan?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know the cos and sin ones

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is it just those over eachother?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do u simply mean sinx/cosx?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1try using tan = sin/cos and see how it works out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright but on the R.S what do i do with the (x +pi/4)

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0\[\tan(x+\frac{\pi}{4})=\frac{\sin(x+\frac{\pi}{4})}{\cos(x+\frac{\pi}{4})}\] Use sum identity for sine and cosine.

phi
 3 years ago
Best ResponseYou've already chosen the best response.1remember sin(a+b)= sin(a)cos(b) + cos(a) sin(b) and cos(a+b)= cos(a)cos(b) sin(a)sin(b)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I plugged in the sum identity but i dont know what do to next I have sin and cosx's which i dont know what to do with

phi
 3 years ago
Best ResponseYou've already chosen the best response.1write down what you have so far

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sinxcospi/4 +cpsxsinpi/4 / cospi/4cosx + sinpi/4sinx

phi
 3 years ago
Best ResponseYou've already chosen the best response.1pi/4 is 45º it is good to have memorized cos(45) and sin(45)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i knew that but can i convert it in the eqqn?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i could right 45deg instead of pi/4?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1btw, notice that you have the wrong sign in the denominator

phi
 3 years ago
Best ResponseYou've already chosen the best response.1pi/4 and 45 mean the same thing.... it matters when using a calculator, or in certain math operations where radians are more useful... but the cos(45º) = cos(pi/4 radians)

phi
 3 years ago
Best ResponseYou've already chosen the best response.1just like 1 foot is the same as 12 inches

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think i got it, gotta head to class now, ill ask teacher if hes free
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