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

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

Mathematics
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tell me if you need clarification i can drop LS and RS
yes your back!
hmm seems i really can't figure out what you mean. maybe it's time to show the picture

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ahh
here'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
sec
I ended up getting (sinx+cosx) (1/cosxsinx) for my left side
woo it worked
How do I prove identities that are like this:
1+tanx/1-tanx = Tan (x+pi/4)
\[\frac{1+\tan(x)}{1-\tan(x)}=\tan(x+\frac{\pi}{4})\] 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)
sum identity for tan?
I know the cos and sin ones
is it just those over eachother?
oops im confused.
do u simply mean sinx/cosx?
  • phi
try using tan = sin/cos and see how it works out
alright but on the R.S what do i do with the (x +pi/4)
\[\tan(x+\frac{\pi}{4})=\frac{\sin(x+\frac{\pi}{4})}{\cos(x+\frac{\pi}{4})}\] Use sum identity for sine and cosine.
  • phi
remember sin(a+b)= sin(a)cos(b) + cos(a) sin(b) and cos(a+b)= cos(a)cos(b)- sin(a)sin(b)
I 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
write down what you have so far
sinxcospi/4 +cpsxsinpi/4 / cospi/4cosx + sinpi/4sinx
  • phi
pi/4 is 45º it is good to have memorized cos(45) and sin(45)
oh i knew that but can i convert it in the eqqn?
so i could right 45deg instead of pi/4?
  • phi
btw, notice that you have the wrong sign in the denominator
  • phi
pi/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
just like 1 foot is the same as 12 inches
alright thanks
I think i got it, gotta head to class now, ill ask teacher if hes free
Thanks for your help

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