Homework Help: trig

Posted by gaby on Thursday, November 6, 2008 at 9:53pm.

Verify that each equation is an identity.

16. 1+tanx/sinx+cosx =secx

ok i have a clue on how to do it. i multiplyed the denominator by sinx-cosx and i also did the top but when i do i get this weird fraction with all these cos and sin and then i get lost...plz help me and explain...

Find a numerical value of one trigonometric function of x.
30. 1+tanx/1+cotx=2
same thing lol..i multiplyed the bottom and top by 1-cotx...then i get stumped...plz explain

trig - Reiny, Thursday, November 6, 2008 at 10:09pm
you should use brackets so it looks like
(1+tanx)/(sinx+cosx) =secx

you are on the right track, after multiplying top and bottom by sinx - cosx you get

LS = (1+tanx)(sinx-cosx)/(sin^2 x - cos^2 x)
= (sinx - cosx + sin^2 x/cosx - sinx)/(sin^2x - cos^2x) after expanding
= (sin^2x - cos^2)/cosx ÷ (sin^2x - cos^2x)
= 1/cosx
= secx
= RS

#30 seems to work the same way.
trig - gaby, Thursday, November 6, 2008 at 10:21pm
i dont understand the second step..did u turn tan into sin/cos..? because im trying to do it and i cant get it

wat i did for the top is
sinx-cosx+tansinx-cosx
and then sinx-cosx+sin^2/cosx-cosx

can the two cos at the end cancel..thats wats screwing me up i think
- trig - Reiny, Thursday, November 6, 2008 at 10:27pm
  here is my multiplication for the top
  
  (1+tanx)(sinx-cosx) or
  (1+ sinx/cosx)(sinx-cosx) or
  sinx - cosx + sin^2x/cosx - sinx/cosx * cosx
  = sinx - cosx + sin^2x/cosx - sinx
  = -cosx + sin^2x/cosx , now take a common denominator
  = (-cos^2x + sin^2x)/cosx
  = (sin^2x - cos^2x)/cosx
  
  now you should be able to follow the rest
trig - gaby, Thursday, November 6, 2008 at 10:31pm
yay thnx!

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