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 sinxcosx 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 1cotx...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)(sinxcosx)/(sin^2 x  cos^2 x)
= (sinx  cosx + sin^2 x/cosx  sinx)/(sin^2x  cos^2x) after expanding
= (sin^2x  cos^2)/cosx รท (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 don't 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
sinxcosx+tansinxcosx
and then sinxcosx+sin^2/cosxcosx
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)(sinxcosx) or
(1+ sinx/cosx)(sinxcosx) 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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