Verify that each equation is an identity.

Question

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

Answer 1

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.

Answer 2

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
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

Answer 3

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

Answer 4

yay thnx!