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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