# Math (Trig) a.s.a.p, thanks!

posted by .

verify the identities:

1.) 2tanx-(1+tanx)^2 = -secx

2.) cos(x-330degrees)=1/2(ã3cosx-sinx)
*the ãis only over the 3*

3.)
1+cos2x/sinx=cotx

any help is greatly appreciated!!

• Math (Trig) typo edit -

ã is supposed to represent square root.
so for number 2
it's 1/2 (square root 3 cosx-sinx)

• Math (Trig) a.s.a.p, thanks! -

1.
LS = 2sinx/cosx - (1+sinx/cosx)^2
= 2sinx/cosx - ((cosx + sinx)/cosx)
= 2sinx/cosx - (cos^2 x + 2sinxcosx + sin^2 x)/cos^2 x

= (2sinxcosx -(1 + 2sinxcosx))/cos^2 x
= -1/cos^2 x
= -sec^2 x

you had -secx on the right side
I think you made a typo

2. LS = cosxcos330 + sinxsin330

but sin 330 = -sin30 = -1/2
and cos 330 = cos30 = √3/2

so above
= (√3/2)cosx - 1/2 sinx
= 1/2(√3 cosx - sinx)
= RS

Why don't you try the third one.
Change everything to sines and cosines
let me know how you did.

• Math (Trig) a.s.a.p, thanks! -

for the third one you had

1+cos2x/sinx=cotx

I tried (1+cos2x)/sinx=cotx and tested with 40º, did not work

I tried 1+cos^2 x/sinx=cotx and tested with 40º, did not work

I tried 1+(cos2x/sinx)=cotx and tested with 40º, did not work

check your typing of the question.

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