Help Proving Trigonometric Identities?

Question

I need to prove that

(1+secx)/(sinx +tanx) = csc x

and that

(1-cosx)/(cosx) = (tan^2 x)/(secx + 1)

These are the only two that I have left and I cannot figure them out! If you could please help me that would be great. Thanks!

Answer 1

1)

(1 + sec x)/(sinx + tanx) = cscx

First I would convert to sines and cosines

(1 + sec x)/(sinx + tanx)
= (1 + 1/cosx) / (sinx/1 + sinx/cosx)

Next we will find common denominators....

= (cosx/cosx + 1/cosx) / ((sinx/1)(cosx/cosx) + (sinx/cosx))
= ((cosx + 1)/(cosx)) / ((sinxcosx)/(cosx) + (sinx/cosx))
= ((cosx + 1)/(cosx)) / ((sinxcosx + sinx)/(cosx))

Now, if you re-write this on real paper you will see that you have two fractions and the denominator of each is cosx, so it cancels out

= (cosx + 1) / (sinxcosx + sinx)

Factor sine out of the denominator

= (cosx + 1) / (sinx(cosx + 1))
= 1/sinx
= cscx



2)

(1-cosx)/(cosx) = (tan^2 x)/(secx + 1)


(tan^2 x)/(secx + 1)
= ((tanx)(tanx)) / (secx + 1)
= (sinx/cosx)(sinx/cosx) / (1/cosx + 1)
= (sin^2 x/cos^2 x) / (1/cosx + cosx/cosx)
= (sin^2 x/cos^2 x) / ((1 + cosx)/(cosx))

Flip the bottom and multiply, which cancels a cos, leaving

= (sin^2 x) / (cosx(1 + cosx))

Now convert to all cosines since that's what you want to end up with

= (1 - cos^2 x) / (cosx(1 + cosx))

You will have to factor the numerator into the difference of squares

= (1 + cosx)(1 - cosx) / (cosx(1 + cosx))

Now the (1 + cosx) will cancel

= (1 - cosx)/(cosx)

Answer 2

substitute all trig fn with sin and cos

(1+1/cos)/(sin + sin/cos) = (cos + 1)/cos * cos / (sin(cos+1)
= 1/sin = csc


use tan^2 = sec^2 -1
(sec^2 - 1)/(sec + 1) = (sec + 1)(sec -1)/(sec+1)
= sec - 1
= 1/cos - 1 = (1 - cos)/ cos

Answer 3

You are right that it is equal to tan(2x) . now tan(2x) = tan(x+x) =(tanx + tanx) /(1 - tanx tanx) =2tanx /(1 - tan(x) tanx or 2tanx /1 - tan^2(x)