Prove the identity (Trig): csc x (1 + cos x) + 1 = cos x (sec x) + (sin x / 1 - cos x)?

Question

Yeah I'm in 10th grade and I have a Trig/Precalc Final on Monday. I kinda forgot how to do this so I was wondering if anyone could solve it by working out the steps so I can tell what your doing. Thanks a bunch.

Answer 1

It looks better for me to work with right side...
........... sec x
cos x (1/cos x) + (sin x / 1 - cos x)

1 + sin x / ( 1-cosx)

lets forget the 1 for a second and multiply the other fraction by 1
or (1+cosx)/(1+cosx)

sin x (1+cosx)
------------------
(1-cosx)(1+cosx)

sinx (1+cosx)
-----------------
1-cos^2 x

sinx(1+cosx)
----------------
sin^2 x

1+cosx
--------
sin x

cscx (1+cosx)

and then we add the 1 and we have proved it!

1 + cscx (1+cosx) = cos x (sec x) + (sin x / 1 - cos x)

QED

Answer 2

For those issues, continuously start up from the more desirable complicated searching aspect. damage down the left aspect like elementary binomials. =>SecxCscx+SinxCosx-SecxSinx-CosxCscx from right here, replace the words from secx to a million/cosx and cscx to a million/sinx => a million/(cosxsinx) + sinxcosx-sinx/cosx - cosx/sinx stumble on a hardship-free denominator for all => a million+sin^2xCos^2x- Sin^2x - Cos^2x --------------------------------------... (cosx)(sinx) from trig identities, a million-sin^2x = cos^2x, so we actually finally end up with Sin^2x * cos^2x on the right and those cancel with the denominator to provide you sinxcosx. Now, simplify the different aspect utilizing the trig identification a million+tan^2x = sec^2x you receives => tanx/sex^2x = sinx/cosx * cos^2x = sinxcosx So LS(left aspect) = RS QED desire this helped

Answer 3

csc x (1 + cos x) + 1
= (1 + cos x) / sin x + 1
= [(1 + cos x)(1 - cos x)] / [sin x(1 - cos x)] + 1
= (1 + cos^2 x) / [sin x(1 - cos x)] + 1
= 1 + sin^2 x / [sin x(1 - cos x)]
= cos x (sec x) + sin x / (1 - cos x)

Answer 4

multiplying the Nr and Dr by (1-cosx)
=cscx(1+cosx)(1-cosx)/(1-cosx)
=cscx(1-cos^2x)/1-cosx
cscx*sin^2x/1-cosx
=sinx/1-cosx