How do I solve thsi identiy? (sinx/1-cotx) + (cosx/1-tanx) = sinx +cosx?

Question

I need help quick! Thanks. This is for adv. alg/ trig.

Answer 1

(sinx/1-cotx) + (cosx/1-tanx) = sinx +cosx?

1-cotx = 1-(cosx/sinx) = (sinx-cosx)/sinx
1-tanx = 1-(sinx/cosx) = (cosx-sinx)/cosx

sinx/(1-cotx) = sinx/[(sinx-cosx)/sinx] = (sinx)^2/(sinx-cosx)
cosx/(1-tanx) = cosx/[(cosx-sinx)/cosx] = (cosx)^2/(cosx-sinx)
=-(cosx)^2/(sinx-cosx)

(sinx/1-cotx) + (cosx/1-tanx) = [(sinx)^2 - (cosx)^2] / (sinx-cosx) = [(sinx-cosx)(sinx+cosx)] / (sinx-cosx)

= sinx+cosx

Answer 2

all the above solutions are incorrect because of the fact branch comes before addition through default. cosx/a million + cotx = sinx/a million + tanx cosx + cotx = sinx + tanx those are no longer equivalent, you may desire to have made a mistake someplace.

Answer 3

(sinx/1-cotx) + (cosx/1-tanx)

multiply top and bottom of the 1st fraction by sinx and the 2nd fraction by cosx

= sin²x / (sinx - cosx) + cos² x / (cosx - sinx)

= sin²x / (sinx - cosx) - cos² x / (sinx - cosx)

= (sin²x - cos² x) / (sinx - cosx)

= (sinx - cosx)(sinx + cosx)/(sinx - cosx)

= sinx + cosx

Answer 4

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

the answer should be this want.Try to count again for the answer if you are not sure. please use the formule.i hav tried again a lot of time but i still cant get sinx+cosx for the last answer...

Answer 5

1 - cot x = sinx/sinx - cosx/sinx = (sinx - cosx)/sinx
sinx/1-cotx = sin^2(x)/(sinx - cosx)
1 - tan x = cosx/cosx - sinx/cosx = (cosx - sinx)/cosx
cosx/1-tan x = cos^2(x)/(cosx - sinx) = -cos^2(x)/(sinx - cosx)

[sin^2(x) - cos^2(x)]/[sinx - cosx]
(sinx + cosx)(sinx-cosx)/(sinx - cosx)
= sinx + cosx

Answer 6

LHS
sinx /(1 - cosx / sin x) + cos x/(1 - sinx / cosx)
sinx/(sinx - cosx)/sinx + cosx/(cosx - sinx)/cosx)
sin²x / (sinx - cosx) + cos²x.(cosx - sinx)
sin²x/ (sinx - cosx) - cos²x.(sinx - cosx)
(sin²x - cos²x) / (sinx - cosx)
(sinx - cosx).(sinx + cosx) / (sinx - cosx)
sinx + cosx
RHS
sinx + cosx
LHS = RHS

Answer 7

(sinx/1-cotx) + (cosx/1-tanx)
= sinx/ (1- cosx/sinx) + cosx/((1 - sinx/cosx)
= sin^2x /(sinx - cosx) + cos^2x / (cosx - sinx)
= - sin^2x /(cosx - sinx) + cos^2x / (cosx - sinx)
= (cos^2x - sin^2x) / (cosx - sinx)
= (cosx - sinx) (cosx + sinx) /(cosx - sinx)
=cosx + sinx
=sinx + cosx

Answer 8

With identities like this, I have found it helpful to break them down into their sine and cosine components, something like this...

( sin x / (1 - cot x) ) + (cos x / ( 1 - tan x) )
= sin x / (1 - cos x / sin x) + cos x / (1 - sin x / cos x)

...then simplifying...

= sin x / ((sin x - cos x) / sin x) + cos x / ((cos x - sin x) / cos x)
= sin^2 x / (sin x - cos x) - cos^2 x / (sin x - cos x)
= (sin^2 x - cos^2 x) / (sin x - cos x)
= (sin x + cos x) * (sin x - cos x) / (sin x - cos x)
= sin x + cos x