1+tanx/1+cotx= secx/cscx?

Question

verify the identity

Answer 1

(1+tanx)/(1+cotx) = secx/cscx

(1 + sinx/cosx)/(1 + cosx/sinx)

(cosx/cosx + sinx/cosx)/(sinx/sinx + cosx/sinx)

[(cosx + sinx)/cosx]/[[sinx + cosx)/sinx]

(1/cosx)/(1/sinx)

secx/cscx

Answer 2

Cscx-cotx Secx 1

Answer 3

L.H.S
=(1+tanx)/(1+cotx)
=(1+tanx)/(1+1/tanx) Because:cotx=1/tanx
=(1+tanx)/{(tanx+1)/tanx}
={(1+tanx)/tanx+1} * tanx
=tanx

R.H.S
=secx/cosecx
=(1/cosx)/(1/sinx)
=sinx/cosx
=tanx

Hence L.H.S=R.H.S
Its proved

Answer 4

i'll do each and every expression separately to start up with (for ease of examining). tan x/(a million - cot )x = (sinx/cosx) / (a million - cos x/sin x) = (sin x/cos x) / [(sin x - cos x)/sin x)} invert and multiply: = (sin x / cos x)*sin x / (sin x - cos x) = sin^2 x /cos x (sin x - cos x) cot x /( a million - tan x) = (cos x / sin x) / (a million - sin x/cos x) = (cos x / sin x) / [(cos x - sin x) / cos x] invert and multiply: = (cos x / sin x)*cos x /(cos x - sin x) = cos^2 x /sin x (cos x - sin x) Now upload the two expressions: sin^2 x /cos x (sin x - cos x) + cos^2 x /sin x (cos x - sin x) = sin^3 x - cos^3 x/sin x cos x (sin x - cos x) factorise distinction of two cubes in good line: = (sin x - cos x) (sin^2 x + sinx cosx + cos^2x)/sin x cos x (sin x - cos x) cancel (sin x - cos x): = (sin^2 x + sinx cosx + cos^2x)/sin x cos x = (a million + sinx cosx)/sinxcosx or a million + (a million/2) sin 2x / (a million/2) sin 2x = (2 + sin2x)/sin 2x

Answer 5

LHS
top line , N
(1 + sinx / cosx )
(cos x + sin x) / ( cos x )
bottom line , D
1 + cot x
1 + cos x / sin x
(sin x + cos x) / (sin x)
N/D = sin x / cos x = tan x

RHS
1 / (cos x) / 1 / sin x
sin x / cos x = tan x

LHS = RHS