Proving these trig identities??-Grade 12

cot x tan x sec x cos x =1

cscx/secx=cotx

secx(1+cosx)=secx+1

and (1-sin^2x)(1+cot^2x)=cot^2x

cot x tan x sec x cos x =1

Left Hand Side:
= cot x tan x sec x cos x
= (cosx/sinx) (sinx/cosx) (1/cosx) (cosx) << they all cancel each other so..
= 1 = Right Hand Side


cscx/secx=cotx

LHS = cscx/secx
= (1/sinx) / (1/cosx)
= (1/sinx) * (cosx/1)
= cotx = RHS


secx(1+cosx)=secx+1

LHS = secx(1+cosx)
= (1/cosx) * (1+cosx)
= (1/cosx) + (cosx/cosx) << Notice that they were multiplied by each other
= secx + 1 = RHS


and (1-sin^2x)(1+cot^2x)=cot^2x

LHS = (1-sin^2x) * (1+cot^2x)
= 1 + cot^2x - sin^2x - (sin^2x * cot^2x)
= 1 + cot^2x - sin^2x - (sin^2x * (cos^2x/sin^2x))
= 1 + cot^2x - sin^2x - cos^2x
= 1 + cot^2x - (sin^2x + cos^2x)
= 1 + cot^2x - (1)
= cot^2x = RHS


Rules I used:
cotx = (cosx/sinx)
tanx = (sinx/cosx)
secx = (1/cosx)
cscx = (1/sinx)
sin^2x + cos^2x = 1

(1) cot x = 1/(tan x), and sec x = 1/(cos x) so you have [tan x]/[tan x] * [cos x]/[cos x] = 1 * 1 = 1
(2) RHS = [cos x]/[sin x] = 1/[sin x] * [cos x] = [csc x]/[sec x] = LHS
(3) secx(1 + cosx) = secx + secx*cosx = secx + (1/cosx)*cosx = secx + 1
(4) LHS = cos^x[1 + cos^x/sin^x] = {cos^2x/sin^2x}[sin^2x + cos^2x) = [cot^2x]*1 = RHS

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

cscx/secx =
(1/sinx)/(1/cosx) =
cosx/sinx =
cotx

secx(1 + cosx) =
secx + secxcosx =
secx + cosx/cosx =
secx + 1

(1 - sin^2x)(1 + cot^2x) =
cos^2x csc^2x =
cos^2x/sin^2x =
cot^2x

Look it up on google