Pls help me prove any of these identities...? 1] tanx/(1-cotx) + cotx/(1-tanx) = 1 + tanx + cotx?

Question

Pls help me prove these identities...?
1] tanx/(1-cotx) + cotx/(1-tanx) = 1 + tanx + cotx

2] 1/(cosx-cotx)² = (1+cosx)/(1-cosx)

3] 1/(cosx-cotx) = (1+cosx)/(1-cosx)

Answer 1

1.)
(tan(x)/(1 - cot(x)) + (cot(x)/(1 - tan(x)) = 1 + tan(x) + cot(x)

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

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

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

((tan(x)^2)/(tan(x) - 1)) - (1/(tan(x))/(tan(x) - 1))) = 1 + tan(x) + cot(x)

((tan(x)^2)/(tan(x) - 1)) - (1/(tan(x) * (tan(x) - 1))) = 1 + tan(x) + cot(x)

Multiply the left side by (tan(x))(tan(x) - 1)

(tan(x)^3 - 1)/((tan(x))(tan(x) - 1)) = 1 + tan(x) + cot(x)

((tan(x) - 1)(tan(x)^2 + tan(x) + 1))/(tan(x) * (tan(x) - 1)) = 1 + tan(x) + cot(x)

(tan(x)^2 + tan(x) + 1)/(tan(x)) = 1 + tan(x) + cot(x)

(tan(x)^2/tan(x)) + (tan(x)/(tan(x)) + (1/(tan(x))) = 1 + tan(x) + cot(x)

tan(x) + 1 + cot(x) = 1 + tan(x) + cot(x)

or in this case

1 + tan(x) + cot(x) = 1 + tan(x) + cot(x)

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2 and 3 are equivalent, so they can't be proved, they can only be disproved.

Answer 2

maybe I can get you started my answers are usually for algebra its just that maybe I see something (1+tanx) / (1+cotx) = (1-tanx) / (cotx-1) the right denominator 1/(cotx-1)=1/(-1+cotx)= -1/(1-cotx) so (1-tanx)/(cotx-1)= -(1-tanx)/(1-cotx) gives you (1+tanx) / (1+cotx)= -(1-tanx)/(1-cotx) now you can make the denominators into 'difference of squares' left side (1+tanx)/(1+cotx)*(1-cotx)/(1-cotx)=(1... right side -(1-tanx)/(1-cotx)*(1+cotx)/(1+cotx)= -(1-tanx)(1+cotx)/(1-cotx^2) the (1-cotx^2)'s cancel leaving (1+tanx)(1-cotx)= -(1-tanx)(1+cotx) maybe you can take it from there? good luck

Answer 3

thereis a typo in your sum
the sign of separator between the firstand scond term should be minus
1.{tan^2x/tanx-1}-{1/tanx(tanx-1)}
={tan^3x-1}/tanx(1-tanx)
=(tanx-1)(tan^2x+tanx+1)/tanx(tanx-1)
=tan^2x+tanx+1/tanx
=tanx+1+cotx

2.let x=pi/6
cosx-cotx
=rt3/2-rt3
=(rt3-2rt3)/2
1/cosx-cotx)^2
=4/(3+12-12)
=4/3
(1+cosx/1-cosx)=1+rt3/2/1-rt3/2
=2+rt3/2-rt3
=(2+rt3)^2/1
=4+4rt3+3
=7+rt3
LHS is notequal to RHS
so the sum is not right

check all the sums and post them again