identities???

Question

how do you prove that (1/cscx-cotx)-(1/cscx+cotx) is equal to (2cotx)

Answer 1

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

(1/cscx-cotx)-(1/cscx+cotx) = 2cotx

LHS
= 1 / (csc x-cot x) - 1 / (csc x+cot x)
= (csc x + cot x - csc x + cot x) / (csc^2 x - cot^2 x)
(used formula (a - b)(a + b) = a^2 - b^2 )

= (2 cot x) / [(1 /sin^2 x) - (cos^2 x / sin^2 x)]
(used formulas:
csc x = 1 / sin x
cot x = cos x / sin x )

= (2 cot x) / [(1 - cos^2 x) / sin^2 x]
= (2 cot x) / [(sin^2 x) / sin^2 x]
(used formula: 1 - cos^2 x = sin^2 x )

= (2 cot x) / [1]
= 2 cot x
= RHS

Answer 2

LHS = 1/cscx - cotx - 1/cscx - cotx
= -2cotx = -RHS
This however is not what you are after.
Have you typed in the question correctly?

Answer 3

You can't.
(1/cscx-cotx)-(1/cscx+cotx)
=1/cscx-cotx -1/cscx-cotx = -2cotx

Answer 4

rewrite cscx as 1/sinx and rewrite cotx as cosx/sinx.
this simplifies to:
sinx-cosx/sinx-sinx-cosx/sinx
which= -2cotx...r u sure 2cotx is the answer instead of -2cotx?