Question

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

Answer 1

Are you supposed to prove that is an identity or something?

Answer 2

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

Answer 3

ok, so we just need to manipulate the left side to prove it = right side,

Answer 4

let us start with LHS

take the LCM and solve to beign with

Answer 5

try this :
\(\dfrac{\sin x}{1-\cos x}\dfrac{1+\cos x}{1+\cos x}+\dfrac{\sin x}{1-\cos x}\dfrac{1+\cos x }{1+\cos x}\)

Answer 6

now you have a common denominator :)
so that you can combine the numerator.

Answer 7

what did u get in numerator ?

Answer 8

\(\dfrac{\sin x}{1-\cos x}\dfrac{1+\cos x}{1+\cos x}+\dfrac{\sin x}{1+\cos x}\dfrac{1-\cos x }{1-\cos x}\)

Answer 9

i corrected it^
did you get it ?

Answer 10

I haven't done it lol

Answer 11

\(\dfrac{\sin x}{1-\cos x}\dfrac{1+\cos x}{1+\cos x}+\dfrac{\sin x}{1+\cos x}\dfrac{1-\cos x }{1-\cos x} \\ = \dfrac{\sin x(1+\cos x)+\sin x(1-\cos x)}{(1-\cos x)(1+\cos x)} \)

Answer 12

so youre left with sin x ?

Answer 13

2 sin x in numerator

denominator = \(1-\cos^2 x = \sin^2x\)

Answer 14

So 1- cos^2 x = sin^2 x

Answer 15

jk so it's
2 sin x over 1 - cos^2 x = sin^2 x ?

Answer 16

yeah,
\(\dfrac{2\cancel {\sin x}}{\sin^\cancel{2}x} = \dfrac{2}{\sin x} = 2 \csc x = Right \: \: side\)

Answer 17

can you write out the full answer, I'm thoroughly confused

Answer 18

\(Left \: \: side =\\\dfrac{\sin x}{1-\cos x}\dfrac{1+\cos x}{1+\cos x}+\dfrac{\sin x}{1+\cos x}\dfrac{1-\cos x }{1-\cos x} \\ = \dfrac{\sin x(1+\cos x)+\sin x(1-\cos x)}{(1-\cos x)(1+\cos x)} \\ = \dfrac{\sin x +\sin x \cos x +\sin x - \sin x \cos x}{1-\cos^2x} \\ =\dfrac{2\sin x }{\sin^2x} =\dfrac{2\cancel {\sin x}}{\sin^\cancel{2}x} = \dfrac{2}{\sin x} = 2 \csc x = Right \: \: side\)

Answer 19

Thank you (:

Answer 20

welcome ^_^