Question

sinx/1-cosx +sinx/1+cosx = 2cscx

Answer 1

Combine the fractions by finding a common denominator.

Answer 2

\[\frac{\sin x}{1-\cos x}+\frac{\sin x}{1+\cos x}=2\csc x,~\text{right?}\]

Answer 3

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

Answer 4

It's just a matter of simplifying now...

Answer 5

ok thank you

Answer 6

i got 2sinx

Answer 7

Nope, try again

Answer 8

$$\frac2{\sin x}=2\frac1{\sin x}=2\csc x$$

Answer 9

how do you get sinx as denomenator

Answer 10

\(\sin^2 x + \cos^2 x = 1\)

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

Answer 11

You have : (1-cos x)(1+cos x) as denominator.

Use : (a+b)(a-b) = a^2 - b^2 identity.

You get : \((1-\cos x)(1 + \cos x) = 1- \cos^2 x\)

As : \(1-\cos^2 x = \sin^2 x \) Therefore, you get denominator as : \(\sin^2 x\)

Answer 12

In the numerator you had : \(2\sin x\) while in the denominator you have : \(\sin^2 x\)

\(\cfrac{2\sin x}{\sin^2 x} \)

Answer 13

Cancel the terms and simplify.

Answer 14

$$(1-\cos x)(1+\cos x)=1-\cos^2x=\sin^2x$$ is what our denominator simplifies to

Answer 15

$$(1+\cos x)\sin x+(1-\cos x)\sin x=(1+\cos x+1-\cos x)\sin x=2\sin x$$is our numerator

Answer 16

^^

Answer 17

$$\frac{2\sin x}{\sin^2 x}=\frac2{\sin x}=2\frac1{\sin x}=2\csc x$$