Question

Prove that `(cscx)/(1+cscx)-(cscx)/(1-cscx)=2sec^2x`?

Answer 1

You need to know that `csc x = 1/sin x`, that `sec x = 1/cos x`, and the formula `sin^2 x + cos^2 x =1`

Explanation:

I assume you meant:
`(csc x) / (1 + csc x) - (csc x) / (1 - csc x) = 2 sec^2 x`

These cosecant (`csc`) everywhere are suspicious,
so let's change them since we know:
`csc x = 1/sin x`

Let's transform our left-hand-side of the equation:
`(csc x) / (1 + csc x) - (csc x) / (1 - csc x)`

Let's first put everything under the same denominator:
`= ( csc x*(1-csc x) - csc x * (1+csc x ))/( (1+csc x) (1-csc x) )`

and simplifying a bit, we get:
`= ( csc x-csc^2 x - csc x - csc^2 x ) / (1 - csc^2 x)`

so
`= (-2csc^2 x)/(1-csc^2 x)`

This then becomes
`= (-2/sin^2 x)/(1-1/sin^2 x)`

Again simplifying, we get:
`= (-2/sin^2 x)/( (sin^2 x -1)/sin^2 x )`

We cancel the `1/sin^2 x` from the numerator and denominator and we get:
`=-2/(sin^2x-1)`

Remember that there is this wonderful formula `sin^2 x + cos^2 x = 1`. Then our expression becomes:
`=-2/(sin^2x-sin^2 x-cos^2 x)`
`=2/cos^2 x`
and since `cos x = 1/sec x`, we have

`=2sec^2 x`

which is exactly the right-hand-side of the equation.

Q.E.D.

Answer 2

Please see below.

Explanation:

`cscx/(1+cscx)-cscx/(1-cscx)`

=`(1/sinx)/(1+1/sinx)-(1/sinx)/(1-1/sinx)`

Multiplying each term by `sinx` we get

`1/(sinx+1)-1/(sinx-1)`

= `1/(1+sinx)+1/(1-sinx)`

= `(1-sinx+1+sinx)/(1-sin^2x)`

= `2/cos^2x`

= `2sec^2x`

Answer 3

Proved in the explanation.

Explanation:

Prove: `(csc(x)) / (1 + csc(x)) - (csc(x)) / (1 - csc(x)) = 2 sec^2(x)`

Multiply the left side of the equation by 1 in the form, `sin(x)/sin(x)`

`sin(x)/sin(x)((csc(x)) / (1 + csc(x)) - (csc(x)) / (1 - csc(x))) = 2 sec^2(x)`

Please observe that, when the sine function is distributed, everywhere there was `csc(x)` becomes 1 and everywhere there was 1 becomes `sin(x)`

`1 / (sin(x) + 1) - 1/ (sin(x) - 1) = 2 sec^2(x)`

Multiply the first term by 1 in the form `(sin(x) - 1)/(sin(x) - 1)`:

`1 / (sin(x) + 1)(sin(x) - 1)/(sin(x) - 1) - 1/ (sin(x) - 1) = 2 sec^2(x)`

This makes the denominator become the difference of two squares:

`(sin(x) - 1)/(sin^2(x) - 1) - 1/ (sin(x) - 1) = 2 sec^2(x)`

Multiply the second term by 1 in the form `(sin(x) + 1)/(sin(x) + 1)`:

`(sin(x) - 1)/(sin^2(x) - 1) - 1/ (sin(x) - 1)(sin(x) + 1)/(sin(x) + 1) = 2 sec^2(x)`

This makes the denominator become the same difference of two squares:

`(sin(x) - 1)/(sin^2(x) - 1) - (sin(x) + 1)/ (sin^2(x) - 1) = 2 sec^2(x)`

Put both numerators over the common denominator:

`(sin(x) - 1 - sin(x) - 1)/ (sin^2(x) - 1) = 2 sec^2(x)`

Combine like terms:

`-2/ (sin^2(x) - 1) = 2 sec^2(x)`

Substitute `-cos^2(x)` for `(sin^2(x) - 1)`:

`-2/-cos^2(x) = 2 sec^2(x)`

Because `1/cos(x) = sec(x)`, the left side becomes the same as the right:

`2sec^2(x) = 2 sec^2(x)`

Q.E.D.

Answer 4

Make a common denominator.

Explanation:

`(csc x) / (1 + cscx)` - `(csc x)/(1-csc x)` = `2 sec^2 x`

Taking the left hand side & multiplying to get a common denominator,

`((csc x)(1-csc x) - (csc x)(1+cscx)) / ((1+csc x)(1-cscx))`

= `(csc x - csc^2 x - csc x - csc^2 x) / (1 - csc x + csc x - csc^2 x)`

= `(-2 csc^2 x)/(1 - csc^2 x)`

= `-2/sin^2 x` / `1 - 1/sin^2 x`

( / is the divide symbol, easier to see when not in fraction form)

= `-2/sin^2 x` / `(sin^x - 1)/sin^2 x`

(Common denominator again by multiplying 1 by `sin^2 x`)

= `-2/sin^2 x` * `sin^2x / (sin^2 x - 1)`

(switch numerator & denominator and cancel out common terms)

= `-2/(sin^2 x-1)`

= `-2/-cos^2 x`

(`sin^2 x + cos ^2 x = 1`) Hence, `sin^2 x - 1 = -cos^2 x`

= `2/cos^2 x`

= `2 sec^2 x`