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

Apr 8, 2017

You need to know that $\csc x = \frac{1}{\sin} x$, that $\sec x = \frac{1}{\cos} x$, and the formula ${\sin}^{2} x + {\cos}^{2} x = 1$

Explanation:

I assume you meant:
$\frac{\csc x}{1 + \csc x} - \frac{\csc x}{1 - \csc x} = 2 {\sec}^{2} x$

These cosecant ($\csc$) everywhere are suspicious,
so let's change them since we know:
$\csc x = \frac{1}{\sin} x$

Let's transform our left-hand-side of the equation:
$\frac{\csc x}{1 + \csc x} - \frac{\csc x}{1 - \csc x}$

Let's first put everything under the same denominator:
$= \frac{\csc x \cdot \left(1 - \csc x\right) - \csc x \cdot \left(1 + \csc x\right)}{\left(1 + \csc x\right) \left(1 - \csc x\right)}$

and simplifying a bit, we get:
$= \frac{\csc x - {\csc}^{2} x - \csc x - {\csc}^{2} x}{1 - {\csc}^{2} x}$

so
$= \frac{- 2 {\csc}^{2} x}{1 - {\csc}^{2} x}$

This then becomes
$= \frac{- \frac{2}{\sin} ^ 2 x}{1 - \frac{1}{\sin} ^ 2 x}$

Again simplifying, we get:
$= \frac{- \frac{2}{\sin} ^ 2 x}{\frac{{\sin}^{2} x - 1}{\sin} ^ 2 x}$

We cancel the $\frac{1}{\sin} ^ 2 x$ from the numerator and denominator and we get:
$= - \frac{2}{{\sin}^{2} x - 1}$

Remember that there is this wonderful formula ${\sin}^{2} x + {\cos}^{2} x = 1$. Then our expression becomes:
$= - \frac{2}{{\sin}^{2} x - {\sin}^{2} x - {\cos}^{2} x}$
$= \frac{2}{\cos} ^ 2 x$
and since $\cos x = \frac{1}{\sec} x$, we have

$= 2 {\sec}^{2} x$

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

Q.E.D.

Apr 8, 2017

Explanation:

$\csc \frac{x}{1 + \csc x} - \csc \frac{x}{1 - \csc x}$

=$\frac{\frac{1}{\sin} x}{1 + \frac{1}{\sin} x} - \frac{\frac{1}{\sin} x}{1 - \frac{1}{\sin} x}$

Multiplying each term by $\sin x$ we get

$\frac{1}{\sin x + 1} - \frac{1}{\sin x - 1}$

= $\frac{1}{1 + \sin x} + \frac{1}{1 - \sin x}$

= $\frac{1 - \sin x + 1 + \sin x}{1 - {\sin}^{2} x}$

= $\frac{2}{\cos} ^ 2 x$

= $2 {\sec}^{2} x$

Apr 8, 2017

Proved in the explanation.

Explanation:

Prove: $\frac{\csc \left(x\right)}{1 + \csc \left(x\right)} - \frac{\csc \left(x\right)}{1 - \csc \left(x\right)} = 2 {\sec}^{2} \left(x\right)$

Multiply the left side of the equation by 1 in the form, $\sin \frac{x}{\sin} \left(x\right)$

$\sin \frac{x}{\sin} \left(x\right) \left(\frac{\csc \left(x\right)}{1 + \csc \left(x\right)} - \frac{\csc \left(x\right)}{1 - \csc \left(x\right)}\right) = 2 {\sec}^{2} \left(x\right)$

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

$\frac{1}{\sin \left(x\right) + 1} - \frac{1}{\sin \left(x\right) - 1} = 2 {\sec}^{2} \left(x\right)$

Multiply the first term by 1 in the form $\frac{\sin \left(x\right) - 1}{\sin \left(x\right) - 1}$:

$\frac{1}{\sin \left(x\right) + 1} \frac{\sin \left(x\right) - 1}{\sin \left(x\right) - 1} - \frac{1}{\sin \left(x\right) - 1} = 2 {\sec}^{2} \left(x\right)$

This makes the denominator become the difference of two squares:

$\frac{\sin \left(x\right) - 1}{{\sin}^{2} \left(x\right) - 1} - \frac{1}{\sin \left(x\right) - 1} = 2 {\sec}^{2} \left(x\right)$

Multiply the second term by 1 in the form $\frac{\sin \left(x\right) + 1}{\sin \left(x\right) + 1}$:

$\frac{\sin \left(x\right) - 1}{{\sin}^{2} \left(x\right) - 1} - \frac{1}{\sin \left(x\right) - 1} \frac{\sin \left(x\right) + 1}{\sin \left(x\right) + 1} = 2 {\sec}^{2} \left(x\right)$

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

$\frac{\sin \left(x\right) - 1}{{\sin}^{2} \left(x\right) - 1} - \frac{\sin \left(x\right) + 1}{{\sin}^{2} \left(x\right) - 1} = 2 {\sec}^{2} \left(x\right)$

Put both numerators over the common denominator:

$\frac{\sin \left(x\right) - 1 - \sin \left(x\right) - 1}{{\sin}^{2} \left(x\right) - 1} = 2 {\sec}^{2} \left(x\right)$

Combine like terms:

$- \frac{2}{{\sin}^{2} \left(x\right) - 1} = 2 {\sec}^{2} \left(x\right)$

Substitute $- {\cos}^{2} \left(x\right)$ for $\left({\sin}^{2} \left(x\right) - 1\right)$:

$- \frac{2}{-} {\cos}^{2} \left(x\right) = 2 {\sec}^{2} \left(x\right)$

Because $\frac{1}{\cos} \left(x\right) = \sec \left(x\right)$, the left side becomes the same as the right:

$2 {\sec}^{2} \left(x\right) = 2 {\sec}^{2} \left(x\right)$

Q.E.D.

Apr 8, 2017

Make a common denominator.

Explanation:

$\frac{\csc x}{1 + \csc x}$ - $\frac{\csc x}{1 - \csc x}$ = $2 {\sec}^{2} x$

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

$\frac{\left(\csc x\right) \left(1 - \csc x\right) - \left(\csc x\right) \left(1 + \csc x\right)}{\left(1 + \csc x\right) \left(1 - \csc x\right)}$

= $\frac{\csc x - {\csc}^{2} x - \csc x - {\csc}^{2} x}{1 - \csc x + \csc x - {\csc}^{2} x}$

= $\frac{- 2 {\csc}^{2} x}{1 - {\csc}^{2} x}$

= $- \frac{2}{\sin} ^ 2 x$ / $1 - \frac{1}{\sin} ^ 2 x$

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

= $- \frac{2}{\sin} ^ 2 x$ / $\frac{{\sin}^{x} - 1}{\sin} ^ 2 x$

(Common denominator again by multiplying 1 by ${\sin}^{2} x$)

= $- \frac{2}{\sin} ^ 2 x$ * ${\sin}^{2} \frac{x}{{\sin}^{2} x - 1}$

(switch numerator & denominator and cancel out common terms)

= $- \frac{2}{{\sin}^{2} x - 1}$

= $- \frac{2}{-} {\cos}^{2} x$

(${\sin}^{2} x + {\cos}^{2} x = 1$) Hence, ${\sin}^{2} x - 1 = - {\cos}^{2} x$

= $\frac{2}{\cos} ^ 2 x$

= $2 {\sec}^{2} x$