Establish the identity. ((cot^2 x)/(csc(x)-1)) = (1+sin(x))/(sin(x)) = ____________________ Could someone explain to me how to solve this?

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Answer 1 · 2018-08-03T22:56:59

Follow proof below.

We have: $\frac{{cot}^{2} (x)}{csc (x) - 1}$ $frac(cot^(2)(x))(csc(x) - 1)$

One of the Pythagorean identities is ${cot}^{2} (x) + 1 = {csc}^{2} (x)$ $cot^(2)(x) + 1 = csc^(2)(x)$ .

We can rearrange it to get:

$\Rightarrow {cot}^{2} (x) = {csc}^{2} (x) - 1$ $Rightarrow cot^(2)(x) = csc^(2)(x) - 1$

Let's apply this rearranged identity to our proof:

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

The numerator is the difference of two squares, and can be factorised as:

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

$= csc (x) + 1$ $= csc(x) + 1$

Now, one of the standard trigonometric identities is $csc (x) = \frac{1}{sin (x)}$ $csc(x) = frac(1)(sin(x))$ .

Applying this, we get:

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

$= \frac{1 + sin (x)}{sin (x)}$ $= frac(1 + sin(x))(sin(x)) " " "$ $Q.E.D.$ $"Q.E.D."$