How do I simplify: [(tan^2x)(csc^2x)-1] / [(cscx)(tan^2x)(sinx)] ?

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Answer 1 · 2017-07-11T00:55:10

$frac(tan^(2)(x) csc^(2)(x) - 1)(csc(x) tan^(2)(x) sin(x)) = 1$

We have: $frac(tan^(2)(x) csc^(2)(x) - 1)(csc(x) tan^(2)(x) sin(x))$

Let's apply the standard trigonometric identities $tan(x) = frac(sin(x))(cos(x))$ and $csc(x) = frac(1)(sin(x)$ :

$= frac(frac(sin^(2)(x))(cos^(2)(x)) cdot frac(1)(sin^(2)(x)) - 1)(csc(x) tan^(2)(x) sin(x))$

$= frac(frac(1)(cos^(2)(x)) - 1)(csc(x) tan^(2)(x) sin(x))$

Then, let's apply another standard trigonometric identity; $sec(x) = frac(1)(cos(x))$ :

$= frac(sec^(2)(x) - 1)(csc(x) tan^(2)(x) sin(x))$

One of the Pythagorean identities is $tan^(2)(x) + 1 = sec^(2)(x)$ .

We can rearrange it to get:

$Rightarrow tan^(2)(x) = sec^(2)(x) - 1$

Let's apply this rearranged identity to our proof:

$= frac(tan^(2)(x))(csc(x) tan^(2)(x) sin(x))$

$= frac(1)(csc(x) sin(x))$

Finally, let's apply the standard trigonometric identity $csc(x) = frac(1)(sin(x))$ :

$= frac(1)(frac(1)(sin(x)) cdot sin(x))$

$= frac(1)(1)$

$= 1$

$therefore frac(tan^(2)(x) csc^(2)(x) - 1)(csc(x) tan^(2)(x) sin(x)) = 1$