We have: frac(tan^(2)(x) csc^(2)(x) - 1)(csc(x) tan^(2)(x) sin(x))tan2(x)csc2(x)−1csc(x)tan2(x)sin(x)
Let's apply the standard trigonometric identities tan(x) = frac(sin(x))(cos(x))tan(x)=sin(x)cos(x) and csc(x) = frac(1)(sin(x)csc(x)=1sin(x):
= frac(frac(sin^(2)(x))(cos^(2)(x)) cdot frac(1)(sin^(2)(x)) - 1)(csc(x) tan^(2)(x) sin(x))=sin2(x)cos2(x)⋅1sin2(x)−1csc(x)tan2(x)sin(x)
= frac(frac(1)(cos^(2)(x)) - 1)(csc(x) tan^(2)(x) sin(x))=1cos2(x)−1csc(x)tan2(x)sin(x)
Then, let's apply another standard trigonometric identity; sec(x) = frac(1)(cos(x))sec(x)=1cos(x):
= frac(sec^(2)(x) - 1)(csc(x) tan^(2)(x) sin(x))=sec2(x)−1csc(x)tan2(x)sin(x)
One of the Pythagorean identities is tan^(2)(x) + 1 = sec^(2)(x)tan2(x)+1=sec2(x).
We can rearrange it to get:
Rightarrow tan^(2)(x) = sec^(2)(x) - 1⇒tan2(x)=sec2(x)−1
Let's apply this rearranged identity to our proof:
= frac(tan^(2)(x))(csc(x) tan^(2)(x) sin(x))=tan2(x)csc(x)tan2(x)sin(x)
= frac(1)(csc(x) sin(x))=1csc(x)sin(x)
Finally, let's apply the standard trigonometric identity csc(x) = frac(1)(sin(x))csc(x)=1sin(x):
= frac(1)(frac(1)(sin(x)) cdot sin(x))=11sin(x)⋅sin(x)
= frac(1)(1)=11
= 1=1
therefore frac(tan^(2)(x) csc^(2)(x) - 1)(csc(x) tan^(2)(x) sin(x)) = 1
