To prove that [tex]\(\frac{\sin ^2 x-\cos ^2 x}{\cos x\left[\sin \left(180^{\circ}-x\right)-\cos x \right]}-1=\tan x\)[/tex], we'll proceed step by step.
First, let's simplify the given expression:
[tex]\[
\frac{\sin ^2 x-\cos ^2 x}{\cos x\left[\sin \left(180^{\circ}-x\right)-\cos x \right]}-1
\][/tex]
We know that:
[tex]\[
\sin (180^\circ - x) = \sin x
\][/tex]
So we can substitute [tex]\(\sin (180^\circ - x)\)[/tex] with [tex]\(\sin x\)[/tex]:
[tex]\[
\frac{\sin ^2 x-\cos ^2 x}{\cos x\left[\sin x - \cos x \right]}-1
\][/tex]
Now we need to simplify the fraction [tex]\(\frac{\sin^2 x - \cos^2 x}{\cos x(\sin x - \cos x)}\)[/tex].
Notice that [tex]\(\sin^2 x - \cos^2 x\)[/tex] can be factored using the difference of squares:
[tex]\[
\sin^2 x - \cos^2 x = (\sin x - \cos x)(\sin x + \cos x)
\][/tex]
Thus, the expression becomes:
[tex]\[
\frac{(\sin x - \cos x)(\sin x + \cos x)}{\cos x(\sin x - \cos x)}
\][/tex]
We see that [tex]\(\sin x - \cos x\)[/tex] appears in both the numerator and the denominator, so they cancel each other out:
[tex]\[
\frac{\sin x + \cos x}{\cos x}
\][/tex]
Simplifying further:
[tex]\[
\frac{\sin x}{\cos x} + \frac{\cos x}{\cos x}
\][/tex]
[tex]\[
= \tan x + 1
\][/tex]
Now we return to the original expression and subtract 1:
[tex]\[
\tan x + 1 - 1 = \tan x
\][/tex]
Thus, we have shown that:
[tex]\[
\frac{\sin ^2 x-\cos ^2 x}{\cos x\left[\sin \left(180^{\circ}-x\right)-\cos x\right]}-1 = \tan x
\][/tex]
Hence, the given identity is proven:
[tex]\[
\boxed{\tan x}
\][/tex]
