a) \( \frac{\tan x-\cot x}{\tan x+\cot x}=\operatorname{sen}^{2} x-\cos ^{2} x \)
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To solve the equation \( \frac{\tan x - \cot x}{\tan x + \cot x} = \sin^2 x - \cos^2 x \), we can first express \(\tan x\) and \(\cot x\) in terms of sine and cosine: \[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \] Substituting these into the left side gives: \[ \frac{\frac{\sin x}{\cos x} - \frac{\cos x}{\sin x}}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}} = \frac{\frac{\sin^2 x - \cos^2 x}{\sin x \cos x}}{\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}} = \frac{\sin^2 x - \cos^2 x}{\sin^2 x + \cos^2 x} \] Given that \(\sin^2 x + \cos^2 x = 1\), we can simplify the left side to: \[ \sin^2 x - \cos^2 x \] Thus, our equation becomes: \[ \frac{\sin^2 x - \cos^2 x}{1} = \sin^2 x - \cos^2 x \] Since both sides are equal, our original identity holds true for \(x\) values where \(\sin x \neq 0\) and \(\cos x \neq 0\). This result showcases that the equation is valid for most angles in their respective ranges, excluding specific multiples of \(\frac{\pi}{2}\) where sine and cosine take on integral values.