ANSWER
See the explanation for the proof.
EXPLANATION
First we have to write the secant and the cotangent in terms of sine and cosine:
[tex]\sec x=\frac{1}{\cos x}[/tex][tex]\cot x=\frac{\cos x}{\sin x}[/tex]Replace into the equation,
[tex]-2\mleft(\frac{\cos x}{\sin x}\mright)^2=\frac{1}{\frac{1}{\cos x}+1}-\frac{1}{\frac{1}{\cos x}-1}[/tex]On the left side distribute the square and on the right side do the addition and subtraction in the denominators,
[tex]-2\frac{\cos^2x}{\sin^2x}^{}=\frac{1}{\frac{1+\cos x}{\cos x}}-\frac{1}{\frac{1-\cos x}{\cos x}}[/tex]1 over a fraction is equivalent to the reciprocal of the fraction,
[tex]-2\frac{\cos^2x}{\sin^2x}^{}=\frac{\cos x}{1+\cos x}-\frac{\cos x}{1-\cos x}[/tex]Now do the subtraction on the right side,
[tex]-2\frac{\cos^2x}{\sin^2x}^{}=\frac{\cos x(1-\cos x)-\cos x(1+\cos x)}{(1+\cos x)(1-\cos x)}[/tex]Note that the denominator is a difference of two squares,
[tex]-2\frac{\cos^2x}{\sin^2x}^{}=\frac{\cos x(1-\cos x)-\cos x(1+\cos x)}{1-\cos ^2x}[/tex]Apply the distributive property on the two terms of the numerator,
[tex]-2\frac{\cos^2x}{\sin^2x}^{}=\frac{\cos x-\cos ^2x-\cos x-\cos ^2x}{1-\cos ^2x}[/tex][tex]-2\frac{\cos^2x}{\sin^2x}^{}=\frac{-\cos ^2x-\cos ^2x}{1-\cos ^2x}[/tex][tex]-2\frac{\cos^2x}{\sin^2x}^{}=\frac{-2\cos ^2x}{1-\cos ^2x}[/tex]Now, for the denominator remember the trigonometric identity
[tex]\sin ^2\alpha+\cos ^2\alpha=1[/tex]If we solve it for sin²α
[tex]\sin ^2\alpha=1-\cos ^2\alpha[/tex]Therefore, the expression we have in the denominator of the right side of the equation is equivalent to sin²x,
[tex]-2\frac{\cos^2x}{\sin^2x}^{}=-2\frac{\cos ^2x}{\sin ^2x}[/tex]Now we have on both sides,
[tex]-2\cot ^2x=-2\cot ^2x[/tex]Hence, we have proven that the two expressions are equivalent.
