Question

How do I solve this problem?Prove the two expressions are equal

Answer 1

See the explanation for the proof.

Step-by-step explanation

First we have to write the secant and the cotangent in terms of sine and cosine:

`\sec x=(1)/(\cos x)`
`\cot x=(\cos x)/(\sin x)`

Replace into the equation,

`-2\mleft((\cos x)/(\sin x)\mright)^2=(1)/((1)/(\cos x)+1)-(1)/((1)/(\cos x)-1)`

On the left side distribute the square and on the right side do the addition and subtraction in the denominators,

`-2(\cos^2x)/(\sin^2x)^{}=(1)/((1+\cos x)/(\cos x))-(1)/((1-\cos x)/(\cos x))`

1 over a fraction is equivalent to the reciprocal of the fraction,

`-2(\cos^2x)/(\sin^2x)^{}=(\cos x)/(1+\cos x)-(\cos x)/(1-\cos x)`

Now do the subtraction on the right side,

`-2(\cos^2x)/(\sin^2x)^{}=(\cos x(1-\cos x)-\cos x(1+\cos x))/((1+\cos x)(1-\cos x))`

Note that the denominator is a difference of two squares,

`-2(\cos^2x)/(\sin^2x)^{}=(\cos x(1-\cos x)-\cos x(1+\cos x))/(1-\cos ^2x)`

Apply the distributive property on the two terms of the numerator,

`-2(\cos^2x)/(\sin^2x)^{}=(\cos x-\cos ^2x-\cos x-\cos ^2x)/(1-\cos ^2x)`
`-2(\cos^2x)/(\sin^2x)^{}=(-\cos ^2x-\cos ^2x)/(1-\cos ^2x)`
`-2(\cos^2x)/(\sin^2x)^{}=(-2\cos ^2x)/(1-\cos ^2x)`

Now, for the denominator remember the trigonometric identity

`\sin ^2\alpha+\cos ^2\alpha=1`

If we solve it for sin²α

`\sin ^2\alpha=1-\cos ^2\alpha`

Therefore, the expression we have in the denominator of the right side of the equation is equivalent to sin²x,

`-2(\cos^2x)/(\sin^2x)^{}=-2(\cos ^2x)/(\sin ^2x)`

Now we have on both sides,

`-2\cot ^2x=-2\cot ^2x`

Hence, we have proven that the two expressions are equivalent.