prove that (sec^2(x)) / (sec^2(x) - 1) - 1 = cot^2(x)

[NEHA]

Prove that (sec^2(x)) / (sec^2(x) - 1) - 1 = cot^2(x)

This is what i have done:

. . .[1 / cos^2(x)] / [1 / cos^2(x) - 1] = cot^2(x)

Then I crossed out 1/cos^2(x) from top and bottom, leaving:

. . .-1 - 1 = cot^2(x)

. . .-2 = cot^2(x)

But isn't this wrong?

 

[soroban]

Re: prove

Hello, NEHA!

Be careful with your cancelling . . .

Prove that: \(\displaystyle \L\,\frac{\sec^2x}{\sec^2x\,-1} \,-\,1\:=\:\cot^2 x\)

We have: \(\displaystyle \L\:\frac{\sec^2x}{\tan^2x}\,-\,1\;=\;\frac{\sec^2x\,-\,\tan^2x}{\tan^2x} \;=\;\frac{1}{\tan^2x}\;=\;\cot^2x\)

 

[stapel]

Is it correct to say that 11/12 equals 1/2, because you can cross off the 1's?

So it is correct to say that [1/cos²(x)] / [1/cos²(x) - 1] is somehow equal to 1 (or -1, or 1/1, or whatever it was you did)?

Try converting the denominator, 1/cos²(x) - 1, to a common denominator, and then combine the terms. Only then can you flip the lower fraction and multiply it against the top fraction.

Eliz.