Prove that $1/(\sin x + 1) - 1/(\sin x - 1) = 2 \sec^2 (x)$

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Can anyone solve this for me?

Prove that $\frac1{\sin x + 1} - \frac1{\sin x - 1} = 2 \sec^2 (x)$.

This is as far as I went:

$$\frac{(sin x - 1) - (sin x + 1)}{(sin x + 1)(sin x - 1)}$$

$$=\frac{sin x -1 - sin x -1}{sin^2 (x) + 1^2 }$$

$$=\frac{-2}{sin^2 (x) + 1}$$

What to do next?

I am new here, and currently in high-school, so don't make it too complicated. Thanks in advance.

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1 Answer

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After finding a common denominator and simplifying the numerator, the LHS becomes

$$ \dfrac{-2}{(\sin(x)-1)(\sin(x)+1)}$$

The denominator is a difference of squares, and using the pythagorean identity will yield the desired result.

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