Question

1/sec (x) - 1 + 1/sec (x)+1=2cos(x)cosec²(x). solve that or prove it​

Answer 1

Proof in explanation.

Explanation:

`(1)/(\sec(x)-1)+(1)/(\sec(x)+1)`

I'm going to find a way to combine the fractions as one.

Multiply the first fraction by
`(\sec(x)+1)/(\sec(x)+1)` and multiply the second fraction by
`(\sec(x)-1)/(\sec(x)-1)`.

`(\sec(x)-1)/((\sec(x)-1)(\sec(x)+1))+(\sec(x)+1)/((\sec(x)+1)(\sec(x)-1))`

`(\sec(x)-1+\sec(x)+1)/((\sec(x)+1)(\sec(x)-1))`

The bottom product can easily be determine since when multiplying the conjugate of
`a+b` which
`a-b`, we only have to multiply first terms and then last terms giving us
`a^2-b^2`.

`(2\sec(x))/(\sec^2(x)-1)`

Recall the Pythagorean Identity:
`1+\tan^2(x)=\sec^2(x)`

So I can replace
`\sec^2(x)-1` with
`\tan^2(x)`:

`(2\sec(x))/(\tan^2(x))`

`2\sec(x) \cdot (1)/(\tan^2(x))`

`2\sec(x) \cdot (\cos^2(x))/(\sin^2(x))`

`2 \sec(x) \cos(x) \cos(x) (1)/(\sin^2(x))`

`2 (\sec(x) \cos(x)) \cos(x) (1)/(\sin^2(x))`

`2(1)\cos(x)\csc^2(x)`

`2\cos(x)\csc^2(x)`