rewrite 1/secx-1 so thta it's not a fraction?

Answer 1

1/[ sec(x) - 1 ]

Multiply the top and bottom by the conjugate of the bottom,
[sec(x) + 1]

[sec(x) + 1] / ( [sec(x) - 1] [sec(x) + 1] )

Now, the denominator becomes a difference of squares.

[sec(x) + 1] / [sec^2(x) - 1]

Apply the identity tan^2(x) = sec^2(x) - 1.

[sec(x) + 1] / tan^2(x)

Write this into two fractions.

sec(x)/tan^2(x) + 1/tan^2(x)

Write the first fraction in terms of sine and cosine.

[1/cos(x)] / [sin^2(x)/cos^2(x)] + 1/tan^2(x)

Multiply the top and bottom by cos^2(x) to eliminate the complex fraction.

[cos(x)] / [sin^2(x)] + 1/tan^2(x)

Split into a product of fractions.

[cos(x)/sin(x)] [1/sin(x)] + 1/tan^2(x)

1/tan^2(x) is equal to cot^2(x) {being reciprocals and all}. The rest can be evaluated by definition.

cot(x) csc(x) + cot^2(x)

Answer 2

I'm assuming that you mean 1/(sec x -1) and not (1/sec x) -1=cos x -1.

Multiply and divide by sec x +1. Then notice that sec^2 x -1=tan^2 x. THis gives [sec x +1]/tan^2 x=csc x cot x +cot^2 x.

Answer 3

I can't tell whether you mean (1/sec x) - 1 or 1/(sec x - 1). Technically it is the first, but that seems a bit too easy. But here are answers for both:

1/sec x - 1 = cos x - 1

1/(sec x - 1) = (sec x + 1) / [(sec x + 1) (sec x - 1)]
= (sec x + 1) / (sec^2 x - 1)
= (sec x + 1) / tan^2 x
= (sec x + 1) cot^2 x.