Question

Question #f4b07

Answer 1

I tried this:

Explanation:

Have a look:

Answer 2

Please see the proofs below.

Explanation:

Verify:

`cos(u)/(1-sin(u))=(1+sin(u))/cos(u)`

Multiply the left side by 1 in the form of `(1 + sin(u))/(1 + sin(u))`:

`(1 + sin(u))/(1 + sin(u))cos(u)/(1-sin(u))=(1+sin(u))/cos(u)`

The denominator becomes the difference of two squares:

`((1 + sin(u))cos(u))/(1-sin^2(u))=(1+sin(u))/cos(u)`

Substitute `cos^2(u)` for `(1-sin^2(u))`:

`((1 + sin(u))cos(u))/cos^2(u)=(1+sin(u))/cos(u)`

The cosine in the numerator cancels one of the cosines in the denominator:

`(1 + sin(u))/cos(u)=(1+sin(u))/cos(u)`

Verified.

Verify:

`tan^2(x)/(sec(x)+1) = sec(x)-1`

Multiply the left side by 1 in the form `(sec(x)-1)/(sec(x)-1)`

`(sec(x)-1)/(sec(x)-1)tan^2(x)/(sec(x)+1) = sec(x)-1`

The numerator becomes the difference of two squares:

`(sec(x)-1)tan^2(x)/(sec^2(x)-1) = sec(x)-1`

Substitute `tan^2(x)` for `sec^2(x)-1`

`(sec(x)-1)tan^2(x)/(sec^2(x)-1) = sec(x)-1`

The fraction becomes 1:

`sec(x)-1 = sec(x)-1`

Verified.