Question

Can someone help me solve this? Thanks!

Answer 1

To verify an identity, one should make changes to only 1 side, until it is identical to the other side; I shall make changes to only the left.

Explanation:

Given:

`sec(theta)/(csc(theta)-cot(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)`

Multiply the first fraction by 1 in the form `(csc(theta)+cot(theta))/(csc(theta)+cot(theta))`

`sec(theta)/(csc(theta)-cot(theta))(csc(theta)+cot(theta))/(csc(theta)+cot(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)`

This makes the denominator become the difference of two squares:

`(sec(theta)(csc(theta)+cot(theta)))/(csc^2(theta)-cot^2(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)`

This particular difference of two squares is the left side of the identity `csc^2(theta)-cot^2(theta) = 1`, therefore, the denominator becomes 1 and disappears:

`sec(theta)(csc(theta)+cot(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)`

Use the identities `sec(theta)csc(theta) =1` and `sec(theta)cot(theta) = csc(theta)`:

`1+csc(theta)-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)`

Multiply the next fraction by 1 in the form `(csc(theta)-cot(theta))/(csc(theta)-cot(theta))`

`1+csc(theta)-sec(theta)/(csc(theta)+cot(theta))(csc(theta)-cot(theta))/(csc(theta)-cot(theta)) = 2csc(theta)`

The produces the same difference of two squares, therefore, we shall merely delete the numerators:

`1+csc(theta)-sec(theta)(csc(theta)-cot(theta)) = 2csc(theta)`

Use the identities `sec(theta)csc(theta) =1` and `sec(theta)cot(theta) = csc(theta)`:

`1+csc(theta)-1+csc(theta) = 2csc(theta)`

Combine like terms:

`2csc(theta) = 2csc(theta)` Q.E.D.

Answer 2

Please see below

Explanation:

`sectheta/(csctheta-cottheta)-sectheta/(csctheta+cottheta)`

= `(sectheta(csctheta+cottheta)-sectheta(csctheta-cottheta))/(csc^2theta-cot^2theta)`

= `sectheta(csctheta+cottheta-csctheta+cottheta)` `->` as `csc^2theta-cot^2theta=1`

= `secthetaxx2cottheta`

= `1/costhetaxx2costheta/sintheta`

= `2/sintheta=2csctheta`