a) You need to remember that the values of sine and cosine functions in the quadrant 3 are negative.

You also should remember that `sec alpha = 1/cos alpha` such that:

`sec alpha + cos alpha = 1/cos alpha + cos alpha `

You need to bring the terms to...

## See

This Answer NowStart your **subscription** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

a) You need to remember that the values of sine and cosine functions in the quadrant 3 are negative.

You also should remember that `sec alpha = 1/cos alpha` such that:

`sec alpha + cos alpha = 1/cos alpha + cos alpha `

You need to bring the terms to a common denominator such that:

`(1 + cos^2 alpha)/cos alpha `

You should use the identity `1 + tan^2 alpha = 1/(cos^2 alpha)`

`cos^2 alpha = 1/(1 + tan^2 alpha)`

`cos^2 alpha = 1/(1 + 1/16) =gt cos^2 alpha = 16/17`

`cos alpha = +-sqrt(16/17)`

You need to remember that since `alpha` is in quadrant 3, you need to keep only the negative value for `cos alpha` such that:

`cos alpha = -sqrt(16/17) =gt cos alpha = -4sqrt17/17`

`(1 + cos^2 alpha)/cos alpha = (1 + 16/17)/(-4sqrt17/17) = -33sqrt17/68`

**Hence, evaluating `sec alpha + cos alpha` yields `sec alpha + cos alpha = -33sqrt17/68.` **

b) `sin alpha + csc alpha = sin alpha + 1/sin alpha`

`sin alpha + csc alpha = (sin^2 alpha + 1)/sin alpha`

You need to use the identity `1/(1 + cot^2 alpha) = sin^2 alpha`

You also should use the identity `cot alpha = 1/tan alpha` , hence `cot alpha = 1/(1/4) = 4` .

`sin^2 alpha = 1/(1 + 16) = 1/17`

`sin alpha = +-sqrt(1/17)`

You need to remember that since `alpha` is in quadrant 3, you need to keep only the negative value for `sin alpha` such that:

`sin alpha = -sqrt(1/17)`

`sin alpha + csc alpha = (1/17 + 1)/(-sqrt17/17)`

`sin alpha + csc alpha = (18/17)/(-sqrt17/17)`

`sin alpha + csc alpha = -18sqrt17/17`

**Hence, evaluating `sec alpha + cos alpha` yields `sin alpha + csc alpha = -18sqrt17/17.` **