Please Help me To solve this Problem

You're given that φ is an angle that terminates in the third quadrant (III). This means that both cos(φ) and sin(φ), and thus sec(φ) and csc(φ), are negative.

Recall the Pythagorean identity,

cos²(φ) + sin²(φ) = 1

Multiply the equation uniformly by 1/cos²(φ),

cos²(φ)/cos²(φ) + sin²(φ)/cos²(φ) = 1/cos²(φ)

1 + tan²(φ) = sec²(φ)

Solve for sec(φ) :

sec(φ) = - √(1 + tan²(φ))

Given that cot(φ) = 1/4, we have tan(φ) = 1/cot(φ) = 1/(1/4) = 4. Then

sec(φ) = - √(1 + 4²) = -√17

7,24

step-by-step explanation:

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)