Can you help to solve these questions...

1)Write tant in terms of sint if t is in quadrant IV:
tan t= sint/Asqrt(B-C^2)
the sign A (input + or -) is...........
the constant B is...........
the function C is .........

2)Write sect in terms of tant if t is in quadrant II:
sec t=Asqrt(B+C^2)
the sign A (input + or -) is................
the constant B is.................
the function C is..............

Okay, just try not to fall for the complex looks of the problem, in fact it's really simple:

1) As the golden rule of trigonometry we know: Sin²θ + Cos²θ = 1 thus:
Cox²θ = 1 - Sin²θ → Cosθ = √(1 - Sin²θ)
And also: Tanθ = Sinθ / Cosθ

→ Tanθ = Sinθ / √(1 - Sin²θ)
or in your term: tan(t)= sin(t)/sqrt(1-Sin^2(t))

So when comparing the above equation with the one in the question we get:
A = +1 → The sign of A is "+"
B = +1 → Constant B = 1
C = Sin(t) → Function C = Sin(t)

2) In case of Sec and Cosec first of all convert them to equivalent equations in the form of respectively 1/Cos and 1/Sin:

Secθ = 1/Cosθ = - √(1/Cos²θ)
Care for the "-" sign in the above statement. As you know in quadrant II the value of Cosθ is negative and so is the value of 1/Cosθ. But when expressing 1/Cosθ in the form of √(1/Cos²θ) as you know the output value is |1/Cosθ|, thus we put a negative sign next to it to state that the angle is in the second quadrant.

So yet we have:
Secθ = - √(1/Cos²θ), and since 1 = Cos²θ + Sin²θ
→ Secθ = - √( (Cos²θ + Sin²θ) / Cos²θ ) = - √( 1 + Sin²θ/Cos²θ )
→ Secθ = - √(1 + Tan²θ)

And by comparing it two the equation in question we get:
A = -1 → The sign of A is "-"
B = +1 → Constant B = 1
C = Tanθ

1) tan t = sin t / cos t = sin t / sqrt(1 - sin^2 t)
So, A = 1, B = 1, C = sin t

2) sec t = - sqrt(1 + tan^2 t)
So, A = -1, B = 1, C = tan t