for 90 < t < 180, both cos and tan are negative

tan t = sin t / cos t = sqrt(1 - cos^2 t) / cos t

cos t tan t = sqrt(1 - cos^2 t)

cos^2 t tan^2 t = 1 - cos^2 t

cos^2 t + cos^2 t tan^2 t = 1

cos^2 t (1 + tan^2 t) = 1

cos^2 t = 1 / (1 + tan^2 t)

cos = -sqrt(1 / (1 + tan^2 t) ==> option A

since sqrt(1) = 1, this could also be written as -1 / sqrt(1 + tan^2 t)

take the negative root, because cos is negative in Q2

1 + tan²(θ) = sec²(θ)

cos²(θ) = 1/(1 + tan²(θ))

cos(θ) = -1/√(1 + tan²(θ))

The negative is placed because cos(θ) < 0 on (90, 180).