In a right triangle with angle θ, sin(θ)=opposite/hypotenuse, cos(θ)=adjacent/hypotenuse, tan(θ)=opposite/adjacent=k. 0-3π/2 covers three quadrants Q1, Q2, Q3. In Q1 and Q3 tan(θ)=k, so we only want to know about Q1 and Q3 for the other trig ratios. In Q1, sine and cosine are both positive, but in Q3 both are negative. tan(θ)=k which can be written k/1, so opposite=k, adjacent=1 and hypotenuse (by Pythagoras)=√(1+k2), making sin(θ)=k/√(1+k2) and cos(θ)=1/√(1+k2). This is Q1.
In Q3 we have, sin(π+θ)=sin(π)cos(θ)+cos(π)sin(θ)=0-sin(θ)=-k/√(1+k2);
cos(π+θ)=cos(π)cos(θ)-sin(π)sin(θ)=-cos(θ)-0=-1/√(1+k2).