What is the definition of the cotangent function in relation to a right triangle?

Exercice 1: D. Ecrirc sous forme trigonométrique : \( z_{1}=2\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right) \) \( \begin{array}{ll}z_{2}=-3\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) \quad z_{3}=2\left(\cos \frac{\pi}{4}+i \sin \frac{3 \pi}{4}\right) \\ z_{4}=\cos \frac{\pi}{6}+i \sin \left(-\frac{\pi}{6}\right) \quad z_{5}=\cos \theta-i \sin \theta \quad z_{5}=\sin \theta+i \cos \theta\end{array} \)

If \( x \) is in the third quadrant, then \( \cot x \) in terms of \( \sec x \) is A) \( \frac{\sqrt{\sec ^{2} x-1}}{\sec ^{2} x-1} \) B) \( -\frac{\sqrt{\sec ^{2} x-1}}{\sec ^{2} x-1} \) C) \( -\frac{\sqrt{\sec ^{2} x+1}}{\sec ^{2} x-1} \)

If \( \sec (3 \theta+15)=\csc (35-\theta) \) where \( \theta \) is an acute angle, then \( \csc \frac{3}{2} \theta=\cdots \) \( \begin{array}{llll}\text { (a) } 2 & \text { (b) } \sqrt{3} & \text { (c) } \sqrt{2} & \text { (d) } \frac{1}{2}\end{array} \)

(6) The value of \( \sec \left(300^{\circ}\right) \sin \left(270^{\circ}-\theta\right)+\tan \left(-45^{\circ}\right) \cos (3 M-9) \) is \( \begin{array}{lll}\text { (a) }-3 \sin \theta & \text { (b) }-3 \cos \theta & \text { (c) } 3 \cos \theta\end{array} \) (7) If the product of two roots of the equation : \( (k-2) x^{2}-6 x+12=0 \) equals 3

The value of \( \sec \left(300^{\circ}\right) \sin \left(270^{\circ}-\theta\right)+\tan \left(-45^{\circ}\right) \cos \left(360^{\circ}-\theta\right) \) is \( \begin{array}{llll}\text { (a) }-3 \sin \theta & \text { (b) }-3 \cos \theta & \text { (c) } 3 \cos \theta & \text { (d) } 0\end{array} \)