إذا كان $\mathrm{\Pi }<\mathrm{x}<3\frac{\mathrm{\Pi }}{2}$ وكان $\tan \mathrm{x}=7/3$ فجد قيمة كل من: $\csc \mathrm{x} , \sec \mathrm{x} , \cot \mathrm{x}$

$\begin{array}{rl}& \tan \mathrm{x}=\frac{7}{3}\to \cot =\frac{3}{7}\\ & {\csc }^2\mathrm{x}=1+{\cot }^2\mathrm{x}\\ & {\csc }^2\mathrm{x}=1+\frac{9}{49}=\frac{49+9}{49}=\frac{58}{49}\\ & \csc \mathrm{x}=\frac{-\sqrt{58}}{7}\\ & {\sec }^2\mathrm{x}=1+{\tan }^2\mathrm{x}=1+\frac{49}{9}=\frac{9+49}{9}=\frac{58}{9}\\ & \sec =\frac{-\sqrt{58}}{3}\\ & \cot \mathrm{x}=1+\tan \mathrm{x}=1÷\frac{7}{3}=1\times \frac{3}{7}=\frac{3}{7}\end{array}$

تمارين (3-4)

(1)- إذا كان $3\frac{\mathrm{\Pi }}{2}<\mathrm{x}<2\mathrm{\Pi }$ وكان $\cos \mathrm{x}=2/3$ فجد قيمة كل من:

$\csc \mathrm{x} , \sec \mathrm{x} , \cot \mathrm{x}$

$\begin{array}{rl}& \sin \mathrm{x}=\frac{-\sqrt{5}}{3}\to \csc =\frac{1}{\sin \mathrm{x}}=\frac{1}{\frac{-\sqrt{5}}{3}}=\frac{-3}{\sqrt{5}}\\ & \cos \mathrm{x}=\frac{2}{3}\to \sec =\frac{1}{\cos \mathrm{x}}=\frac{1}{2}=\frac{3}{2}\\ & \cot \mathrm{x}=\frac{\cos \mathrm{x}}{\sin \mathrm{x}}=\frac{2}{3}÷\frac{-\sqrt{5}}{3}=\frac{2}{3}\times \frac{-3}{\sqrt{5}}=\frac{-2}{\sqrt{5}}\end{array}$

(2)- إذا كان $\mathrm{\Pi }<\mathrm{x}<3\frac{\mathrm{\Pi }}{2}$ وكان $\tan \mathrm{x}=7/3$ فجد قيمة كل من:

$\csc \mathrm{x} , \sec \mathrm{x} , \cot \mathrm{x}$

$\begin{array}{rl}& \tan \mathrm{x}=\frac{7}{3}\to \cot =\frac{3}{7}\\ & {\csc }^2\mathrm{x}=1+{\cot }^2\mathrm{x}\\ & {\csc }^2\mathrm{x}=1+\frac{9}{49}=\frac{49+9}{49}=\frac{58}{49}\\ & \csc \mathrm{x}=\frac{-\sqrt{58}}{7}\\ & {\sec }^2\mathrm{x}=1+{\tan }^2\mathrm{x}=1+\frac{49}{9}=\frac{9+49}{9}=\frac{58}{9}\\ & \sec =\frac{-\sqrt{58}}{3}\\ & \cot \mathrm{x}=1+\tan \mathrm{x}=1÷\frac{7}{3}=1\times \frac{3}{7}=\frac{3}{7}\end{array}$

(3)- أثبت صحة المتطابقات الآتية:

أ- $\tan \mathrm{x}=\sin x \sec \mathrm{x}$

$\begin{array}{rl}\text{right side}& =\sin \mathrm{x} \sec \mathrm{x}\\ & =\sin \mathrm{x}\cdot \frac{1}{\cos \mathrm{x}}=\frac{\sin \mathrm{x}}{\cos \mathrm{x}}=\tan \mathrm{x}=\text{lift side}\end{array}$

ب-${\sec }^2\mathrm{x}=\frac{{\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}}{1-{\sin }^2\mathrm{x}}$

$\text{right side}=\frac{1}{{\cos }^2\mathrm{x}}={\left(\frac{1}{\cos \mathrm{x}}\right)}^2={\sec }^2\mathrm{x}=\text{lift side}$

ج- $(1-{\sin }^2\mathrm{x})(1+{\tan }^2\mathrm{x})=1$

$\text{right side}=(1-{\sin }^2\mathrm{x})(1+{\tan }^2\mathrm{x})={\cos }^{2}x {\sec }^2\mathrm{x}={\cos }^2\mathrm{x}\frac{1}{{\cos }^2\mathrm{x}}=1=\text{L.S}$

د- $\frac{1-{\cos }^2\mathrm{x}}{\tan \mathrm{x}}=\sin \mathrm{xcos}\mathrm{x}$

$\text{right side}=\frac{1-{\cos }^2\mathrm{x}}{\tan \mathrm{x}}=\frac{{\sin }^2\mathrm{x}}{\sin \mathrm{x}}={\sin }^2\mathrm{x}\cdot \frac{\cos \mathrm{x}}{\sin \mathrm{x}}=\sin \mathrm{xcos}\mathrm{x}=1.5$

هـ- $\frac{1+\sin \mathrm{x}-{\sin }^2\mathrm{x}}{\cos \mathrm{x}}=\cos \mathrm{x}+\tan \mathrm{x}$

$\begin{array}{rl}\text{lift side}& =\frac{1+\sin \mathrm{x}-{\sin }^2\mathrm{x}}{\cos \mathrm{x}}=\frac{1-{\sin }^2\mathrm{x}+\sin \mathrm{x}}{\cos \mathrm{x}}=\frac{{\cos }^2\mathrm{x}+\sin \mathrm{x}}{\cos \mathrm{x}}\\ & =\frac{{\cos }^{\mathrm{z}}\mathrm{x}}{\cos \mathrm{x}}+\frac{\sin \mathrm{x}}{\cos \mathrm{x}}=\cos \mathrm{x}+\tan \mathrm{x}=\text{right side}\end{array}$