Complex Number : Part (1)

ကိန်းစစ်မျဉ်း (real number line) ပေါ်တွင် ဖေါ်ပြနိုင်သော ကိန်းအားလုံးကို ကိန်းစစ် (real number) ဟုခေါ်သည်။ ဤသို့ဆိုလျှင် ကိန်းစစ်မျဉ်းပေါ်တွင် မဖော်ပြနိုင်သော ကိန်းရှိပါသလား။ ရှိပါသည်။ အောက်ပါ ဥပမာကိုလေ့လာကြည့်မည်။

$x^2+1=0$

အထက်ဖော်ပြပါ ညီမျှခြင်းကို ဖြေရှင်းနိုင်ပါသလား။

$\begin{array}{l} x^2+1=0\\\\ x^2=-1\\\\ x=\sqrt{-1} \end{array}$

မည်သည့်ကိန်းစစ်ကို မဆို နှစ်ထပ် (စုံထပ်ကိန်း) တင်လျှင် အပါင်းကိန်းသာ ရမည်ဖြစ်သည်။ အနုတ်ကိန်းမရနိုင်ပါ။ ထို့ကြောင့် $x^2+1=0$ ကိုပြေလည်စေသော ကိန်းစစ်အဖြေမရှိပါ။ ထို့ကြောင့် ကိန်းစစ်နယ်ပယ်ကိုသာ လေ့လာခဲ့စဉ်က အဆိုပါညီမျှခြင်းကို ပြေလည်စေသော အဖြေမရှိဟု သတ်မှတ်ပြီး ဖြေရှင်းခြင်းကို ရပ်ဆိုင်းခဲ့ကြသည်။ သို့သော် သင်္ချာပညာရှင်များက အဆိုပါညီမျှခြင်းမျိုးကို ဖြေရှင်းနိုင်ရန် ကိန်းစနစ်ကို ချဲ့ထွင်ခဲ့ကြသည်။

Imaginary Number

ကိန်းစစ်နယ်ပယ်တွင် ဖြေရှင်းနိုင်းခြင်းမရှိသော $\sqrt{-1}$ ကို imaginary unit (သင်္ကေတအားဖြင့် $i$) ဟုသတ်မှတ်ခဲ့ကြသည်။ ထို့ကြောင့် $i^2= \left( \sqrt{-1}\right)^2=-1$ ဟု သတ်မှတ်နိုင်ပါသည်။ အောက်ပါ ဥပမာများကို ဆက်လက်လေ့လာကြည့်ပါ။

$\begin{array}{l} \sqrt{-4}=\sqrt{4}\sqrt{-1}=2i\\\\ \sqrt{-9}=\sqrt{9}\sqrt{-1}=3i\\\\ \sqrt{-10}=\sqrt{10}\sqrt{-1}=\sqrt{10}i \end{array}$
ဤနေရာတွင် square root of positive number များ အတွက် principal root (positive root) ကို သာယူပါသည်။

Complex Number

Definition
A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written a + bi where ais the real part and b is the imaginary part.Complex numbers are usually denoted by the symbol z.
z = a + bi where z = complex number
$\quad\quad$ a = Re (z)= Real part of z
$\quad\quad$ b = Im (z)= imaginary part of z

Imaginary and Complex Numbers
A complex number is a number of the form a + bi where

• a is the real part of the complex number.
• bi is the imaginary part of the complex number.
• If b = 0, then a + bi is a real number.
• If a = 0 and b is not equal to 0, the complex number is called an imaginary number.
• An imaginary number is an even root of a negative number.

complex number တစ်ခုကို $a + bi$ ပုံစံဖြင့် ဖော်ပြသည်။ $a$ ကို complex number တစ်ခု၏ ကိန်းစစ်အပိုင်း (real part)ဟု ခေါ်ပြီး၊ $b$ ကို complex number တစ်ခု၏ ကိန်းယောင် အပိုင်း (imaginary part) ဟုခေါ်သည်။ အကယ်၍ $b = 0$ ဖြစ်လျှင် imaginary part မရှိတော့ပါ။ ထိုအခါ complex number သည် real number ဖြစ်သွားမည်။ အကယ်၍ $a = 0$ နှင့် $b \ne 0$ ဖြစ်လျှင် real part မရှိတော့ပဲ imaginary part သာ ရှိတော့မည်။ ထိုအခါ complex number သည် imaginary number ဖြစ်သွားမည်။ အနုတ်ကိန်းတစ်ခု၏ စုံထပ်ကိန်းရင်တိုင်းသည် imaginary number ဖြစ်သည်။

Example 1 Express $\sqrt{-16}$ in the form of $a+bi$. Solution $\sqrt{-16}= \sqrt{16}\sqrt{-1}=4i=0+4i$

Plotting a Complex Number on the Complex Plane

Complex number များသည် real part + imaginary part ဖြစ်သောကြောင့် real number line တွေ နေရာမသတ်မှတ်နိုင်ပါ။ သို့ရာတွင် real dimension (real axis) နှင့် imaginary dimension (imaginary axis)တို့ကိုသုံး၍ နေရာသတ်မှတ်နိုင်သည်။ အဆိုပါ real axis နှင့် imaginary axis နှစ်ခုဖြင့်ဖွဲ့စည်းထားသော two-dimenssional plane ကို complex plane ဟုခေါ်သည်။ Complex plane တစ်ခုတွင် horizontal axis ($x$-axis) ကို real axis ဟု သတ်မှတ်ပြီး vertical axis ($y$-axis) ကို imaginary axis ဟု သတ်မှတ်သည်။

Complex Plane
In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis as shown in figure.

Fig 1: Complex Plane

Example 2 Plot the complex number $z=3 + 4i$ on the complex plane. Solution Fig 2: $z=3+4i$

Equality of Complex Numbers

Two complex numbers $a + bi$ and $c + di$, written in standard form, are equal to each other as
$a + bi = c + di$
if and only if $a = c$ and $b = d$.

Adding and Subtracting Complex Numbers

Complex number များ ပေါင်းခြင်း နုတ်ခြင်းတွင်

• real part အချင်းချင်း ပေါင်း/နုတ် ရသည်။
• imaginary part အချင်းချင်း ပေါင်း/နုတ် ရသည်။

For example, if $z_1=a + bi$ and $z_2=c + di$, then

$\begin{array}{l} z_1+z_2=(a + bi)+(c+di)=(a+c)+(b+d)i\\\\ z_1-z_2=(a + bi)-(c+di)=(a-c)+(b-d)i \end{array}$

Example 3 If $z_1=3 + 4i$ and $z_2=5 - 12i$, find (a) $z_1+z_2$ (b) $z_1-z_2$ Solution $z_1=3 + 4i$ and $z_2=5 - 12i$ (a) $z_1+z_2 = (3 + 4i) + (5 - 12i) = 8 - 8i$ (b) $z_1-z_2 = (3 + 4i) - (5 - 12i) = -2 + 16i$

Multiplying Complex Numbers

Complex Numbers များမြှောက်ခြင်းတွင် Real Numbers များ မြှောက်ခြင်းမှာကဲ့သို့ပင် ဖြန့်ဝေရဂုဏ်သတ္တိ (distributive law) ကို သုံးရသည်။

Multiplying a Complex Number by a Real Number

If $z =a + bi$ and $k$ is a real number, then

$\begin{array}{l} kz=k(a + bi)=ka + kb i\\\\ \dfrac{1}{k}z=\dfrac{1}{k}(a + bi)=\dfrac{a}{k}+\dfrac{b}{k}i \end{array}$

Example 4 If $z_1=3 - 12i$ and $z_2=4 + i$, find (a) $2z_1$ (b) $\dfrac{1}{3}z_1$ (c) $z_1 + 3z_2$ (d) $\dfrac{1}{2}(z_1 - 2z_2)$ Solution $\quad\quad z_{1}=3-12 i$ $\quad\quad z_{2}=4+i$ $\begin{aligned}\text{(a)}\quad 2 z_{1}&=2(3-12 i)\\\\ &=6-24 i \end{aligned}$ $\begin{aligned} \text{(b)}\quad \dfrac{1}{3} z_{1} &=\frac{1}{3}(3-12 i) \\\\ &= 1-4 i \end{aligned}$ $\begin{aligned} \text{(c)}\quad z_1+z_2 &=3-12 i+12+3 i \\\\ &=15-9 i \end{aligned}$ $\begin{aligned} \text{(d)}\quad \dfrac{1}{2}(z_1 - 2z_2) &=\dfrac{1}{2}\left(3-12 i - 2(4+i)\right)\\\\ &=\dfrac{1}{2}\left(3-12 i - 8-2i\right)\\\\ &=\dfrac{1}{2}\left(-5-14 i\right)\\\\ &=-\dfrac{5}{2}-7 i \end{aligned}$

Multiplying a Complex Number by a Complex Number

If $z_1 =a + bi$ and $z_2 =c + di$, then

$\begin{array}{l} z_1\cdot z_2=(a+b i)(c+d i)=a c+a d i+b c i+b d i^{2}\\\\ z_1\cdot z_2=(a+b i)(c+d i)=a c+a d i+b c i-b d\quad (\because\ i^2=-1)\\\\ z_1\cdot z_2=(a+b i)(c+d i)=(a c-b d)+(a d+b c) i \end{array}$

Example 5 If $z_1=3 + 2i$ and $z_2=2 -5 i$, find (a) $z_1\cdot z_2$ (b) ${z_1}^2$ Solution $z_1=3 + 2i$ $z_2=2 -5 i$, $\begin{aligned} \text{(a)}\quad z_1\cdot z_2 &= (3 + 2i)(2 -5 i) \\\\ &= 6 - 15i + 4i -10i^2\\\\ &= 6 - 11i +10 \quad(\because\quad i^2=-1)\\\\ &= 16 - 11i \end{aligned}$ $\begin{aligned} \text{(b)}\quad {z_1}^2 &= (3 + 2i)^2 \\\\ &= 9+12i+4i^2\\\\ &= 9 +12i-4 \quad(\because\quad i^2=-1)\\\\ &= 5+12i \end{aligned}$

Conjugate of a Complex Number

Complex Number တစ်ခု၏ conjugate ဆိုသည်မှာ ပေးထားသော complex number ၏ imaginary part ကို ဆန့်ကျင်ဘက် လက္ခဏာသို့ ပြောင်းပေးလိုက်ခြင်းပင် ဖြစ်သည်။ $z=a+bi$ ၏ conjugate မှာ $\overline {z}= a-bi$ ဖြစ်သည်။

Complex Conjugate
The complex conjugate of a complex number which is obtained by changing the sign of the imaginary part. So if $z=a+b i$, its complex conjugate, $\bar{z}$, is defined by
$\bar{z}=\overline{a+b i}=a-bi$

• When a complex number is multiplied by its complex conjugate, the result is a real number.
• When a complex number is added to its complex conjugate, the result is a real number.

Example 6 Find the comblex conjugate of each number (a) $z=3+5i$ (b) $w=-2i$ Solution $\begin{array}{lll} \text{(a)}& \text{conjugate of}\ z & = \overline{z} \\\\ & & = \overline{3+5i} \\\\ & & = 3-5i\\\\ \text{(b)} &\quad\quad\quad\quad\quad w & = -2i\\\\ &\therefore \quad\quad\quad\quad w & = 0-2i\\\\ &\text{conjugate of}\ w & = \overline{w} \\\\ & & = \overline{0-2i} \\\\ & & = 0+2i\\\\ & & = 2i \end{array}$

Example 7 If $z=2 + i$ , find (a) $\overline{z}$ (b) $z\cdot \overline{z}$ (b) $z + \overline{z}$ Solution $z=2 + i$ $\begin{array}{lll} \text{(a)}& \quad\quad\overline{z} & = \overline{2 + i} \\\\ & & = 2-i\\\\ \text{(b)}& z\cdot \overline{z} & = (2+i)(2-i)\\\\ & & = 4-i^2\\\\ & & = 4-(-1)\\\\ & & = 5\\\\ \text{(c)}&z + \overline{z} &= (2+i)+(2-i)\\\\ & & = 4 \end{array}$

Dividing Complex Numbers

Complex number များစားသည့်အခါ စားကိန်း (ပိုင်းခြေ) ကို real number ဖြစ်အောင် ပြုလုပ်ပေးရမည်။ ပိုင်းခြေသည် complex number ဖြစ်နေပါက ပိုင်းဝေနှင့် ပိုင်းခြေ နှစ်ခုလုံးကို conjugate ဖြင့် မြှောက်ပေးခြင်းအားဖြင့် ပိုင်းခြေကို real number ဖြစ်အောင် ပြောင်းပေးနိုင်သည်။

If $z_1=a+bi$ and $z_2=c+di$, then $\begin{aligned} \dfrac{z_1}{z_2} =&\dfrac{a+bi}{c+di}\\\\ =&\dfrac{a+bi}{c+di}\times\dfrac{c-di}{c-di}\\\\ =&\dfrac{ac-adi+bci-bdi^2}{c^2-d^2i^2}\\\\ =&\dfrac{(ac+bd)+ (bc-ad)i}{c^2+d^2} \end{aligned}$

Example 8 Express $\dfrac{2+3i}{4-2i}$ in standard form (in the form of $a+bi$). Solution $\begin{aligned} \dfrac{2+3i}{4-2i} =&\dfrac{2+3i}{4-2i}\times\dfrac{4+2i}{4+2i}\\\\ =&\dfrac{8+4i+12i+6i^2}{16-4i^2}\\\\ =&\dfrac{8+16i-6}{16+4}\quad (\because\quad i^2=-1)\\\\ =&\dfrac{2+16i}{20}\\\\ =&\dfrac{2}{20}+\dfrac{16i}{20}\\\\ =&\dfrac{1}{10}+\dfrac{4}{5}i \end{aligned}$

Exercise Find real numbers $a$ and $b$ for each of the following equations. (a) $a+b i=9+8 i$ (b) $a+b i=10-5 i$ (c) $(a-2)+(b+1) i=6+5 i$ (d) $(a+2)+(b-3) i=4+7 i$ Perform the operation and write the result in standard form. (a) $(5+i)+(2+3 i) $ (b) $(9-i)-(8-i) $ (c) $(-2+\sqrt{-8})+(5-\sqrt{-50})$ (d) $(8+\sqrt{-18})-(4+3 \sqrt{2} i)$ (e) $13 i-(14-7 i)$ (f) $25+(-10+11 i)+15 i$ (g) $(13-2 i)+(-5+6 i)$ (h) $(3+2 i)-(6+13 i)$ Simplify the following and write the result in standard form. (a) $(1+i)(3-2 i)$ (b) $(7-2 i)(3-5 i)$ (c) $12 i(1-9 i)$ (d) $-8 i(9+4 i)$ (e) $(\sqrt{2}+3 i)(\sqrt{2}-3 i)$ (f) $(4+\sqrt{7} i)(4-\sqrt{7} i)$ (g) $(6+7 i)^{2}$ (h) $(5-4 i)^{2}$ Write down the complex conjugate of each of the following the complex numbers. Then multiply the number by its complex conjugate. (a) $9+2 i$ (b) $8-10 i$ (c) $-1-\sqrt{5} i$ (d) $-3+\sqrt{2} i$ (e) $\sqrt{-20}$ (f) $\sqrt{-15}$ (g) $\sqrt{6}$ (h) $1+\sqrt{8}$ Express the quotient in standard form. (a) $\dfrac{2}{4-5 i}$ (b) $\dfrac{13}{1-i}$ (c) $\dfrac{5+i}{5-i}$ (d) $\dfrac{6-7 i}{1-2 i}$ (e) $\dfrac{9-4 i}{i}$ (e) $\dfrac{8+16 i}{2 i}$ (f) $\dfrac{3 i}{(4-5 i)^{2}}$ (g) $\dfrac{5 i}{(2+3 i)^{2}}$ Perform the operation and write the result in standard form. (a) $\dfrac{2}{1+i}-\dfrac{3}{1-i}$ (b) $\dfrac{2 i}{2+i}+\dfrac{5}{2-i}$ (c) $\dfrac{i}{3-2 i}+\dfrac{2 i}{3+8 i}$ (d) $\dfrac{1+i}{i}-\dfrac{3}{4-i}$ Express each of the following complex numbers in standard form. (a) $\sqrt{-6} \sqrt{-2}$ (b) $\sqrt{-5} \sqrt{-10}$ (c) $(\sqrt{-15})^{2}$ (d) $(\sqrt{-75})^{2}$ (e) $\sqrt{-8}+\sqrt{-50}$ (f) $\sqrt{-45}-\sqrt{-5}$ (g) $(3+\sqrt{-5})(7-\sqrt{-10})$ (h) $(2-\sqrt{-6})^{2}$

Miscellaneous Exercise : Proofs of Trigonometric Identities

1. Prove that $\dfrac{\tan \theta+\cot \theta}{\sec \theta+\operatorname{cosec} \theta}=\dfrac{1}{\sin \theta+\cos \theta}$





$\begin{aligned} &\dfrac{\tan \theta+\cot \theta}{\sec \theta+\csc \theta}\\\\ = &\dfrac{\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\cos \theta}+\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}}{\dfrac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}}\\\\ = &\dfrac{1}{\sin \theta \cos \theta} \times \dfrac{\sin \theta \cos \theta}{\sin \theta+\cos \theta}\\\\ = &\dfrac{1}{\sin \theta+ \cos \theta} \end{aligned}$



2. Prove that $\quad \sec x \csc x-\cot x=\tan x$.





$\begin{aligned} & \sec x \csc x-\cot x \\\\ = & \dfrac{1}{\cos x} \dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x} \\\\ = & \dfrac{1-\cos ^{2} x}{\sin x \cos x} \\\\ = & \dfrac{\sin ^{2} x}{\sin x \cos x} \\\\ = & \dfrac{\sin x}{\cos x} \\\\ = & \tan x \end{aligned}$



3. Prove that $\dfrac{1}{1-\cos x}+\dfrac{1}{1+\cos x}=2 \csc^{2} x$





$\begin{aligned} & \dfrac{1}{1-\cos x}+\dfrac{1}{1+\cos x} \\\\ = & \dfrac{1+\cos x+1-\cos x}{(1-\cos x)(1+\cos x)} \\\\ = & \dfrac{2}{1-\cos ^{2} x} \\\\ = & \dfrac{2}{\sin ^{2} x} \\\\ = & 2 \csc ^{2} x \end{aligned}$



4. Show that $\dfrac{\tan \theta+\cot \theta}{\csc \theta}=\sec \theta$





$\begin{aligned} &\dfrac{\tan \theta+\cot \theta}{\csc \theta}\\\\ = &\dfrac{\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}}{\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{1}{\sin \theta \cos \theta} \times \sin \theta \\\\ = &\sec \theta \end{aligned}$



5. Show that $\sqrt{\sec ^{2} \theta-1}+\sqrt{\csc^{2} \theta-1}=\sec \theta \csc \theta$





$\begin{aligned} & \sqrt{\sec ^{2} \theta-1}+\sqrt{\csc ^{2} \theta-1} \\\\ =& \sqrt{\tan ^{2} \theta}+\sqrt{\cot ^{2} \theta} \\\\ =& \tan \theta+\cot \theta \\\\ =& \dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta} \\\\ =& \dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta} \\\\ =& \dfrac{1}{\sin \theta \cos \theta} \\\\ =& \sec \theta \csc \theta \end{aligned}$



6. Prove that $\sec ^{2} x+\csc^{2} x=\sec ^{2} x \csc^{2} x$





$\begin{aligned} & \sec ^{2} x+\csc ^{2} x \\\\ =& \dfrac{1}{\cos ^{2} x}+\dfrac{1}{\sin ^{2} x} \\\\ =& \dfrac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} \\\\ =& \dfrac{1}{\sin ^{2} x \cos ^{2} x} \\\\ =& \dfrac{1}{\cos ^{2} x} \cdot \dfrac{1}{\sin ^{2} x}\\\\ =&\sec ^{2} x \csc ^{2} x \end{aligned}$



7. Show that $\dfrac{1}{\csc \theta-1}-\dfrac{1}{\csc \theta+1}=2 \tan ^{2} \theta$





$\begin{aligned} & \dfrac{1}{\csc \theta-1}-\dfrac{1}{\csc \theta+1} \\\\ =& \dfrac{(\csc \theta+1)-(\csc \theta-1)}{(\csc \theta-1)(\csc \theta+1)} \\\\ =& \dfrac{2}{\csc ^{2} \theta-1} \\\\ =& \dfrac{2}{\cot ^{2} \theta} \\\\ =& 2 \tan ^{2} \theta \end{aligned}$



8. Show that $\dfrac{\csc \theta}{\csc \theta-\sin \theta}=\sec ^{2} \theta$.





$\begin{aligned} &\dfrac{\csc \theta}{\csc \theta-\sin \theta} \\\\ =& \dfrac{\dfrac{1}{\sin \theta}}{\dfrac{1}{\sin \theta}-\sin \theta} \\\\ =& \dfrac{1}{\sin \theta} \times \dfrac{\sin \theta}{\cos ^{2} \theta}\\\\ =&\sec ^{2} \theta \end{aligned}$



9. Prove that $\dfrac{\cos x}{1+\tan x}-\dfrac{\sin x}{1+\cot x}=\cos x-\sin x$.





$\begin{aligned} &\dfrac{\cos x}{1+\tan x}-\dfrac{\sin x}{1+\cot x} \\\\ =&\dfrac{\cos x}{1+\dfrac{\sin x}{\cos x}}-\dfrac{\sin x}{1+\dfrac{\cos x}{\sin x}} \\\\ =&\dfrac{\cos x+\sin x}{\cos x}-\dfrac{\sin x}{\dfrac{\sin x+\cos u}{\sin x}}\\\\ =&\cos x\left(\dfrac{\cos x}{\cos x+\sin x}\right)-\sin x\left(\dfrac{\sin x}{\sin x+\cos x}\right)\\\\ =&\dfrac{\cos ^{2} x-\sin ^{2} x}{\cos x+\sin x}\\\\ =&\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x+\sin x}\\\\ =&\cos x-\sin x \end{aligned}$



10. Show that $\csc \theta-\sin \theta=\cot \theta \cos \theta$.





$\begin{aligned} & \csc \theta-\sin \theta \\\\ =& \dfrac{1}{\sin \theta}-\sin \theta \\\\ =& \dfrac{1-\sin ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{\cos ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{\cos \theta}{\sin \theta} \cos \theta \\\\ =& \cot \theta \cos \theta \end{aligned}$



11. Show that $\dfrac{\tan ^{2} \theta+\sin ^{2} \theta}{\cos \theta+\sec \theta}=\tan \theta \sin \theta$.





$\begin{aligned} & \dfrac{\tan ^{2} \theta+\sin ^{2} \theta}{\cos \theta+\sec \theta} \\\\ =& \dfrac{\dfrac{\sin ^{2} \theta}{\cos ^{2} \theta}+\sin ^{2} \theta}{\cos \theta+\dfrac{1}{\cos \theta}} \\\\ =& \dfrac{\sin ^{2} \theta\left(1+\cos ^{2} \theta\right)}{\cos ^{2} \theta} \times \dfrac{\cos \theta}{\cos ^{2} \theta+T}\\\\ =&\dfrac{\sin ^{2} \theta}{\cos \theta} \\\\ =&\dfrac{\sin \theta}{\cos \theta} \sin \theta \\\\ =&\tan \theta \sin \theta \end{aligned}$



12. Prove that $\sin x(\cot x+\tan x)=\sec x$





$\begin{aligned} & \sin x(\cot x+\tan x) \\ =& \sin x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right) \\\\ =& \sin x\left(\dfrac{\cos 2 x+\sin ^{2} x}{\sin x \cos x}\right) \\\\ =& \dfrac{1}{\cos x} \\\\ =& \sec x \end{aligned}$



13. Show that $\dfrac{\sec \theta}{\cot \theta+\tan \theta}=\sin \theta$.





$\begin{aligned} &\dfrac{\sec \theta}{\cot \theta+\tan \theta}\\\\ =&\dfrac{\dfrac{1}{\cos \theta}}{\dfrac{\cos \theta}{\sin \theta}+\dfrac{\sin \theta}{\cos \theta}}\\\\ =&\dfrac{\dfrac{1}{\cos \theta}}{\dfrac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}}\\\\ =&\dfrac{1}{\cos \theta} \times \dfrac{\sin \theta \cos \theta}{1}\\\\ =&\sin \theta \end{aligned}$



14. Show that $\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}=2 \csc x$.





$\begin{aligned} & \dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x}{1+\cos x} \times \dfrac{1-\cos x}{1-\cos x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x(1-\cos x)}{1-\cos ^{2} x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x(1-\cos x)}{\sin ^{2} x}+\dfrac{1+\cos x}{\sin x}\\\\ =&\dfrac{1-\cos x}{\sin x}+\dfrac{1+\cos x}{\sin x} \\\\ =&\dfrac{1-\cos x+1+\cos x}{\sin x} \\\\ =&\dfrac{2}{\sin x} \\\\ =&2 \csc x \end{aligned}$



15. Show that $\dfrac{(1-\sin A)(1+\sin A)}{\sin A \cos A}=\cot A$.





$\begin{aligned} & \dfrac{(1-\sin A)(1+\sin A)}{\sin A \cos A} \\\\ =& \dfrac{1-\sin ^{2} A}{\sin A \cos A} \\\\ =& \dfrac{\cos ^{2} A}{\sin A \cos A} \\\\ =& \dfrac{\cos A}{\sin A}\\\\ =&\cot A \end{aligned}$



16. Show that $\cos \theta \cot \theta+\sin \theta=\csc \theta$.





$\begin{aligned} & \cos \theta \cot \theta+\sin \theta \\\\ =& \cos \theta \cdot \dfrac{\cos \theta}{\sin \theta}+\sin \theta \\\\ =& \dfrac{\cos ^{2} \theta}{\sin \theta}+\sin \theta \\\\ =& \dfrac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{1}{\sin \theta} \\\\ =& \csc \theta \end{aligned}$



17. Show that $2 \cos x \cot x+1=\cot x+2 \cos x$ can be written in the form $(a \cos x-b)(\cos x-\sin x)=0$, where $a$ and $b$ are constants to be found.





$\begin{aligned} &\quad 2 \cos x \cot x+1=\cot x+2 \cos x \\\\ &\quad 2 \cos x \dfrac{\cos x}{\sin x}+1=\dfrac{\cos x}{\sin x}+2 \cos x \\\\ &\quad \text { Multiplying both sides with}\ \sin x, \\\\ &\quad 2 \cos ^{2} x+\sin x=\cos x+2 \sin x \cos x \\\\ &\quad 2 \cos ^{2} x-\cos x+\sin x-2 \sin x \cos x=0 \\\\ &\quad\cos x(2 \cos x-1)+\sin x(1-2 \cos x)=0\\\\ &\quad\cos x(2 \cos x-1)-\sin x(2 \cos x-1)=0 \\\\ &\quad(2 \cos x-1)(\cos x-\sin x)=0 \\\\ &\therefore(a \cos x-b)(\cos x-\sin x)=(2 \cos x-1)(\cos x-\sin x) \\\\ &\therefore a=2\quad \text { and }\quad b=1 \end{aligned}$

Exercise (8.4): Solutions - Angle Bisector Theorem

Angle Bisector Theorem
A	The bisector of an interior angle of a triangle divides the opposite side internally into a ratio equal to the ratio of the other two sides of the triangle. တြိဂံတစ်ခု၏ အတွင်းထောင့်တစ်ခုကို ထက်ဝက်ပိုင်းသော မျဉ်းသည် မျက်နှာချင်းဆိုင်အနားကို အတွင်းပိုင်းမှ ပိုင်းဖြတ်ရာ ပိုင်းဖြတ်လိုက်သော (အတွင်းပိုင်း) အချိုးသည် ကျန်အနားနှစ်ဖက် အချိုးနှင့် ညီသည်။
B	The bisector of an exterior angle of a triangle divides the opposite side externally into a ratio equal to the ratio of the other two sides of the triangle. တြိဂံတစ်ခု၏ အပြင်ထောင့်တစ်ခုကို ထက်ဝက်ပိုင်းသော မျဉ်းသည် မျက်နှာချင်းဆိုင် အနားကို အပြင်ပိုင်းမှ ပိုင်းဖြတ်ရာ ပိုင်းဖြတ်လိုက်သော (အပြင်ပိုင်း) အချိုးသည် ကျန်အနားနှစ်ဖက် အချိုးနှင့် ညီသည်။

1. Which of the following proportions follow from the fact that $AE$ bisects $\angle WAV$ in $\triangle WAV$ ?
   
   (a) $ \displaystyle \frac{W E}{E V}=\displaystyle \frac{W A}{A V}$
   
   (b) $ \displaystyle \frac{W E}{E V}=\displaystyle \frac{V A}{A W}$
   
   (c) $ \displaystyle \frac{W E}{W A}=\displaystyle \frac{E V}{A V}$
   
   (d) $ \displaystyle \frac{A V}{A W}=\displaystyle \frac{V E}{E W}$
   
   
   
   

   $\begin{array}{l} \quad\quad\text{If}\ AE\ \text{bisects}\ \angle WAV,\\\\ \quad\quad\dfrac{WE}{EV}=\dfrac{WA}{AV}\\\\ \Rightarrow\quad \dfrac{WE}{WA}=\dfrac{EV}{AV}\\\\ \Rightarrow\quad \dfrac{AV}{AW}=\dfrac{VE}{EW}\\\\ \therefore\quad \text{(a), (c), (d) are true and (b) is false.} \end{array}$

   
   


2. $AX$ bisects $\angle CAB$. Complete the following statements:
   
   (a) $A C: A B=\ldots$
   
   (b) $A B: A C=\ldots$
   
   (c) $X C: X B=\ldots$
   
   
   
   

   $\begin{array}{l} \text{(a)}\ \dfrac{AC}{AB}=\dfrac{CX}{XB}\\\\ \text{(b)}\ \dfrac{AB}{AC}=\dfrac{BX}{XC}\\\\ \text{(c)}\ \dfrac{XC}{XB}=\dfrac{AC}{AB} \end{array}$

   
   


3. $PT$ bisects $\angle RPS$. Complete the following statements:
   
   (a) $P Q: P R=\ldots$
   
   (b) $T R: P R=\ldots$
   
   (c) $Q R: T R=\ldots$
   
   
   
   

   $\begin{array}{l} \text{(a)}\ \dfrac{PQ}{PR}=\dfrac{TQ}{TR}\\\\ \text{(b)}\ \dfrac{TR}{PR}=\dfrac{TQ}{PQ}\\\\ \text{(b)}\ \dfrac{QR}{TR}=\dfrac{TQ-TR}{TR}=\dfrac{PQ-PR}{PR} \end{array}$

   
   


4. What can you say about the rays $AD$, $BE$ and $CF$?
   
   

   $\begin{array}{ll} \quad\quad AB = 6, AC = 8, BC = 10\\\\ \quad\quad\dfrac{AB}{AC}=\dfrac{6}{8}=\dfrac{3}{4}\\\\ \quad\quad\dfrac{AC}{BC}=\dfrac{8}{10}=\dfrac{4}{5}\\\\ \quad\quad\dfrac{BC}{AB}=\dfrac{10}{6}=\dfrac{5}{3}\\\\ \quad\quad\dfrac{BD}{DC}=\dfrac{\frac{30}{7}}{\frac{40}{7}}=\dfrac{3}{4}\\\\ \quad\quad\dfrac{AF}{BF}=\dfrac{\frac{8}{3}}{\frac{10}{3}}=\dfrac{4}{5}\\\\ \quad\quad\dfrac{CE}{AE}=\dfrac{5}{3}\\\\ \therefore\quad \dfrac{AB}{AC}=\dfrac{BD}{DC}\Rightarrow AD\ \text{bisects}\ \angle BAC\\\\ \therefore\quad \dfrac{AC}{BC}=\dfrac{AF}{BF}\Rightarrow CF\ \text{bisects}\ \angle ACB\\\\ \therefore\quad \dfrac{BC}{AB}=\dfrac{CE}{AE}\Rightarrow BE\ \text{bisects}\ \angle ABC \end{array}$

   
   


5. If $AD$ and $AE$ are bisectors of the interior and exterior angles at $A$ of $\triangle ABC$, then which of the following are true?
   
   (a) $\angle D A E=90^{\circ}$
   
   (b) $B D: D C=B C: C E$
   
   (c) $B D: D C=B E: C E$
   
   (d) $A D: A E=D C: C E$
   
   

   $\begin{array}{ll} \quad\quad (\bullet) + (\bullet) + (*) + (*) = 180^{\circ}\\\\ \quad\quad 2(\bullet) + 2(*) = 180^{\circ}\\\\ \quad\quad (\bullet) + (*) = 90^{\circ}\\\\ \therefore\quad \angle DAE = 90^{\circ}\\\\ \therefore\quad \text{(a) is true.}\\\\ \quad\quad AD\ \text {bisects}\ \angle BAC.\\\\ \therefore\quad \dfrac{BD}{DC}=\dfrac{AB}{AC}\\\\ \quad\quad AE\ \text {bisects}\ \angle CAF.\\\\ \therefore\quad \dfrac{BE}{CE}=\dfrac{AB}{AC}\\\\ \therefore\quad \dfrac{BD}{DC}=\dfrac{BE}{CE}\\\\ \therefore\quad \text{(b) is false and (c) is true.}\\\\ \quad\quad \text {We cannot say that}\ (\bullet) = (*)\\\\ \therefore\quad \text{We cannot say that}\ AC\ \text {bisects}\ \angle DAE.\\\\ \therefore\quad \text{We cannot say that}\ \dfrac{AD}{AE}=\dfrac{DC}{CE}.\\\\ \therefore\quad \text{(d) is false.} \end{array}$

   
   


6. Find the value of 𝑥 in each of the following figures.
   
   (a)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{3}=\dfrac{10}{6}\\\\ \therefore\quad x=\dfrac{10}{6}\times 3=5 \end{array}$

   
   (b)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{9}=\dfrac{13}{8}\\\\ \therefore\quad x=\dfrac{13}{8}\times 9=\dfrac{117}{8}=14.625 \end{array}$

   
   (c)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{4-x}=\dfrac{8}{6}\\\\ \therefore\quad 3x=16-4x\\\\ \quad\quad 7x=16\\\\ \quad\quad x=\dfrac{16}{7}=2.286 \end{array}$

   
   (d)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{6-x}=\dfrac{4}{3}\\\\ \therefore\quad 3x=24-4x\\\\ \quad\quad 7x=24\\\\ \quad\quad x=\dfrac{24}{7}=3.429 \end{array}$

   
   (e)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{x-2}=\dfrac{7}{3}\\\\ \therefore\quad 3x=7x-14\\\\ \quad\quad 4x=14\\\\ \quad\quad x=\dfrac{7}{2}=3.5 \end{array}$

   
   (f)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{x+2.3}=\dfrac{2.3}{3.8}\\\\ \therefore\quad \dfrac{10x}{10x+23}=\dfrac{23}{38}\\\\ \quad\quad 380x=230x+529\\\\ \quad\quad 150x=529\\\\ \quad\quad x=\dfrac{529}{150}=3.527 \end{array}$

   
   


7. Find the unknown marked lengths in the figure.
   
   

   $\begin{array}{l} \quad\quad CD\ \text{bisects}\ \angle ACB\\\\ \therefore\quad \dfrac{AD}{DB}=\dfrac{AC}{BC}\\\\ \therefore\quad \dfrac{x}{24}=\dfrac{20}{30}\\\\ \therefore\quad x=16\\\\ \quad\quad AE\ \text{bisects}\ \angle CAF\\\\ \therefore\quad \dfrac{CE}{BE}=\dfrac{AC}{AB}\\\\ \quad\quad \dfrac{y}{30+y}=\dfrac{20}{x+24}\\\\ \quad\quad \dfrac{y}{30+y}=\dfrac{20}{40}\\\\ \quad\quad \dfrac{y}{30+y}=\dfrac{1}{2}\\\\ \quad\quad 2y=30+y\\\\ \therefore\quad y=30\\\\ \end{array}$

   
   


8. $A B=12$ cm, $B C=9$ cm $C A=7$ cm. $B D$ bisects $\angle B$ and $A G=A D$, $C H=C D .$ Calculate $B G, B H$. Does $G H \parallel A C$ ?
   
   

   $\begin{array} \quad\quad \text{Let}\ AG=AD=y\ \text{and}\ CH=CD=x.\\\\ \therefore\quad x+y=7\Rightarrow y=7-x\\\\ \quad\quad \text{Since}\ BD\ \text{bisects}\ \angle ABC\\\\ \therefore\quad \dfrac{AD}{DC}=\dfrac{AB}{BC}\\\\ \quad\quad \dfrac{y}{x}=\dfrac{12}{9}\\\\ \quad\quad \dfrac{7-x}{x}=\dfrac{4}{3}\\\\ \therefore\quad 21-3x = 4x\\\\ \quad\quad 7x=21\\\\ \therefore\quad x = 3\\\\ \therefore\quad y = 7-x=7-3=4 \end{array}$

   
   


9. In $\triangle A B C, D E \parallel B C$, $A D=2.7$ cm, $D B=1.8$ cm and $B C=3$ cm. Prove that $B E$ bisects $\angle A B C$.
   
   

   $\begin{array} \quad\quad DE\parallel BC\\\\ \therefore\quad \dfrac{AE}{EC}=\dfrac{AD}{DB}\\\\ \quad\quad \dfrac{AE}{EC}=\dfrac{2.7}{1.8}=\dfrac{3}{2}\\\\ \quad\quad \text{But}\ \dfrac{AB}{BC}=\dfrac{2.7+1.8}{3}=\dfrac{3}{2}\\\\ \therefore\quad \dfrac{AE}{EC}=\dfrac{AB}{BC}\\\\ \therefore\quad BE\ \text{bisects}\ \angle ABC. \end{array}$

   
   


10. In a parallelogram $A B C D$, $A B=3.6$ cm, $B C=2.7$ cm, $A X=3.2$ cm, $X C=2.4$ cm. Prove that $\triangle B C Y$ is isosceles.
    
    

    $\begin{array}{l} \quad\quad \dfrac{AX}{XC}=\dfrac{3.2}{2.4}=\dfrac{4}{3}\\\\ \quad\quad \dfrac{AB}{BC}=\dfrac{3.6}{2.7}=\dfrac{4}{3}\\\\ \therefore\quad \dfrac{AX}{XC}=\dfrac{AB}{BC}\\\\ \therefore\quad BY\ \text{bisects}\ \angle ABC.\\\\ \therefore\quad \alpha = \beta\\\\ \quad\quad \text{Since}\ DC\parallel AB,\\\\ \quad\quad \alpha = \gamma\quad (\text{alternate}\ \angle\text{s})\\\\ \therefore\quad \beta =\gamma\\\\ \therefore\quad \triangle BCY \text{is isosceles.} \end{array}$ Angle bisector theorem အောက်တွင် မေးထားသော မေးခွန်း ဖြစ်သောကြောင့် Angle bisector theorem ဖြင့် သက်သေပြခြင်း ဖြစ်သော်လည်း သဏ္ဌာန်တူ ဥပဒေသကို သုံး၍လည်း သက်သေပြနိုင်ပါသည်။ $\begin{array}{l} \quad\quad \text{Since}\ DC\parallel AB,\\\\ \quad\quad \triangle AXB \sim \triangle CXY\\\\ \therefore\quad \dfrac{AB}{CY}=\dfrac{AX}{CX}\\\\ \therefore\quad \dfrac{3.6}{CY}=\dfrac{3.2}{2.4}\\\\ \quad\quad CY=\dfrac{3}{4} \times 3.6 = 2.7\ \text{cm}\\\\ \therefore\quad BC = CY\\\\ \therefore\quad \triangle BCY \text{is isosceles.} \end{array}$

    
    


11. Calculate $𝐵𝐷$ and $𝐷𝐶$ in terms of $𝑎, 𝑏, 𝑐$.
    
    

    $\begin{array} \quad\quad BD + DC = a\Rightarrow DC = a -BD\\\\ \quad\quad BY\ \text{bisects}\ \angle ABC.\\\\ \therefore\quad \dfrac{BD}{DC}=\dfrac{AB}{AC}\\\\ \therefore\quad \dfrac{BD}{a - BD}=\dfrac{c}{b}\\\\ \quad\quad b\cdot BD=ac-c\cdot BD\\\\ \quad\quad b\cdot BD + c\cdot BD=ac\\\\ \quad\quad (b+c) BD =ac\\\\ \therefore\quad BD=\dfrac{ac}{b+c}\\\\ \therefore\quad DC=a-\dfrac{ac}{b+c}=\dfrac{ab}{b+c} \end{array}$

    
    


12. Given : $A H$ bisects $\angle B A C$ in $\triangle A B C$. $E H \parallel A C$
    
    Prove : $\displaystyle \frac{B E}{E A}=\displaystyle \frac{B A}{A C}$
    
    

    $\begin{array} \quad\quad \text{Since}\ AH\ \text{bisects}\ \angle ABC,\\\\ \quad\quad \dfrac{BA}{AC}=\dfrac{BH}{HC}\\\\ \quad\quad \text{Since}\ EH\parallel AC,\\\\ \quad\quad \dfrac{BE}{EA}=\dfrac{BH}{HC}\\\\ \therefore\quad \dfrac{BE}{EA}=\dfrac{BA}{AC} \end{array}$

    
    


13. Given : In $\triangle A B C, B M=M C$.
    
                $M X$ bisects $\angle A M B$ .
    
                $M Y $ bisects $\angle A M C$.
    
    Prove : $X Y \parallel B C$ .
    
    

    $\begin{array} \quad\quad \text{Let}\ AM=x\ \text{and}\ BM=MC=y,\\\\ \quad\quad \text{Since}\ MX\ \text{bisects}\ \angle AMB,\\\\ \quad\quad \dfrac{AX}{XB}=\dfrac{AM}{BM}=\dfrac{x}{y}\\\\ \quad\quad \text{Similarly}\ MY\ \text{bisects}\ \angle AMC,\\\\ \quad\quad \dfrac{AY}{YC}=\dfrac{AM}{CM}=\dfrac{x}{y}\\\\ \therefore\quad \dfrac{AX}{XB}=\dfrac{AY}{YC}\\\\ \therefore\quad XY\parallel BC \end{array}$

    
    


14. Given : In $\triangle A B C$, $\angle A=2 \angle C$,
    
                $A D$ bisects $\angle B A C$ and
    
                $D E$ bisects $\angle A D B$.
    
    Prove $: \displaystyle \frac{B E}{E A}=\displaystyle \frac{B A}{A C}$
    
    

    $\begin{array}{l} \quad\quad \text{Since}\ \angle A=2\angle C,\\\\ \quad\quad \angle C=\dfrac{1}{2}\angle A\\\\ \therefore\quad \angle C=\angle DAC\\\\ \therefore\quad \triangle DCA\ \text{is isosceles with}\ DC = DA\\\\ \quad\quad \text{Since}\ AD\ \text{bisects}\ \angle BAC,\\\\ \quad\quad \dfrac{BD}{DC}=\dfrac{BA}{AC}\Rightarrow \dfrac{BD}{DA}=\dfrac{BA}{AC}\quad (\because DC=DA)\\\\ \quad\quad \text{Similarly}\ DE\ \text{bisects}\ \angle ADB,\\\\ \therefore\quad \dfrac{BE}{EA}=\dfrac{BD}{DA}\\\\ \therefore\quad \dfrac{BE}{EA}=\dfrac{BA}{AC} \end{array}$