$\tan (x+y) = \frac{\sin (x+y)}{\cos (x+y)}$

$\quad \quad \quad \quad = \frac{\sin x \ \cos y + \cos x \ \sin y}{\cos x \ \cos y - \sin x \ \sin y}$

$\quad \quad \quad \quad = \frac{\frac{\sin x \ \cos y }{\cos x \ \cos y} + \frac{\cos x \ \sin y }{\cos x \ \cos y}}{\frac{\cos x \ \cos y }{\cos x \ cos y} - \frac{\sin x \ \sin y }{\cos x \ \cos y}}$

$\tan (x+y)= \frac{\tan x + \tan y }{1- \tan x \ \tan y}$

$\tan (x-y)= \tan ( x +(-y))$

$\tan(x-y)= \frac{\tan x + \tan (-y) }{1- \tan x \ \tan(-y)}$

$\tan (x-y) = \frac{\tan x - \tan y }{1 + \tan x \ \tan y}$

$\tan (2x) = \tan (x+x) \\ \quad \quad \quad = \frac{\tan x + \tan x}{1-\tan x.\tan x} \\ \quad \quad \quad = \frac{2 \ \tan x}{1-\tan ^2x}$

$\tan (2x) = \frac{2 \ \tan x}{1-\tan ^2x}$

$\biggl(\text{where,} \ 2x \neq \text{odd \ multiple \ of} \ \frac{\pi}{2} \biggl)$

$\tan (3x) = \tan (x+2x) = \frac{\tan x + \tan 2x}{1-\tan x \ \tan 2x}$

$\quad \quad \quad = \frac{\tan x + \frac{2\tan x}{1-\tan ^2x}}{1 - \tan x \frac{\ 2\tan x}{1-\tan ^2x}}$

$\tan (3x) = \frac{\tan x(1-\tan ^2x) + 2\tan x}{1-\tan ^2x - 2 \tan ^2x}$

$\tan (3x) = \frac{3 \tan x - \tan ^3x}{1-3\tan ^2x}$

$\biggl(\text{where,} \ 3x \neq \text{odd \ multiple \ of} \ \frac{\pi}{2}\biggl)$

$\cot (x+y) = \frac{1}{\tan (x+y)} \quad [\because \cot x = \frac{1}{\tan x}]$

$\tan (x+y) = \frac{\tan x + \tan y}{1-\tan x \ \tan y}$

$\cot (x+y) = \frac{1-\tan x \ \tan y}{\tan x + \tan y}$

$\quad \quad \quad \quad \quad = \frac{\frac{1}{\tan x \ \tan y} -1}{\frac{\tan x}{\tan x \ \tan y} + \frac{\tan y}{\tan x \ \tan y}}$

$\cot (x+y) = \frac{\cot x \cot y-1}{\cot x + \cot y}$

$\biggl(\text{where,} \ (x+y) \neq \text{\ multiple \ of} \ \pi \biggl)$

$\cot (x-y) = \frac{\cot x \cot (-y) -1}{\cot x + \cot (-y)}$

$\cot (x-y) = \frac{1+\cot x \cot y}{\cot y - \cot x}$

Again,

$\cot (x+y) = \frac{\cot x \cot y -1 }{\cot x+\cot y}$

Put $y=x$
$\therefore$ $\ \cot (2x) = \frac{\cot ^2x -1}{2 \cot x}$

$\cot (3x) = \cot (x+2x) = \frac{\cot x \ \cot 2x - 1}{\cot x + \cot 2x}$

$\quad \quad \quad \quad = \frac{\cot x \frac{\cot ^2x-1}{2\cot x}-1}{\cot x + \frac{\cot ^2x -1}{2\cot x}}$

$\cot (3x) = \frac{\cot ^3x - 3\cot x}{3\cot ^2x - 1}$
$\biggl(\text{where,} \ 3x \neq \text{\ multiple \ of} \ \pi \biggl)$

$\cosec (x) = \frac{1}{\sin x},$

$\text{Domain}$ : $\R$

$|\sin x| \leqslant 1$

$|\cosec x| \geqslant 1$

$\text{Range}$ : $(-\infty, -1]\cup[1,\infty)$

$\sec (x) = \frac{1}{\cos (x)},$

$\text{Domain}$ : $\R$

$|\cos x| \leqslant 1$

$|\sec x| \geqslant 1$

$\text{Range}$ : $(-\infty, -1]\cup[1,\infty)$

We know that

$2 \cos x \ \cos y = \cos (x+y) + \cos (x-y)$

Put,$x = \frac{A+B}{2}, y = \frac{A-B}{2}$

$\therefore$ $2 \cos \frac{(A+B)}{2} \cos \frac{(A-B)}{2} = \cos A + \cos B$

Show that: $\quad \frac{\cos 7x + \cos 3x}{\sin 7x - \sin 3x} = \cot 2x$

$\because \cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}$

$\therefore \cos 7x + \cos 3x = 2 \cos 5x \cos 2x$

$\because \sin A - \sin B = 2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$

$\therefore \sin 7x - \sin 3x = 2 \cos 5x \sin 2x$
$\quad$

$\text{L.H.S.} =\frac{ 2 \cos 5x \cos 2x}{2 \cos 5x \sin 2x } = \cot 2x = \text{R.H.S.}$

$\text{Show \ that:}$ $\quad \frac{\cos 3x + \cos 4x + \cos 5x}{\sin 3x + \sin 4x + \sin 5x} = \cot 4x$

$\because \ \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$

$\therefore \ \cos 3x + \cos 5x = 2 \cos 4x \cos x$

$\text{Numerator(L.H.S.)} = 2 \cos 4x \cos x + \cos 4x$

$\quad \quad \quad \quad \quad \quad \quad = \cos 4x(1+2 \cos x)$

$\because \ \sin A + \sin B = 2\sin \frac{A+B}{2} \cos\frac{A-B}{2}$

$\therefore \ \sin 3x + \sin 5x = 2 \sin 4x \cos x$

$\text{Denominator(L.H.S.)} = \sin 4x(1+2 \cos x)$

$\quad$

$\text{L.H.S} = \frac{\cos 3x + \cos 4x + \cos 5x}{\sin 3x + \sin 4x + \sin 5x} \\ \quad \quad = \frac{\cos 4x(1+2 \cos x)}{\sin 4x(1+2 \cos x)} \\ \quad \quad = \cot 4x$

$\text{Compute:} \ \sin (18^{\circ})$ $\quad$

Solution:

$\because \sin x = \cos (\frac{\pi}{2} - x)$

$\sin 36^{\circ} = \cos 54^{\circ},$
$\text{We know that,}$

$\sin 2 \theta = 2 \sin \theta \ cos \theta$
$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$
$\text{For} \ \theta = 18^{\circ}, \sin 2 \theta = \cos 3 \theta$

$\therefore 2 \sin \theta \cos \theta - (4 \cos^3 \theta - 3 \cos \theta) = 0$

$\cos \theta (2 \sin \theta - 4 \cos^2 \theta + 3) = 0$
For $\theta = 18^{\circ}, \cos \theta \neq 0$

$\therefore \quad (2 \sin \theta - 4 \cos^2 \theta + 3) = 0$

$\Rightarrow 4 \cos^2 \theta - 2 \sin \theta -3 = 0$

$\Rightarrow 4-4 \sin^2 \theta - 2 \sin \theta -3 = 0$

$\Rightarrow 4 \sin^2 \theta +2 \sin \theta -1 = 0 \\ \quad z \triangleq \sin \theta$

$4z^2 +2z -1 = 0$

$z = \frac {-2 \pm \sqrt{20}}{8}$

$\sin 18^{\circ} = \frac {-2 +\sqrt{20}}{8} = \frac{ \sqrt{5} -1}{4}$

$\text{Prove \ that:} \ \sqrt{ \frac{1- \cos x}{1 + \cos x}} = \cosec x - \cot x$

$\quad$
Solution:

$\text{R.H.S.}=\frac{1}{\sin x} - \frac{ \cos x}{ \sin x}$

$\quad \quad \quad =\frac{1- \cos x}{\sin x}$

$\text{L.H.S.} = \sqrt{ \frac{1- \cos x}{1 + \cos x}} \\\quad \quad = \frac{\sqrt{1- \cos x}}{\sqrt{1- \cos x}} \times \frac{\sqrt{1- \cos x}}{\sqrt{1+ \cos x}} = \frac{1- \cos x}{\sqrt{1- \cos^2 x}}$

$\quad \quad =\frac{1- \cos x}{\sin x} = \text{R.H.S.}$

$\text{Find \ the \ value \ of} \ \ \cos 40^{\circ} - \cos 20^{\circ} + \cos 80^{\circ} .$ $\quad$
Solution:
$\because \ \cos A + \cos B = 2 \cos \Biggl( \frac{A+B}{2}\Biggl) \cos \Biggl( \frac{A-B}{2}\Biggl)$

$\therefore \ \cos 40^{\circ} + \cos 80^{\circ} = 2 \cos 60^{\circ} \cos 20^{\circ}$

$\quad \quad \quad \quad \quad \quad \quad \quad \ = \cos 20^{\circ} \quad [\because \cos 60^{\circ} = \frac {1}{2}]$

$\cos 40^{\circ} - \cos 20^{\circ} + \cos 80^{\circ} \\ = \cos 20^{\circ} - \cos 20^{\circ} \\ =0$

$\text{Prove \ that:} \ \tan \theta + 2 \tan 2 \theta + 4 \tan 4 \theta + 8 \cot 8 \theta = \cot \theta$
$\quad$
Solution:

$\because \tan 2x = \frac{2 \tan x}{1-\tan ^2x}$

$x=4 \theta$

$\tan 8\theta = \frac{2 \tan4 \theta}{1- \tan ^2 4 \theta}$
Now,

$8 \cot 8\theta + 4 \tan 4 \theta = \frac{8(1- \tan ^2 4 \theta)}{2 \tan 4 \theta} + 4 \tan 4 \theta$
$\quad \quad \quad \quad = 4\Biggl[\frac{1}{\tan 4 \theta}\Biggl]$

$\text{L.H.S.} = \tan \theta + 2 \tan 2\theta + \frac{4}{\tan 4 \theta}$

$\text{Again}, \ \tan 4 \theta = \frac{2\tan 2 \theta}{1-\tan ^2 2 \theta}$

$\frac{4}{\tan 4 \theta} + 2 \tan 2 \theta = \frac{4(1-\tan ^2 \theta)}{2\tan 2 \theta} + 2 \tan 2 \theta$

$\quad \quad \quad \quad \quad = \frac{2}{\tan 2 \theta}$

$\text{L.H.S.} = \tan \theta + \frac{2}{\tan 2 \theta}$

$\quad = \tan \theta + \frac{2(1- \tan ^2 \theta)}{2 \tan \theta}$

$\quad = \frac{\tan ^2 \theta +1 - \tan ^2 \theta}{ \tan \theta}$

$\quad = \frac{1}{\tan \theta} \\ \quad = \cot \theta = \text{R.H.S.}$

$\text{Show \ that:} \ \cos A \cos 2A \cos 4A = \frac{ \sin 8A}{8 \sin A}$
$\quad$

Solution:

$\because \ \sin 2 \theta = 2 \sin \theta \cos \theta$

$\theta = 4A$

$\therefore \ \sin 8A = 2 \sin 4A \cos 4A$

$\text{R.H.S.} = \frac{\sin8A}{8 \sin A} = \frac{2 \sin 4A}{8 \sin A} \cos4A$

$\sin 4A = 2\sin 2A \cos 2A$

$\text{We have to show}$

$\frac{2 \sin 4A}{8 \sin A} = \cos A \cos2A \quad$

$\because \sin 4A = 2 \sin 2A \cos 2A$

$\therefore \ \frac{2 \sin 4A}{8 \sin A} = 2 \ \frac{2 \sin 2A \cos 2A}{8 \sin A}$

$\quad \quad \quad \quad = \frac{2 \sin A \cos A}{2 \sin A} \cos 2A \quad \{\because \sin 2A = 2 \sin A \cos A\}$

$\frac{2 \sin 4A}{8 \sin A} = \cos A \cos 2A$

Thank you