<div style='float: left; width:100%;'> <h2 id="successtrigonometric-functions--br-lecture---4success"><Success>Trigonometric functions <br> lecture - 4</Success></h2> </div>

##### <Success>$\text{Formula of} \ \tan (x+y)$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\tan (x+y) = \frac{\sin (x+y)}{\cos (x+y)}$</p> <p>$\quad \quad \quad \quad = \frac{\sin x \ \cos y + \cos x \ \sin y}{\cos x \ \cos y - \sin x \ \sin y}$</p> <p>$\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}} $</p> <p><span style="color:blue">$\tan (x+y)= \frac{\tan x + \tan y }{1- \tan x \ \tan y} $</span></p> </div>

##### <Success>$\text{Formula of} \ \tan (x-y)$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \tan (x-y)= \tan ( x +(-y)) $</p> <p>$ \tan(x-y)= \frac{\tan x + \tan (-y) }{1- \tan x \ \tan(-y)} $</p> <p><span style="color:blue">$\tan (x-y) = \frac{\tan x - \tan y }{1 + \tan x \ \tan y} $</span></p> </div>

##### <Success>$\text{Formula of} \ \tan 2x$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\tan (2x) = \tan (x+x) \\ \quad \quad \quad = \frac{\tan x + \tan x}{1-\tan x.\tan x} \\ \quad \quad \quad = \frac{2 \ \tan x}{1-\tan ^2x}$</p> <p><span style="color:blue">$\tan (2x) = \frac{2 \ \tan x}{1-\tan ^2x}$</span></p> <p>$\biggl(\text{where,} \ 2x \neq \text{odd \ multiple \ of} \ \frac{\pi}{2} \biggl)$</p> </div>

##### <Success>$\text{Formula of} \ \tan 3x$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\tan (3x) = \tan (x+2x) = \frac{\tan x + \tan 2x}{1-\tan x \ \tan 2x} $</p> <p>$\quad \quad \quad = \frac{\tan x + \frac{2\tan x}{1-\tan ^2x}}{1 - \tan x \frac{\ 2\tan x}{1-\tan ^2x}} $</p> <p>$\tan (3x) = \frac{\tan x(1-\tan ^2x) + 2\tan x}{1-\tan ^2x - 2 \tan ^2x}$</p> <p><span style="color:blue">$\tan (3x) = \frac{3 \tan x - \tan ^3x}{1-3\tan ^2x}$</span></p> <p>$\biggl(\text{where,} \ 3x \neq \text{odd \ multiple \ of} \ \frac{\pi}{2}\biggl)$</p> </div>

###### <Success>$\text{Formula of} \ \cot (x+y)$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \cot (x+y) = \frac{1}{\tan (x+y)} \quad [\because \cot x = \frac{1}{\tan x}]$</p> <p>$ \tan (x+y) = \frac{\tan x + \tan y}{1-\tan x \ \tan y} $</p> <p>$ \cot (x+y) = \frac{1-\tan x \ \tan y}{\tan x + \tan y} $</p> <p>$ \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}} $</p> <p><span style="color:blue">$\cot (x+y) = \frac{\cot x \cot y-1}{\cot x + \cot y}$</span></p> <p>$\biggl(\text{where,} \ (x+y) \neq \text{\ multiple \ of} \ \pi \biggl)$</p> <p>======</p> <h5 id="successtextformula-of--cot-x-y----cot-2xsuccess"><Success>$\text{Formula of} \ \cot (x-y) \ \& \ \cot 2x$</Success></h5> <div style='float: left; width:100%;font-size:30px;'> <p>$\cot (x-y) = \frac{\cot x \cot (-y) -1}{\cot x + \cot (-y)}$</p> <p><span style="color:blue">$\cot (x-y) = \frac{1+\cot x \cot y}{\cot y - \cot x}$</span></p> <p>Again,</p> <p>$\cot (x+y) = \frac{\cot x \cot y -1 }{\cot x+\cot y}$</p> <p>Put $ y=x$<br> $\therefore$ <span style="color:blue">$\ \cot (2x) = \frac{\cot ^2x -1}{2 \cot x}$</span></p> </div>

##### <Success>$\text{Formula of} \ \cot 3x$</Success> <div style='float: left; width:100%;font-size:30px;;'> <p>$ \cot (3x) = \cot (x+2x) = \frac{\cot x \ \cot 2x - 1}{\cot x + \cot 2x}$</p> <p>$ \quad \quad \quad \quad = \frac{\cot x \frac{\cot ^2x-1}{2\cot x}-1}{\cot x + \frac{\cot ^2x -1}{2\cot x}}$</p> <p><span style="color:blue">$ \cot (3x) = \frac{\cot ^3x - 3\cot x}{3\cot ^2x - 1}$</span><br> $\biggl(\text{where,} \ 3x \neq \text{\ multiple \ of} \ \pi \biggl)$</p> </div>

#### <Success>Domain & range of cosecant function</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \cosec (x) = \frac{1}{\sin x},$</p> <p><span style="color:blue">$\text{Domain}$</span> : $\R$</p> <p>$|\sin x| \leqslant 1$</p> <p>$|\cosec x| \geqslant 1$</p> <p><span style="color:blue">$\text{Range}$</span> : $(-\infty, -1]\cup[1,\infty)$</p> </div>

##### <Success>Domain & range of secant function</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\sec (x) = \frac{1}{\cos (x)},$</p> <p><span style="color:blue">$\text{Domain}$</span> : $\R$</p> <p>$|\cos x| \leqslant 1$</p> <p>$|\sec x| \geqslant 1$</p> <p><span style="color:blue">$\text{Range}$</span> : $(-\infty, -1]\cup[1,\infty)$</p> </div>

#### <Success>$\text{Proof \ that} \ \cos A + \cos B = 2 \cos \frac{(A+B)}{2} \cos \frac{(A-B)}{2}$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>We know that</p> <p>$2 \cos x \ \cos y = \cos (x+y) + \cos (x-y)$</p> <p>Put,$x = \frac{A+B}{2}, y = \frac{A-B}{2}$</p> <p>$\therefore$ $ 2 \cos \frac{(A+B)}{2} \cos \frac{(A-B)}{2} = \cos A + \cos B$</p> </div>

##### <Success>Problem 1</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Show that: $ \quad \frac{\cos 7x + \cos 3x}{\sin 7x - \sin 3x} = \cot 2x$</p> </div>

#### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\because \cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}$</p> <p>$\therefore \cos 7x + \cos 3x = 2 \cos 5x \cos 2x$</p> <p>$ \because \sin A - \sin B = 2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$</p> <p>$\therefore \sin 7x - \sin 3x = 2 \cos 5x \sin 2x$ <br> $\quad$</p> <p>$\text{L.H.S.} =\frac{ 2 \cos 5x \cos 2x}{2 \cos 5x \sin 2x } = \cot 2x = \text{R.H.S.}$</p> </div>

### <Success>Problem 2</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\text{Show \ that:}$ $\quad \frac{\cos 3x + \cos 4x + \cos 5x}{\sin 3x + \sin 4x + \sin 5x} = \cot 4x $</p> <p>======</p> <h3 id="successsolutionsuccess"><Success>Solution</Success></h3> <div style='float: left; width:100%;font-size:30px;'> <p>$\because \ \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} $</p> <p>$\therefore \ \cos 3x + \cos 5x = 2 \cos 4x \cos x $</p> <p>$\text{Numerator(L.H.S.)} = 2 \cos 4x \cos x + \cos 4x$</p> <p>$ \quad \quad \quad \quad \quad \quad \quad = \cos 4x(1+2 \cos x)$</p> </div> <p>======</p> <h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3> <div style='float: left; width:100%;font-size:30px;'> <p>$\because \ \sin A + \sin B = 2\sin \frac{A+B}{2} \cos\frac{A-B}{2}$</p> <p>$\therefore \ \sin 3x + \sin 5x = 2 \sin 4x \cos x$</p> <p>$\text{Denominator(L.H.S.)} = \sin 4x(1+2 \cos x)$</p> <p>$\quad$</p> <p>$\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 $</p> </div>

##### <Success>Problem 3</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\text{Compute:} \ \sin (18^{\circ})$ $\quad$</p> <p><strong>Solution:</strong></p> <p>$\because \sin x = \cos (\frac{\pi}{2} - x) $</p> <p>$ \sin 36^{\circ} = \cos 54^{\circ},$ <br> $\text{We know that,}$</p> <p>$ \sin 2 \theta = 2 \sin \theta \ cos \theta$<br> $ \cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$<br> $\text{For} \ \theta = 18^{\circ}, \sin 2 \theta = \cos 3 \theta$</p> <p>$\therefore 2 \sin \theta \cos \theta - (4 \cos^3 \theta - 3 \cos \theta) = 0$</p> </div>

##### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \cos \theta (2 \sin \theta - 4 \cos^2 \theta + 3) = 0$<br> For $\theta = 18^{\circ}, \cos \theta \neq 0$</p> <p>$\therefore \quad (2 \sin \theta - 4 \cos^2 \theta + 3) = 0$</p> <p>$\Rightarrow 4 \cos^2 \theta - 2 \sin \theta -3 = 0$</p> <p>$\Rightarrow 4-4 \sin^2 \theta - 2 \sin \theta -3 = 0 $</p> <p>$\Rightarrow 4 \sin^2 \theta +2 \sin \theta -1 = 0 \\ \quad z \triangleq \sin \theta$</p> </div> ====== ##### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$4z^2 +2z -1 = 0$</p> <p>$z = \frac {-2 \pm \sqrt{20}}{8}$</p> <p>$ \sin 18^{\circ} = \frac {-2 +\sqrt{20}}{8} = \frac{ \sqrt{5} -1}{4}$</p> </div>

##### <Success>Problem 4</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\text{Prove \ that:} \ \sqrt{ \frac{1- \cos x}{1 + \cos x}} = \cosec x - \cot x $</p> <p>$\quad$<br> <strong>Solution:</strong></p> <p>$\text{R.H.S.}=\frac{1}{\sin x} - \frac{ \cos x}{ \sin x}$</p> <p>$\quad \quad \quad =\frac{1- \cos x}{\sin x}$</p> <p>$\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}}$</p> <p>$\quad \quad =\frac{1- \cos x}{\sin x} = \text{R.H.S.}$</p> </div>

##### <Success>Problem 5</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \text{Find \ the \ value \ of} \ \ \cos 40^{\circ} - \cos 20^{\circ} + \cos 80^{\circ} .$ $\quad$<br> <strong>Solution:</strong><br> $\because \ \cos A + \cos B = 2 \cos \Biggl( \frac{A+B}{2}\Biggl) \cos \Biggl( \frac{A-B}{2}\Biggl)$</p> <p>$\therefore \ \cos 40^{\circ} + \cos 80^{\circ} = 2 \cos 60^{\circ} \cos 20^{\circ}$</p> <p>$\quad \quad \quad \quad \quad \quad \quad \quad \ = \cos 20^{\circ} \quad [\because \cos 60^{\circ} = \frac {1}{2}]$</p> <p>$\cos 40^{\circ} - \cos 20^{\circ} + \cos 80^{\circ} \\ = \cos 20^{\circ} - \cos 20^{\circ} \\ =0$</p> </div>

### <Success>Problem 6</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \text{Prove \ that:} \ \tan \theta + 2 \tan 2 \theta + 4 \tan 4 \theta + 8 \cot 8 \theta = \cot \theta $ <br> $\quad$<br> <strong>Solution:</strong></p> <p>$\because \tan 2x = \frac{2 \tan x}{1-\tan ^2x}$</p> <p>$x=4 \theta$</p> <p>$ \tan 8\theta = \frac{2 \tan4 \theta}{1- \tan ^2 4 \theta}$<br> Now,</p> <p>$8 \cot 8\theta + 4 \tan 4 \theta = \frac{8(1- \tan ^2 4 \theta)}{2 \tan 4 \theta} + 4 \tan 4 \theta$<br> $\quad \quad \quad \quad = 4\Biggl[\frac{1}{\tan 4 \theta}\Biggl]$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\text{L.H.S.} = \tan \theta + 2 \tan 2\theta + \frac{4}{\tan 4 \theta}$</p> <p>$\text{Again}, \ \tan 4 \theta = \frac{2\tan 2 \theta}{1-\tan ^2 2 \theta}$</p> <p>$\frac{4}{\tan 4 \theta} + 2 \tan 2 \theta = \frac{4(1-\tan ^2 \theta)}{2\tan 2 \theta} + 2 \tan 2 \theta$</p> <p>$\quad \quad \quad \quad \quad = \frac{2}{\tan 2 \theta}$</p> <p>======</p> <h3 id="successsolutionsuccess-2"><Success>Solution</Success></h3> <div style='float: left; width:100%;font-size:30px;'> <p>$\text{L.H.S.} = \tan \theta + \frac{2}{\tan 2 \theta}$</p> <p>$\quad = \tan \theta + \frac{2(1- \tan ^2 \theta)}{2 \tan \theta}$</p> <p>$ \quad = \frac{\tan ^2 \theta +1 - \tan ^2 \theta}{ \tan \theta}$</p> <p>$ \quad = \frac{1}{\tan \theta} \\ \quad = \cot \theta = \text{R.H.S.}$</p> </div>

### <Success>Problem 7</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\text{Show \ that:} \ \cos A \cos 2A \cos 4A = \frac{ \sin 8A}{8 \sin A}$<br> $\quad$</p> <p><strong>Solution:</strong></p> <p>$\because \ \sin 2 \theta = 2 \sin \theta \cos \theta$</p> <p>$ \theta = 4A $</p> <p>$\therefore \ \sin 8A = 2 \sin 4A \cos 4A$</p> <p>$\text{R.H.S.} = \frac{\sin8A}{8 \sin A} = \frac{2 \sin 4A}{8 \sin A} \cos4A$</p> <p>$\sin 4A = 2\sin 2A \cos 2A$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\text{We have to show}$</p> <p>$\frac{2 \sin 4A}{8 \sin A} = \cos A \cos2A \quad $</p> <p>$\because \sin 4A = 2 \sin 2A \cos 2A$</p> <p>$\therefore \ \frac{2 \sin 4A}{8 \sin A} = 2 \ \frac{2 \sin 2A \cos 2A}{8 \sin A}$</p> <p>$ \quad \quad \quad \quad = \frac{2 \sin A \cos A}{2 \sin A} \cos 2A \quad \{\because \sin 2A = 2 \sin A \cos A\}$</p> <p>$ \frac{2 \sin 4A}{8 \sin A} = \cos A \cos 2A $</p> </div>

<div> <h2><Success>Thank you</Success></h2> </div>