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Prove that,
$\frac{ \cos4x + \cos3x + \cos2x}{ \sin4x + \sin3x + \sin2x } = \cot3x$
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Answer 1

Given :

• $\dfrac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x}$

To Prove :

• $\dfrac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x }=\cot3x$

Solution :

Using Identities :

• cos x + cos y = 2cos x+y/2 cos x-y/2
• sin x + sin y = 2sin x+y/2 cos x-y/2

$\longmapsto\tt{\dfrac{2cos\dfrac{4x+2x}{2}\:cos\dfrac{4x-2x}{2}+cos\:3x}{2sin\dfrac{4x+2x}{2}\:cos\dfrac{4x-2x}{2}}}$

$\longmapsto\tt{\dfrac{2cos\dfrac{{\not{6}}x}{{\not{2}}}\:cos\dfrac{{\not{2}}x}{{\not{2}}}+cos\:3x}{2sin\dfrac{{\not{6x}}}{{\not{2}}}\:cos\dfrac{{\not{2}}x}{{\not{2}}}}}$

$\longmapsto\tt{\dfrac{2\:cos\:3x\:cosx+cos\:3x}{2\:sin\:3x\:cosx+sin\:3x}}$

By Taking Common :

$\longmapsto\tt{\dfrac{cos\:3x{\cancel{(2\:cosx+1)}}}{sin\:3x{\cancel{(2\:cosx+1)}}}}$

$\longmapsto\tt{\dfrac{cos\:3x}{sin\:3x}}$

$\longmapsto\tt\bf{cot\:3x}$

Hence Proved!

Answer 2

Answer:

Question :-

$\mapsto \bf{Prove\: that\: :\: \dfrac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x} =\: cos3x}$

Solution :-

$\bigstar\: \: \purple{\bold{L.H.S =\: \dfrac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x}}}$

$\implies \sf \dfrac{(cos4x + cos2x) + cos3x}{(sin4x + sin2x) + sin3x}$

As we know that :

$\clubsuit\: \: \sf\bold{\pink{cos A + cos B =\: 2cos\bigg\lgroup \dfrac{A + B}{2}\bigg\rgroup cos\bigg\lgroup \dfrac{A - B}{2}\bigg\rgroup}}$

$\clubsuit\: \: \: \sf\bold{\pink{sin A + sin B =\: 2sin\bigg\lgroup \dfrac{A + B}{2}\bigg\rgroup cos\bigg\lgroup \dfrac{A - B}{2}\bigg\rgroup}}$

Then, we can write as :

$\implies \sf \dfrac{2cos\bigg\lgroup \dfrac{4x + 2x}{2}\bigg\rgroup cos\bigg\lgroup \dfrac{4x - 2x}{2}\bigg\rgroup + cos3x}{2sin\bigg\lgroup \dfrac{4x + 2x}{2}\bigg\rgroup cos\bigg\lgroup \dfrac{4x - 2x}{2}\bigg\rgroup + sin3x}$

$\implies \sf \dfrac{2\: cos\: 3x\: cos x\: + cos3x}{2\: sin\: 3x\: cos\: x\: + sin 3x}$

$\implies \sf \dfrac{cos\: 3x\cancel{(2\: cos\: x + 1)}}{sin\: 3x\cancel{(2\: cos\: x + 1)}}$

$\implies \sf \dfrac{cos\: 3x}{sin\: 3x}$

As we know that :

$\clubsuit \: \: \sf\bold{\pink{cot A =\: \dfrac{cos A}{Sin A}}}$

$\implies \sf\bold{\red{cot 3x}}$

Again,

$\bigstar\: \: \sf\purple{\bold{R.H.S =\: cot\: 3x}}$

$\implies \sf\bold{\red{cot\: 3x}}$

Hence,

$\leadsto \sf\bold{L.H.S =\: R.H.S}$

$\large\green{\underline{{\boxed{\textbf{Hence\: Proved.}}}}}$

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EXTRA INFORMATION :-

❒ Quotient Identities :

$\clubsuit\: \: \sf\bold{\pink{tan A =\: \dfrac{sin A}{cos A}}}$

$\clubsuit\: \: \sf\bold{\pink{cot A =\: \dfrac{cos A}{sin A}}}$

❒ Reciprocal Identities :

$\clubsuit\: \: \sf\bold{\pink{cot A =\: \dfrac{1}{tan A}}}$

$\clubsuit\: \: \sf\bold{\pink{cosec A =\: \dfrac{1}{sin A}}}$

$\clubsuit\: \: \sf\bold{\pink{sec A =\: \dfrac{1}{cos A}}}$

$\clubsuit\: \: \sf\bold{\pink{sin A =\: \dfrac{1}{cosec A}}}$

$\clubsuit\: \: \sf\bold{\pink{cos A =\: \dfrac{1}{sec A}}}$

$\clubsuit\: \: \sf\bold{\pink{tan A =\: \dfrac{1}{cot A}}}$

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