Question

Cos4x +cos3x +cos2x/sin4x +sin3x +sin2x = cot3x

Answer 1

Answer:

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Answer 2

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

❒ 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 :

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

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

❒ Then, we can write as :

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

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

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

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

❒ As we know that :

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

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

$\begin{gathered}\\\end{gathered}$

❒ Again,

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

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

❒ Hence,

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

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