Question

Show that cot 2x cot x-cot 3x cot2x-cot3x cot x=1​

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

$\rm \: 3x = 2x + x$

So,

$\rm \: cot3x = cot(2x + x)$

We know,

$\boxed{ \rm{ \:cot(x + y) = \frac{cotx \: coty \: - \: 1}{coty + cotx} \: }}$

So, using this result, above can be rewritten as

$\rm \: cot3x = \dfrac{cot2x \: cotx \: - \: 1 }{cot2x + cotx}$

$\rm \: cot3xcot2x + cot3xcotx = cot2xcotx - 1$

On transposition, we get

$\rm \: cotxcot2x - cot2xcot3x - cot3xcotx = 1$

Hence,

$\rm\implies \boxed{ \rm{cotxcot2x - cot2xcot3x - cot3xcotx = 1}}$

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Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(x - y) = sinx \: cosy \: - \: siny \: cosx}\\ \\ \bigstar \: \bf{sin(x + y) = sinx \: cosy \: + \: siny \: cosx}\\ \\ \bigstar \: \bf{cos(x + y) = cosx \: cosy \: - \: sinx \: siny}\\ \\ \bigstar \: \bf{cos(x - y) = cosx \: cosy \:+\: siny \: sinx}\\ \\ \bigstar \: \bf{tan(x + y) = \dfrac{tanx + tany}{1 - tanx \: tany} }\\ \\ \bigstar \: \bf{tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

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$\begin{array}{l} \displaystyle \rm\cot 3 x=\cot (2 x+x)=\frac{\cot 2 x \cot x-1}{\cot x+\cot 2 x} \qquad\qquad \left[\because \cot (A+B)=\frac{\cot A \cot B-1}{\cot B+\cot A}\right] \\ \\ \displaystyle \rm \Rightarrow \quad \quad \cot 3 x(\cot x+\cot 2 x)=\cot 2 x \cot x-1 \\ \\ \displaystyle \rm \Rightarrow \quad \cot 3 x \cot x+\cot 3 x \cot 2 x=\cot 2 x \cot x-1 \\ \\ \displaystyle \rm \Rightarrow \quad \cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1 \end{array}$