Question :-
What is
cot2x cot4x - cot4x cot6x - cot6x cot2x =
(a) - 1
(b) 0
(c) 1
(d) 2
[tex]\large\underline{\sf{Solution-}}[/tex]
We know,
[tex]\rm \: 6x = 4x + 2x \\ [/tex]
[tex]\rm \: \therefore \: cot6x = cot(4x + 2x) \\ [/tex]
[tex]\boxed{ \rm{ \:cot(x + y) = \frac{cotxcoty - 1}{coty + cotx} \: }} \\ [/tex]
So, using this result, we get
[tex]\rm \: cot6x = \dfrac{cot4x \: cot2x \: - 1}{cot2x + cot4x} \\ [/tex]
On cross multiplication we have
[tex]\rm \: cot6x(cot4x + cot2x) = cot4x \: cot2x \: - 1 \\ [/tex]
[tex]\rm \: cot6x \: cot4x +cot6x \: cot2x = cot4x \: cot2x \: - 1 \\ [/tex]
On transposition, we get
[tex]\rm \: cot4x \: cot2x - \: cot6x \: cot4x +cot6x \: cot2x = 1 \\ [/tex]
Hence,
[tex]\rm\implies \:\rm \: cot4x \: cot2x - \: cot6x \: cot4x +cot6x \: cot2x = 1 \\ [/tex]
So, option (c) is correct.
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
[tex]\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}[/tex]
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Verified answer
Question :-
What is
cot2x cot4x - cot4x cot6x - cot6x cot2x =
(a) - 1
(b) 0
(c) 1
(d) 2
[tex]\large\underline{\sf{Solution-}}[/tex]
We know,
[tex]\rm \: 6x = 4x + 2x \\ [/tex]
[tex]\rm \: \therefore \: cot6x = cot(4x + 2x) \\ [/tex]
We know,
[tex]\boxed{ \rm{ \:cot(x + y) = \frac{cotxcoty - 1}{coty + cotx} \: }} \\ [/tex]
So, using this result, we get
[tex]\rm \: cot6x = \dfrac{cot4x \: cot2x \: - 1}{cot2x + cot4x} \\ [/tex]
On cross multiplication we have
[tex]\rm \: cot6x(cot4x + cot2x) = cot4x \: cot2x \: - 1 \\ [/tex]
[tex]\rm \: cot6x \: cot4x +cot6x \: cot2x = cot4x \: cot2x \: - 1 \\ [/tex]
On transposition, we get
[tex]\rm \: cot4x \: cot2x - \: cot6x \: cot4x +cot6x \: cot2x = 1 \\ [/tex]
Hence,
[tex]\rm\implies \:\rm \: cot4x \: cot2x - \: cot6x \: cot4x +cot6x \: cot2x = 1 \\ [/tex]
So, option (c) is correct.
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
[tex]\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}[/tex]