Question :- Prove that
[tex]\sf \: \dfrac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x} = cot3x \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Consider,
[tex]\sf \: \dfrac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{(cos4x + cos2x) + cos3x}{(sin4x + sin2x) + sin3x} \\ \\ [/tex]
We know,
[tex] \:\boxed{\begin{aligned}& \qquad \:\sf \:cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\qquad \: \\ \\& \qquad \:\sf \: sinx + siny=2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\end{aligned}} \qquad \\ \\ [/tex]
So, using these Identities, we get
[tex]\sf \: = \: \dfrac{2cos\bigg(\dfrac{4x + 2x}{2} \bigg)cos\bigg(\dfrac{4x - 2x}{2} \bigg) + cos3x }{2sin\bigg(\dfrac{4x + 2x}{2} \bigg)cos\bigg(\dfrac{4x - 2x}{2} \bigg) + sin3x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2cos\bigg(\dfrac{6x}{2} \bigg)cos\bigg(\dfrac{2x}{2} \bigg) + cos3x }{2sin\bigg(\dfrac{6x}{2} \bigg)cos\bigg(\dfrac{2x}{2} \bigg) + sin3x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2cos3xcosx + cos3x}{2sin3xcosx + sin3x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{cos3x(2cosx + 1)}{sin3x(2cosx + 1)} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{cos3x}{sin3x} \\ \\ [/tex]
[tex]\sf \: = \: cot3x \\ \\ [/tex]
Hence,
[tex]\sf\implies \bf \: \dfrac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x} = cot3x \\ \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional information
[tex] \:\boxed{\begin{aligned}& \qquad \:\sf \:cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\qquad \: \\ \\& \qquad \:\sf \: sinx + siny=2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg) \\ \\ & \qquad \:\sf \:cosx - cosy = - 2sin\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)\qquad \: \\ \\ & \qquad \:\sf \:sinx - siny = 2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)\qquad \:\end{aligned}} \qquad \\ \\ [/tex]
[tex] \frac{cos4x + cos2x + cos3x}{sin4x + sin2x + sin3x} \\ \\ = \frac{2cos \frac{6x}{2} cos \frac{2x}{2} + cos3x}{2sin \frac{6x}{2}cos \frac{2x}{2} + sin3x } \\ \\ = \frac{2cos3xcosx + cos3x}{2sin3xcosx + sin3x} \\ \\ = \frac{cos3x(2cosx + 1)}{sin3x(2cosx + 1)} \\ \\ = cot3x[/tex]
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Answers & Comments
Question :- Prove that
[tex]\sf \: \dfrac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x} = cot3x \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Consider,
[tex]\sf \: \dfrac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{(cos4x + cos2x) + cos3x}{(sin4x + sin2x) + sin3x} \\ \\ [/tex]
We know,
[tex] \:\boxed{\begin{aligned}& \qquad \:\sf \:cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\qquad \: \\ \\& \qquad \:\sf \: sinx + siny=2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\end{aligned}} \qquad \\ \\ [/tex]
So, using these Identities, we get
[tex]\sf \: = \: \dfrac{2cos\bigg(\dfrac{4x + 2x}{2} \bigg)cos\bigg(\dfrac{4x - 2x}{2} \bigg) + cos3x }{2sin\bigg(\dfrac{4x + 2x}{2} \bigg)cos\bigg(\dfrac{4x - 2x}{2} \bigg) + sin3x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2cos\bigg(\dfrac{6x}{2} \bigg)cos\bigg(\dfrac{2x}{2} \bigg) + cos3x }{2sin\bigg(\dfrac{6x}{2} \bigg)cos\bigg(\dfrac{2x}{2} \bigg) + sin3x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2cos3xcosx + cos3x}{2sin3xcosx + sin3x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{cos3x(2cosx + 1)}{sin3x(2cosx + 1)} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{cos3x}{sin3x} \\ \\ [/tex]
[tex]\sf \: = \: cot3x \\ \\ [/tex]
Hence,
[tex]\sf\implies \bf \: \dfrac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x} = cot3x \\ \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional information
[tex] \:\boxed{\begin{aligned}& \qquad \:\sf \:cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\qquad \: \\ \\& \qquad \:\sf \: sinx + siny=2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg) \\ \\ & \qquad \:\sf \:cosx - cosy = - 2sin\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)\qquad \: \\ \\ & \qquad \:\sf \:sinx - siny = 2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)\qquad \:\end{aligned}} \qquad \\ \\ [/tex]
Verified answer
[tex] \frac{cos4x + cos2x + cos3x}{sin4x + sin2x + sin3x} \\ \\ = \frac{2cos \frac{6x}{2} cos \frac{2x}{2} + cos3x}{2sin \frac{6x}{2}cos \frac{2x}{2} + sin3x } \\ \\ = \frac{2cos3xcosx + cos3x}{2sin3xcosx + sin3x} \\ \\ = \frac{cos3x(2cosx + 1)}{sin3x(2cosx + 1)} \\ \\ = cot3x[/tex]