Question

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$ \Large \tt \green{Question :}$
(1 + cos(x) + cos(2x)) / (sin(x) + sin(2x)) = cot(x)

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Answer 1

Given :-

$\sf{\frac{(1 + cos(x) + cos(2x))}{ (sin(x) + sin(2x)) }= cot(x)}$

To Find :-

Here, according to your question, It is given to prove the Identity.

Solution :-

We can prove the Identity by using the methods below,

$\tt{ \frac{(1 + cos(x) + cos(2x))}{ (sin(x) + sin(2x)) }= cot(x)}$

$ \large \mathcal{LHS :}$

$\mapsto\tt{ \frac{1 + cos(x) + cos(2x)}{sin(x) + sin(2x)}}$

$\mapsto \tt{ \frac{ 1 + cosx + 2cos^{2}x - 1 }{sinx + 2sinxcosx}}$

$\mapsto \tt{ \frac{cosx(1 + 2cosx) }{sinx(1 + 2cosx) }}$

• Therefore,

$\mapsto \tt{ \frac{ (1 + cosx + cos2x)}{(sinx + sin2x)} }$

$\mapsto \tt{ \frac{cosx(1 + 2cosx)}{ sinx(1 + 2cosx) }}$

$\mapsto \tt{ \frac{ cosx}{sinx}}$

$\mapsto \tt{ cotx }$

$ \large \mathcal{: RHS}$

• Hence, Proved!!!

Thanks!!! :D

Answer 2

Given :-

• L.H.S = (1 + cos(x) + cos(2x))/(sin(x) + sin(2x))
• R.H.S = Cot(x)

To Show :-

• Left hand side = Right hand side :-]

Solution :-

• To proof L.H.S = R.H.S by simplifying L.H.S by applying trigonometry formula.

Calculation for L.H.S :-

↠ (1 + cos(x) + cos(2x))/(sin(x) + sin(2x))

• Cos2x = 2Cos²x - 1 , sin(2x) = 2sin(x).cos(x)

↠ (1 + cos(x) + 2cos²(x) - 1))/(sin(x) + 2sin(x).cos(x))

↠ (cos(x) + 2cos²(x))/(sin(x) + 2sin(x).cos(x))

• Taking common cos(x) at top. :-]
• Taking common sin(x) at bottom :-]

↠ {cos(x)(1 + 2cos(x))/{sin(x)(1 + 2cos(x))

• Cancelling (1 + 2cos(x)) here we get :-]

↠ cos(x)/sin(x) ↠ cot(x)

↠(1 + cos(x) + cos(2x))/(sin(x) + sin(2x)) = cot(x)

Hence, proved , L.H.S = R.H.S :-