MHB How Do You Solve $\cot x \cdot \cot(x+y)$ Given $\cos y = 17\cos(2x+y)$?

[anemone]

Here is this week's POTW:

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Evaluate $\cot x \cdot \cot(x+y)$ , if $\cos y = 17\cos(2x+y)$.

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[anemone]

Congratulations to the following members for their correct solution:

1. greg1313
2. kaliprasad

Solution from greg1313:

Spoiler

$$\cos y=17\cos(2x+y)$$

$$\cos y=17(\cos2x\cos y-\sin2x\sin y)$$

$$1=17(\cos2x-\sin2x\tan y)$$

$$\tan y=\dfrac{17\cos2x-1}{17\sin2x}$$

$$\cot x\cdot\cot(x+y)=\cot x\cdot\dfrac{1-\tan x\tan y}{\tan x+\tan y}=\dfrac{\cot x-\dfrac{17\cos2x-1}{17\sin2x}}{\tan x+\dfrac{17\cos2x-1}{17\sin2x}}$$

$$=\dfrac{\dfrac{17\sin2x\cos x-17\cos2x\sin x+\sin x}{17\sin2x\sin x}}{\dfrac{17\sin2x\sin x+17\cos2x\cos x-\cos x}{17\sin2x\cos x}}$$

$$=\dfrac{\dfrac{18}{17\sin2x}}{\dfrac{16}{17\sin2x}}=\dfrac98$$

Alternate solution from kaliprasad:

Spoiler

We have $\dfrac{\cos\,y}{\cos(2x+y)} = \dfrac{17}{1}$
using componendo dividendo we get
$\dfrac{\cos\,y+ \cos(2x+y)}{\cos\,y- \cos(2x+y)} = \dfrac{17+1}{17-1}$
or $\dfrac{ 2 \cos(x+y) \cdot \cos\,x}{2\sin(x+y)\cdot \sin\,x} = \dfrac{18}{16}=\dfrac{9}{8} $
or $\cot(x+y)\cdot \cot\,x= \dfrac{9}{8} $