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Question: If \(m\cos(\theta + \alpha) = n\cos(\theta - \alpha)\), then \(\cot\theta\cot\alpha\)equal to...

If mcos⁡(θ+α)=ncos⁡(θ−α)m\cos(\theta + \alpha) = n\cos(\theta - \alpha), then cot⁡θcot⁡α\cot\theta\cot\alphaequal to

A

m+nm−n\frac{m + n}{m - n}

B

m−nm+n\frac{m - n}{m + n}

C

m+nn−m\frac{m + n}{n - m}

D

n−mn+m\frac{n - m}{n + m}

Answer

m+nn−m\frac{m + n}{n - m}

Explanation

Solution

mn=cos⁡(θ−α)cos⁡(θ+α)\frac{m}{n} = \frac{\cos(\theta - \alpha)}{\cos(\theta + \alpha)}

By componendo and dividendo rule,

m+nm−n=cos⁡(θ−α)+cos⁡(θ+α)cos⁡(θ−α)−cos⁡(θ+α)\frac{m + n}{m - n} = \frac{\cos(\theta - \alpha) + \cos(\theta + \alpha)}{\cos(\theta - \alpha) - \cos(\theta + \alpha)} ⇒ m+nm−n=2cos⁡θcos⁡α−2sin⁡θsin⁡α\frac{m + n}{m - n} = \frac{2\cos\theta\cos\alpha}{- 2\sin\theta\sin\alpha}

cot⁡θcos⁡α=m+nn−m\cot\theta\cos\alpha = \frac{m + n}{n - m}.