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Mathematics Question on Properties of Inverse Trigonometric Functions

If sin (θ−ϕ)=n sin(θ−ϕ),n≠1,\left(\theta-\phi\right) = n \, sin (\theta - \phi),n \ne1, then the value of tan⁡θtan⁡ϕ\frac{\tan\theta}{\tan\phi} is equal to

A

nn−1\frac{n}{n-1}

B

n+1n−1\frac{n+1}{n-1}

C

n1−n\frac{n}{1-n}

D

n−1n+1\frac{n-1}{n+1}

Answer

n+1n−1\frac{n+1}{n-1}

Explanation

Solution

sin⁡(θ+ϕ)=nsin⁡(θ−ϕ)\sin (\theta+\phi)=n \sin (\theta-\phi)
⇒sin⁡(θ+ϕ)sin⁡(θ−ϕ)=n\Rightarrow \frac{\sin (\theta+\phi)}{\sin (\theta-\phi)}=n
Applying componendo and dividendo, we get
sin⁡(θ+ϕ)+sin⁡(θ−ϕ)sin⁡(θ+ϕ)−sin⁡(θ−ϕ)=n+1n−1\frac{\sin (\theta+\phi)+\sin (\theta-\phi)}{\sin (\theta+\phi)-\sin (\theta-\phi)}=\frac{n+1}{n-1}
⇒[2sin⁡(θ+ϕ+θ−ϕ2)cos⁡(θ+ϕ−θ+ϕ2)][2cos⁡(θ+ϕ+θ−ϕ2)sin⁡(θ+ϕ−θ+ϕ2)]\Rightarrow \frac{\left[2 \sin \left(\frac{\theta+\phi+\theta-\phi}{2}\right) \cos \left(\frac{\theta+\phi-\theta+\phi}{2}\right)\right]}{\left[2 \cos \left(\frac{\theta+\phi+\theta-\phi}{2}\right) \sin \left(\frac{\theta+\phi-\theta+\phi}{2}\right)\right]}
=n+1n−1⇒sin⁡θcos⁡ϕcos⁡θsin⁡ϕ=n+1n−1=\frac{n+1}{n-1} \Rightarrow \frac{\sin \theta \cos \phi}{\cos \theta \sin \phi}=\frac{n+1}{n-1}
⇒tan⁡θtan⁡ϕ=n+1n−1\Rightarrow \frac{\tan \theta}{\tan \phi}=\frac{n+1}{n-1}