Solveeit Logo
Login

Question

Question: If \(\tan\theta = \frac{x\sin\varphi}{1 - x\cos\varphi}\) and \(\tan\varphi = \frac{y\sin\theta}{1 -...

If tan⁡θ=xsin⁡φ1−xcos⁡φ\tan\theta = \frac{x\sin\varphi}{1 - x\cos\varphi} and tan⁡φ=ysin⁡θ1−ycos⁡θ\tan\varphi = \frac{y\sin\theta}{1 - y\cos\theta}, then xy=\frac{x}{y} =

A

sin⁡φsin⁡θ\frac{\sin\varphi}{\sin\theta}

B

sin⁡θsin⁡φ\frac{\sin\theta}{\sin\varphi}

C

sin⁡φ1−cos⁡θ\frac{\sin\varphi}{1 - \cos\theta}

D

sin⁡θ1−cos⁡φ\frac{\sin\theta}{1 - \cos\varphi}

Answer

sin⁡θsin⁡φ\frac{\sin\theta}{\sin\varphi}

Explanation

Solution

We have tan⁡θ=xsin⁡φ1−xcos⁡φ\tan\theta = \frac{x\sin\varphi}{1 - x\cos\varphi}

⇒1xtan⁡θ−tan⁡θcos⁡φ=sin⁡φ\Rightarrow \frac{1}{x}\tan\theta - \tan\theta\cos\varphi = \sin\varphi

⇒1x=sin⁡φ+cos⁡φtan⁡θtan⁡θ\Rightarrow \frac{1}{x} = \frac{\sin\varphi + \cos\varphi\tan\theta}{\tan\theta} and tan⁡φ=ysin⁡θ1−ycos⁡θ\tan\varphi = \frac{y\sin\theta}{1 - y\cos\theta}

⇒tan⁡φ=sin⁡θ1y−cos⁡θ⇒1ytan⁡φ−tan⁡φcos⁡θ=sin⁡θ\Rightarrow \tan\varphi = \frac{\sin\theta}{\frac{1}{y} - \cos\theta} \Rightarrow \frac{1}{y}\tan\varphi - \tan\varphi\cos\theta = \sin\theta

⇒1ytan⁡φ=sin⁡θ+tan⁡φcos⁡θ\Rightarrow \frac{1}{y}\tan\varphi = \sin\theta + \tan\varphi\cos\theta

∴1y=sin⁡θ+tan⁡φcos⁡θtan⁡φ\therefore\frac{1}{y} = \frac{\sin\theta + \tan\varphi\cos\theta}{\tan\varphi}

Now xy=[tan⁡θsin⁡φ+cos⁡φtan⁡θ]×[sin⁡θ+tan⁡φcos⁡θtan⁡φ]\frac{x}{y} = \left\lbrack \frac{\tan\theta}{\sin\varphi + \cos\varphi\tan\theta} \right\rbrack \times \left\lbrack \frac{\sin\theta + \tan\varphi\cos\theta}{\tan\varphi} \right\rbrack

=tan⁡θtan⁡φ[sin⁡θ+cos⁡θsin⁡φcos⁡φsin⁡φ+cos⁡φsin⁡θcos⁡θ]=tan⁡θcos⁡θtan⁡φcos⁡φ=sin⁡θsin⁡φ= \frac{\tan\theta}{\tan\varphi}\left\lbrack \frac{\sin\theta + \cos\theta\frac{\sin\varphi}{\cos\varphi}}{\sin\varphi + \cos\varphi\frac{\sin\theta}{\cos\theta}} \right\rbrack = \frac{\tan\theta\cos\theta}{\tan\varphi\cos\varphi} = \frac{\sin\theta}{\sin\varphi}

Aliter : xsin⁡φ=tan⁡θ−xcos⁡φtan⁡θx\sin\varphi = \tan\theta - x\cos\varphi\tan\theta

⇒x=tan⁡θsin⁡φ+cos⁡φtan⁡θ\Rightarrow x = \frac{\tan\theta}{\sin\varphi + \cos\varphi\tan\theta} =sin⁡θcos⁡θsin⁡φ+cos⁡φsin⁡θ=sin⁡θsin⁡(θ+φ)= \frac{\sin\theta}{\cos\theta\sin\varphi + \cos\varphi\sin\theta} = \frac{\sin\theta}{\sin(\theta + \varphi)}

Similarly, y=sin⁡φsin⁡(θ+φ)y = \frac{\sin\varphi}{\sin(\theta + \varphi)}; ∴xy=sin⁡θsin⁡φ.\therefore\frac{x}{y} = \frac{\sin\theta}{\sin\varphi}.