Question

If $\tan \theta =\dfrac{x\sin \phi }{1-x\cos \phi }$and $\tan \phi =\dfrac{y\sin \theta }{1-y\cos \theta }$, then find $\dfrac{x}{y}$

Answer 1

Hint: To solve the above question we will first find the value of x and y by using the cross multiplication method for the given expressions. Then we will divide x by y and then we will simplify it to get a simpler answer. For simplifying, we will use the formula $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.

Complete step-by-step solution
Since we have to find the value of $\dfrac{x}{y}$ so we will first find the value of x and y.
Since, from question we know that $\tan \theta =\dfrac{x\sin \phi }{1-x\cos \phi }$
After cross multiplication we will get:
$\tan \theta -x\cos \phi \tan \theta =x\sin \phi $
$\Rightarrow \tan \theta =x\left( \cos \phi \tan \theta +\sin \phi \right)$
Now, we will take common take $\cos \phi $ common from the Right-hand side, then we will get:
$\Rightarrow x\cos \phi \left( \tan \theta +\dfrac{\sin \phi }{\cos \phi } \right)=\tan \theta $
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, so we will put it in above equation:
$\Rightarrow x\cos \phi \left( \tan \theta +\tan \phi \right)=\tan \theta $
$\therefore x=\dfrac{\tan \theta }{\cos \phi \left( \tan \theta +\tan \phi \right)}..............\left( 1 \right)$
Now, we will find the value of y using the equation $\tan \phi =\dfrac{y\sin \theta }{1-y\cos \theta }$
After cross multiplication we will get:
$\tan \phi -y\cos \theta \tan \phi =y\sin \theta $
$\Rightarrow \tan \phi =y\left( \cos \theta \tan \phi +\sin \theta \right)$
Now, we will take common take $\cos \theta $ common from the Right-hand side, then we will get:
$\Rightarrow y\cos \theta \left( \tan \phi +\dfrac{\sin \theta }{\cos \theta } \right)=\tan \phi $
We know that $\tan \phi =\dfrac{\sin \phi }{\cos \phi }$, so we will put it in above equation
$\Rightarrow y\cos \theta \left( \tan \theta +\tan \phi \right)=\tan \phi $
$\therefore y=\dfrac{\tan \phi }{\cos \theta \left( \tan \theta +\tan \phi \right)}..............\left( 2 \right)$
Now, we will divide equation (1) and (2), then we will get:
$\Rightarrow \dfrac{x}{y}=\dfrac{\dfrac{\tan \theta }{\cos \phi \left( \tan \theta +\tan \phi \right)}}{\dfrac{\tan \phi }{\cos \theta \left( \tan \theta +\tan \phi \right)}}$
Now, after cancelling the term $\left( \tan \theta +\tan \phi \right)$ we will get:
$\Rightarrow \dfrac{x}{y}=\dfrac{\dfrac{\tan \theta }{\cos \phi }}{\dfrac{\tan \phi }{\cos \theta }}$
$\Rightarrow \dfrac{x}{y}=\dfrac{\tan \theta \cos \theta }{\tan \phi \cos \phi }$
Since, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ so, $\tan \theta \cos \theta $ is equal to $\cos \theta \times \dfrac{\sin \theta }{\cos \theta }=\sin \theta $ . Similarly,$\tan \phi \cos \phi =\sin \phi $
Hence, $\dfrac{x}{y}=\dfrac{\sin \theta }{\sin \phi }$
This is our required solution.

Note: Students are required to note that whenever in the question we are asked to find the value and we are getting the value in any function or variable then we have to simplify that function or variable as much as possible to get simpler form other students will not get full marks in the examination. Another approach is to directly divide the given expressions and then apply the simplifications so as to cancel out similar terms and get the simplest form.