Question

If $\tan \theta = \dfrac{p}{q}$ show that $\dfrac{{p\sin \theta - q\cos \theta }}{{p\sin \theta + q\cos \theta }} = \dfrac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}}$.

Answer 1

Hint: Here we will proceed the solution by dividing the LHS part with $\cos \theta $ then we have to show that LHS is equal to RHS.

Complete Step-by-Step Solution:-
Here we have to prove that LHS=RHS
So, let us consider
$\dfrac{{p\sin \theta - q\cos \theta }}{{p\sin \theta + q\cos \theta }}$$ \to 1$
Now let us divide equation $1$ with $\cos \theta $
We know that $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $
So equation $1$ turns to $\dfrac{{p\tan \theta - q}}{{p\tan \theta + q}}$$ \to 2$
Given that $\tan \theta = \dfrac{p}{q}$
Now let us substitute $\tan $ value in equation $2$
$ \Rightarrow $$\dfrac{{p\left( {\dfrac{p}{q}} \right) - q}}{{p\left( {\dfrac{p}{q}} \right) + q}}$
Let us simplify the above equation by taking LCM
$ \Rightarrow \dfrac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}}$
Hence we have proved that LHS=RHS.

NOTE: In this problem without making the solution lengthy we have divided the LHS with $\cos \theta $ where the equation has turned into tan form and tan value has been given for substitution which is very simple to get the answer.