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Question: If \(\tan \theta = \dfrac{p}{q}\) show that \(\dfrac{{p\sin \theta - q\cos \theta }}{{p\sin \theta +...

If tan⁡θ=pq\tan \theta = \dfrac{p}{q} show that psin⁡θ−qcos⁡θpsin⁡θ+qcos⁡θ=p2−q2p2+q2\dfrac{{p\sin \theta - q\cos \theta }}{{p\sin \theta + q\cos \theta }} = \dfrac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}}.

Explanation

Solution

Hint: Here we will proceed the solution by dividing the LHS part with cos⁡θ\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
psin⁡θ−qcos⁡θpsin⁡θ+qcos⁡θ\dfrac{{p\sin \theta - q\cos \theta }}{{p\sin \theta + q\cos \theta }} →1 \to 1
Now let us divide equation 11 with cos⁡θ\cos \theta
We know that sin⁡θcos⁡θ=tan⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta
So equation 11 turns to ptan⁡θ−qptan⁡θ+q\dfrac{{p\tan \theta - q}}{{p\tan \theta + q}} →2 \to 2
Given that tan⁡θ=pq\tan \theta = \dfrac{p}{q}
Now let us substitute tan⁡\tan value in equation 22
⇒\Rightarrow p(pq)−qp(pq)+q\dfrac{{p\left( {\dfrac{p}{q}} \right) - q}}{{p\left( {\dfrac{p}{q}} \right) + q}}
Let us simplify the above equation by taking LCM
⇒p2−q2p2+q2\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⁡θ\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.