Question 12

Maths Question: 

$\text{Find the values in terms of }p\text{ and }q\text{ of }\frac{p\cos \theta +q\sin \theta }{p\cos \theta -q\sin \theta }\text{ where }\cot \theta =\frac{p}{q}$

Maths Solution: 

$\begin{align}  & \text{Divide both numerator and denominator by }\sin \theta  \\ & \frac{p\cos \theta +q\sin \theta }{p\cos \theta -q\sin \theta }=\frac{\tfrac{p\cos \theta +q\sin \theta }{\sin \theta }}{\tfrac{p\cos \theta -q\sin \theta }{\sin \theta }} \\ & \frac{p\cos \theta +q\sin \theta }{p\cos \theta -q\sin \theta }=\frac{\tfrac{p\cos \theta }{\sin \theta }+\tfrac{q\sin \theta }{\sin \theta }}{\tfrac{p\cos \theta }{\sin \theta }-\tfrac{q\sin \theta }{\sin \theta }}=\frac{p\cot \theta +q}{p\cot \theta -q} \\ & \frac{p\cos \theta +q\sin \theta }{p\cos \theta -q\sin \theta }=\frac{p(\tfrac{p}{q})+q}{p(\tfrac{p}{q})-q}=\frac{\tfrac{{{p}^{2}}}{q}+p}{\tfrac{{{q}^{2}}}{q}-q}=\frac{\tfrac{{{p}^{2}}+{{q}^{2}}}{q}}{\tfrac{{{p}^{2}}-{{q}^{2}}}{q}} \\ & \frac{p\cos \theta +q\sin \theta }{p\cos \theta -q\sin \theta }=\frac{{{p}^{2}}+{{q}^{2}}}{{{p}^{2}}-{{q}^{2}}} \\\end{align}$

University mathstopic: