Question

If $p\cot \theta =q$ examine whether $\frac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\frac{p^2-q^2}{p^2+q^2}$

Answer 1

If $pcot\theta =q$

$\cot \theta =\frac{q}{p}$ .....(1)

Now solving $LHS,$

$\frac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }$

since $\cot \theta =\frac{\cos \theta }{\sin \theta }$

Therefore,

$=\frac{\sin \theta (p-q\cot \theta )}{\sin \theta (p+q\cot \theta )}$

$=\frac{p-q\cot \theta }{p+q\cot \theta }$

Substituting the value of $\cot \theta$ , We get

$=\frac{p-q\times \frac{q}{p}}{p+q\times \frac{q}{p}}$

$=\frac{p^2-q^2}{p^2+q^2}$

$=RHS$

Hence proved.