Question

If $\tan \theta =\dfrac{p}{q}$, then find the value of the expression $\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }$

Answer 1

Hint: Use the fact that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. Hence assume that $\sin \theta =pk,\cos \theta =qk$. Hence find the value of $p\sin \theta +q\cos \theta $ and the value of $p\sin \theta -q\cos \theta $. Divide the two expressions and hence find the value of $\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }$. Alternatively divide the numerator and the denominator of the expression $\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }$ by $\cos \theta $ and use the fact that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and hence find the value of the given expression.

Complete step by step answer:
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Hence, we have
$\dfrac{\sin \theta }{\cos \theta }=\dfrac{p}{q}$
Let $\sin \theta =pk$ and $\cos \theta =qk$
Hence, we have
$p\sin \theta +q\cos \theta =p\left( pk \right)+q\left( qk \right)={{p}^{2}}k+{{q}^{2}}k$
Taking k common from the terms on RHS, we get
$p\sin \theta +q\cos \theta =k\left( {{p}^{2}}+{{q}^{2}} \right)$
Similarly, we have
$p\sin \theta -q\cos \theta =p\left( pk \right)-q\left( qk \right)=\left( {{p}^{2}}-{{q}^{2}} \right)k$
Hence, we have
$\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\dfrac{{{p}^{2}}-{{q}^{2}}}{{{p}^{2}}+{{q}^{2}}}$

Hence the value of $\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }$ is $\dfrac{{{p}^{2}}-{{q}^{2}}}{{{p}^{2}}+{{q}^{2}}}$

Note: Alternative Solution: Best Method
We can solve the above question using componendo-dividendo which states that if $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Hence, we have
$\dfrac{\sin \theta }{\cos \theta }=\dfrac{p}{q}$
Multiplying both sides by $\dfrac{p}{q}$, we get
$\dfrac{p\sin \theta }{q\cos \theta }=\dfrac{{{p}^{2}}}{{{q}^{2}}}$
Applying componendo-dividendo, we get
$\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\dfrac{{{p}^{2}}-{{q}^{2}}}{{{p}^{2}}+{{q}^{2}}}$, which is the same as obtained above
[2] Alternative Solution:
Dividing numerator and denominator of the expression $\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }$ by $\cos \theta $, we get
$\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\dfrac{\dfrac{p\sin \theta -q\cos \theta }{\cos \theta }}{\dfrac{p\sin \theta +q\cos \theta }{\cos \theta }}$
We know that $\dfrac{a\pm b}{c}=\dfrac{a}{c}\pm \dfrac{b}{c}$
Using the above identity, we get
$\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\dfrac{p\dfrac{\sin \theta }{\cos \theta }-q}{p\dfrac{\sin \theta }{\cos \theta }+q}$
We know that $\tan \theta =\dfrac{p}{q}$
Hence, we have
$\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\dfrac{p\left( \dfrac{p}{q} \right)-q}{p\left( \dfrac{p}{q} \right)+q}$
Multiplying numerator and denominator of RHS by q, we get
$\dfrac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\dfrac{{{p}^{2}}-{{q}^{2}}}{{{p}^{2}}+{{q}^{2}}}$, which is the same as obtained above.