Question

If $p\sin \theta +q\cos \theta =a$ and $p\cos \theta -q\sin \theta =b$ then $\frac{p+a}{q+b}+\frac{q-b}{p-a}$ is equal to

• A. $1$
• B. $2$$
• C. $0$
• D. $noneofthese$

Answer 1

The correct option is D $noneofthese$

$\frac{p+a}{q+b}+\frac{q-b}{p-a}$

$=\frac{p^2-a^2+q^2-b^2}{(q+b)(p-a)}$ .......$\left(1\right)$

$a^2=p^2{\sin }^2\theta +q^2{\cos }^2\theta +2pq\sin \theta \cos \theta$

$p^2-a^2=p^2-p^2{\sin }^2\theta -q^2{\cos }^2\theta -2pq\sin \theta \cos \theta$

$p^2-a^2=p^2(1-{\sin }^2\theta )-q^2{\cos }^2\theta -2pq\sin \theta \cos \theta$

$p^2-a^2=p^2{\cos }^2\theta -q^2{\cos }^2\theta -2pq\sin \theta \cos \theta$

$p^2-a^2=(p^2-q^2){\cos }^2\theta -2pq\sin \theta \cos \theta$

$b^2=p^2{\sin }^2\theta +q^2{\cos }^2\theta -2pq\sin \theta \cos \theta$

$q^2-b^2=q^2-q^2{\sin }^2\theta -q^2{\cos }^2\theta +2pq\sin \theta \cos \theta$

$q^2-b^2=q^2(1-{\sin }^2\theta )-q^2{\cos }^2\theta +2pq\sin \theta \cos \theta$

$q^2-b^2=q^2{\cos }^2\theta -q^2{\cos }^2\theta +2pq\sin \theta \cos \theta$

$q^2-b^2=(p^2-q^2){\cos }^2\theta +2pq\sin \theta \cos \theta$

Now,

$p^2-a^2+q^2-b^2$

$=(p^2-q^2){\cos }^2\theta -2pq\sin \theta \cos \theta +(p^2-q^2){\cos }^2\theta +2pq\sin \theta \cos \theta$

$=2(p^2-q^2){\cos }^2\theta$