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

Answer 1

We have,

$p\sin \theta +q\cos \theta =a$ ...(1)

$p\cos \theta -q\sin \theta =b$ ...(2)

Squaring (1) and (2), and then adding, we get

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

$⟹(p^{2}si{n}^2\theta +q^{2}co{s}^2\theta +2pqsin\theta cos\theta )+(p^{2}co{s}^2\theta +q^{2}si{n}^2\theta -2pqsin\theta cos\theta )=a^2+b^2$

$⟹p^2(si{n}^2\theta +co{s}^2\theta )+q^2(co{s}^2\theta +si{n}^2\theta )=a^2+b^2$

$⟹p^2\left(1\right)+q^2\left(1\right)-a^2-b^2=0$

$⟹(p^2-a^2)+(q^2-b^2)=0$

$⟹(p+a)(p-a)+(q+b)(q-b)=0$

Dividing $(p-a)(q+b)$ on both sides.

$⟹\frac{p+a}{q+b}+\frac{q-b}{p-a}=0$