If psinα=q sin(120°+α)=r sin(240°+α) prove that pq+qr+rp=0
If psinα=q sin(120°+α)=r sin(240°+α) prove that pq+qr+rp=0
Solution:
P sinα=q sin(120°+α)=r sin(240°+α)
or, p sinα = q(sin120° cosα+cos120° sinα) = r (sin240° cosα +cos240° sinα)
or, p sinα=q{sin(180°-60°) cosα + cos(180°-60°) sinα}= r {sin(270°-30°) cosα + cos(270°-30°) sinα}
or, p sinα = q(sin 60° cosα –cos60° sinα) = r (-cos30° cosα –sin30° sinα)
or, p sinα =q( $\frac{\sqrt{3}}{2}$ cosα – $\frac{1}{2}$ sinα) = r (-$\frac{\sqrt{3}}{2}$ cosα – $\frac{1}{2}$ sinα)
or, p sinα = $\frac{q}{2}$ ( $\sqrt{3}$cosα – sinα) = – $\frac{r}{2}$ ($\sqrt{3}$ cosα +sinα)
Let, p sinα = $\frac{q}{2}$ ($\sqrt{3}$ cosα – sinα) = – $\frac{r}{2}$ ($\sqrt{3}$ cosα +sinα) =K(say)
∴ P= $\frac{k}{\sin \alpha }$ , q = $\frac{2k}{\sqrt{3}cos\alpha – \sin \alpha }$ and
r = -$\frac{2k}{\sqrt{3}cos\alpha +\sin \alpha }$
∴ pq+qr+rp
= $\frac{2k^2}{\sin \alpha \left(\sqrt{3}\cos \alpha – sin\alpha \right)} – \frac{4k^2}{3cos^2\alpha – sin^2\alpha } – \frac{2k^2}{\sin \alpha \left(\sqrt{3}\cos \alpha + sin\alpha \right)}$
= $\frac{2k^2\left(\sqrt{3}\cos \alpha + sin\alpha \right) – 4k^2\sin \alpha – 2k^2\left(\sqrt{3}\cos \alpha – sin\alpha \right)}{\sin \alpha \left(3cos^2\alpha – sin^2\alpha \right)} $
= $\frac{2\sqrt{3}k^2\cos \alpha + 2k^2sin\alpha – 4k^2\sin \alpha – 2\sqrt{3}k^2\cos \alpha + 2k^2\sin \alpha }{\sin \alpha \left(3cos^2\alpha – sin^2\alpha \right)}$
= $\frac{4k^2\sin \alpha – 4k^2\sin \alpha }{\sin \alpha \left(3cos^2\alpha – sin^2\alpha \right)} $
= 0 [Proved]