Question

If $a=bcos\frac{2\pi }{3}=ccos\frac{4\pi }{3}$, then write the value of ab+bc+ca.

Answer 1

Let,<br/>
$a=bcos\frac{2\pi }{3}=ccos\frac{4\pi }{3}=k$<br/>
$\Rightarrow a=k,b=\frac{k}{cos\frac{2\pi }{3}}$ and $c=\frac{k}{cos\frac{4\pi }{3}}$ <br/>
Now, $ab+bc+ca=k\times \frac{k}{cos\frac{2\pi }{3}}+\frac{k}{cos\frac{2\pi }{3}}\times \frac{k}{cos\frac{4\pi }{3}}+\frac{k}{cos\frac{4\pi }{3}}\times k$<br/>
$=k^{2}sec\frac{2\pi }{3}+k^{2}sec\frac{2\pi }{3}sec\frac{4\pi }{3}+k^{2}sec\frac{4\pi }{3}$<br/>
$=k^{2}sec(\frac{\pi }{2}+\frac{\pi }{6})+k^{2}sec(\frac{\pi }{2}+\frac{\pi }{6})sec(\pi +\frac{\pi }{3})+k^{2}sec(\pi +\frac{\pi }{3})$<br/>
$=-k^{2}cosec\frac{\pi }{6}+k^{2}cosec\frac{\pi }{6}sec\frac{\pi }{3}-k^{2}sec\frac{\pi }{3}$<br/>
$=-k^2\times 2+k^2\times 2\times 2-k^2\times 2$<br/>
$=-4k^2+4k^2=0$<br/>
$\therefore ab+bc+ca=0$