Question

let a, b, c are non zero constant number then ${lim}_{r\to \infty }\frac{cos\frac{a}{r}-cos\frac{b}{r}cos\frac{c}{r}}{sin\frac{b}{r}sin\frac{c}{r}}$ equals to

• A. $\frac{a^2+b^2-c^2}{2bc}$
• B. $\frac{c^2+a^2-b^2}{2bc}$
• C. $\frac{b^2+c^2-a^2}{2bc}$
• D. independent of a, b and c

Answer 1

Answer: (C)

$\frac{b^2+c^2-a^2}{2bc}$

$\underset{r\to \infty }{lim}\frac{\cos \frac{a}{r}-\cos \frac{b}{r}\cos \frac{c}{r}}{\sin \frac{b}{r}\sin \frac{c}{r}}$
$=\underset{r\to \infty }{lim}\frac{\cos \frac{a}{r}-\frac{1}{2}(\cos \left(\frac{b+c}{r}\right)+\cos \left(\frac{b-c}{r}\right))}{\frac{1}{2}\left(\cos \left(\frac{b-c}{r}\right)-\cos \left(\frac{b+c}{r}\right)\right)}$
By Applying L Hospital,s Rule

$=\underset{r\to \infty }{lim}\frac{-a\sin \frac{a}{r}-\frac{1}{2}(-(b+c)\sin \left(\frac{b+c}{r}\right)-(b-c)\sin \left(\frac{b-c}{r}\right))}{\frac{-1}{2}(b-c)\sin \left(\frac{b-c}{r}\right)+\frac{1}{2}(b+c)\sin \left(\frac{b+c}{r}\right)}$ $(here\frac{-1}{r^2}$

$wouldcanceloutfromnumeratoranddenominator)$
$=\underset{r\to \infty }{lim}\frac{{-a}^2\cos \frac{a}{r}+\frac{1}{2}\left({(b+c)}^2\cos \left(\frac{b+c}{r}\right)+{(b-c)}^2\cos \left(\frac{b-c}{r}\right)\right)}{\frac{-1}{2}{(b-c)}^2\cos \left(\frac{b-c}{r}\right)+\frac{1}{2}{(b+c)}^2\cos \left(\frac{b+c}{r}\right)}$
$=\frac{b^2+c^2-a^2}{2bc}$