Question

Let $a,b,c$ are non zero constant number then the value of

$\underset{r\to \infty }{lim}\frac{\cos \frac{a}{r}-\cos \frac{b}{r}\cos \frac{c}{r}}{\sin \frac{b}{r}\sin \frac{c}{r}}$ is?

• 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

The correct option is C $\frac{b^2+c^2-a^2}{2bc}$

${lim}_{r\to \infty }\frac{\cos \left(\frac{a}{r}\right)-\cos \left(\frac{b}{r}\right)\cos \left(\frac{c}{r}\right)}{\sin \left(\frac{b}{r}\right)\sin \left(\frac{c}{r}\right)}=\frac{\cos \left(\frac{a}{r}\right)-\cos \left(\frac{b+c}{r}\right)-\sin \left(\frac{b}{r}\right)\sin \left(\frac{c}{r}\right)}{\sin \left(\frac{b}{r}\right)\sin \left(\frac{c}{r}\right)}=\frac{-2\sin \left(\frac{a+b+c}{2r}\right)\sin \left(\frac{a-b-c}{2r}\right)-\sin \left(\frac{b}{r}\right)\sin \left(\frac{c}{r}\right)}{\sin \left(\frac{b}{r}\right)\sin \left(\frac{c}{r}\right)}$

${lim}_{r\to \infty }\frac{-2\sin \left(\frac{a+b+c}{2r}\right)\sin \left(\frac{a-b-c}{2r}\right)-\sin \left(\frac{b}{r}\right)\sin \left(\frac{c}{r}\right)}{\sin \left(\frac{b}{r}\right)\sin \left(\frac{c}{r}\right)}=\frac{-2\times \frac{a+b+c}{2r}\times \frac{a-b-c}{2r}-\frac{b}{r}\times \frac{c}{r}}{\frac{b}{r}\times \frac{c}{r}}=\frac{b^2+c^2-a^2+bc-bc}{2bc}=\frac{b^2+c^2-a^2}{2bc}$

we have used

${lim}_{x\to 0}\frac{\sin x}{x}=1$

$\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

Hence, option 'C' is correct.