Question

If $a<c$ in $\Delta ABC,$ then sum to infinite terms of the series $n\frac{a}{c}\sin B+\frac{n(n+1)}{2!}\frac{a^2}{c^2}\sin 2B+.....\infty$ is equal to

• A. $\frac{c^n}{b^n}\sin A$
• B. $\frac{a^n}{b^n}\sin nA$
• C. $\frac{c^n}{b^n}\sin nA$
• D. $\sin nA$

Answer 1

The correct option is C $\frac{c^n}{b^n}\sin nA$

Let$C=1+n\frac{a}{c}\cos B+\frac{n(n+1)}{2!}\frac{a^2}{c^2}\cos 2B+...\infty S=n\frac{a}{c}\sin B+\frac{n(n+1)}{2!}\frac{a^2}{c^2}\sin 2B+...\infty$
$\therefore C+iS=1+n\frac{a}{c}{e}^{iB}+\frac{n(n+1)}{2!}\frac{a^2}{c^2}{e}^{i2B}+...\infty ={(a-\frac{a}{c}{e}^{iB})}^{-n}$
Now let $a=k\sin Ac=k\sin C$
$C+iS={(1-\frac{\sin A}{\sin C}(\cos B+i\sin B))}^{-n}={\sin }^{n}C{(\sin C-\sin A\cos B-i\sin A\sin B)}^{-n}$
U\sin g $A+B+C=180\sin (A+B)=\sin (180-C)=\sin C\sin (A+B)=\sin A\cos B+\cos A\sin B$
$C+iS={\sin }^{n}C{(\sin B\cos A-i\sin A\sin B)}^{-n}=\frac{{\sin }^{n}C}{{\sin }^{n}B}{(\cos A-i\sin A)}^{-n}$
Resubstituting $a=k\sin Ac=k\sin C$ and from De-moiver"s theorem, We get
$C+iS=\frac{c^n}{b^n}(\cos nA+i\sin nA)$
Equating imaginary part
$S=\frac{c^n}{b^n}\sin nA$
We get
$n\frac{a}{c}\sin B+\frac{n(n+1)}{2!}\frac{a^2}{c^2}\sin 2B+...\infty =\frac{c^n}{b^n}\sin nA$