If the value of $\frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}}$ is

Question

If the value of $\frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}}$ is $\frac{a \sqrt{5}-b}{c}$, where $a, b, c$ are natural numbers and $\operatorname{gcd}(a, c)=1$, then $a+b+c$ is equal to :\

Professor PC Thomas Classes · Answer

(a) $: \frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}}$ $$ \begin{aligned} & =\frac{3\left[\frac{1+\sqrt{5}}{4}ight]+5\left[\frac{\sqrt{5}-1}{4}ight]}{5\left[\frac{1+\sqrt{5}}{4}ight]-3\left[\frac{\sqrt{5}-1}{4}ight]} \ & {\left[\because \sin 18^{\circ}=\frac{\sqrt{5}-1}{4} 	ext { and } \cos 36^{\circ}=\frac{1+\sqrt{5}}{4}ight]} \ & =\frac{3+3 \sqrt{5}+5 \sqrt{5}-5}{5+5 \sqrt{5}-3 \sqrt{5}+3}=\frac{8 \sqrt{5}-2}{2 \sqrt{5}+8}=\frac{4 \sqrt{5}-1}{4+\sqrt{5}} \ & =\frac{(4 \sqrt{5}-1)(4-\sqrt{5})}{(\sqrt{5}+4)(4-\sqrt{5})}=\frac{16 \sqrt{5}-20-4+\sqrt{5}}{16-5} \ & =\frac{17 \sqrt{5}-24}{11}=\frac{a \sqrt{5}-b}{c} \ & 	herefore \quad a=17, b=24, c=11 \ & 	herefore \quad a+b+c=17+24+11=52 \end{aligned} $$