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 :

Question

Aicharya · Accepted Answer

$\begin{aligned} & \frac{\frac{3(\sqrt{5}+1)}{4}+5\left(\frac{\sqrt{5}-1}{4}ight)}{5\left(\frac{\sqrt{5}+1}{4}ight)-3\left(\frac{\sqrt{5}-1}{4}ight)}=\frac{8 \sqrt{5}-2}{2 \sqrt{5}+8} \ & =\frac{4 \sqrt{5}-1}{\sqrt{5}+4} 	imes \frac{\sqrt{5}-4}{\sqrt{5}-4}\end{aligned}$ $\begin{aligned} & =\frac{20-16 \sqrt{5}-\sqrt{5}+4}{-11} \ & =\frac{17 \sqrt{5}-24}{11} \Rightarrow a=17, b=27, c=11 \ & a+b+c=52\end{aligned}$

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

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