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 : [JEE Main 2024, 08 Apr (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: d) 52. $\frac{3\cos 36^\circ +5\sin 18^\circ }{5\cos 36^\circ -3\sin 18^\circ }\ =\frac{\frac{3(\sqrt{5}+1)}{4}+5\left(\frac{\sqrt{5}-1}{4}\right)}{5\left(\frac{\sqrt{5}-1}{4}\right)-3\left(\frac{\sqrt{5}-1}{4}\right)}$ $=\frac{8\sqrt{5}-2}{2\sqrt{5}+8}$ $=\frac{4\sqrt{5}-1}{\sqrt{5}+4}\times \frac{\sqrt{5}-4}{\sqrt{5}-4}$ $=\frac{20-16\sqrt{5}-\sqrt{5}+4}{-11}$ $=\frac{17\sqrt{5}-24}{11}\ \Rightarrow a=17,b=24,c=11$ $a+b+c=52$

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…

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