If the value of frac{3 cos 36^{circ}+5 sin 18 | vedclass

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(C) We know that $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.Substituting these values into the expression:$

Vedclass · Accepted Answer

$52$

Vedclass · Answer

(C) We know that $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.Substituting these values into the expression:$\frac{3(\frac{\sqrt{5}+1}{4}) + 5(\frac{\sqrt{5}-1}{4})}{5(\frac{\sqrt{5}+1}{4}) - 3(\frac{\sqrt{5}-1}{4})} = \frac{3\sqrt{5}+3+5\sqrt{5}-5}{5\sqrt{5}+5-3\sqrt{5}+3} = \frac{8\sqrt{5}-2}{2\sqrt{5}+8} = \frac{4\sqrt{5}-1}{\sqrt{5}+4}$.Rationalizing the denominator:$\frac{4\sqrt{5}-1}{\sqrt{5}+4} 	imes \frac{4-\sqrt{5}}{4-\sqrt{5}} = \frac{16\sqrt{5}-20-4+\sqrt{5}}{16-5} = \frac{17\sqrt{5}-24}{11}$.Comparing this with $\frac{a\sqrt{5}-b}{c}$,we get $a=17, b=24, c=11$.Since $\gcd(17, 11)=1$,the values are valid.Thus,$a+b+c = 17+24+11 = 52$.

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

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