If the value of is where are natural numbers and then is equal to Trigonometery

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Solution

Q.
If the value of $\frac{3 \cos 36^{o} + 5 \sin 18^{o}}{5 \cos 36^{o} - 3 \sin 18^{o}}$ is $\frac{a \sqrt{5} - b}{c},$ where $a, b, c$ are natural numbers and $g c d (a, c) = 1,$ then $a + b + c$ is equal to : [2024]

1 52

2 50

3 54

4 40

Ans.
(1)

$\frac{3 \cos 36^{o} + 5 \sin 18^{o}}{5 \cos 36^{o} - 3 \sin 18^{o}}$ $= \frac{3 [\frac{1 + \sqrt{5}}{4}] + 5 [\frac{\sqrt{5} - 1}{4}]}{5 [\frac{1 + \sqrt{5}}{4}] - 3 [\frac{\sqrt{5} - 1}{4}]}$ $[∵ \sin 18^{o} = \frac{\sqrt{5} - 1}{4} and \cos 36^{o} = \frac{1 + \sqrt{5}}{4}]$

   $= \frac{3 + 3 \sqrt{5} + 5 \sqrt{5} - 5}{5 + 5 \sqrt{5} - 3 \sqrt{5} + 3} = \frac{8 \sqrt{5} - 2}{2 \sqrt{5} + 8} = \frac{4 \sqrt{5} - 1}{4 + \sqrt{5}}$

   $= \frac{(4 \sqrt{5} - 1) (4 - \sqrt{5})}{(\sqrt{5} + 4) (4 - \sqrt{5})} = \frac{16 \sqrt{5} - 20 - 4 + \sqrt{5}}{16 - 5}$

   $= \frac{17 \sqrt{5} - 24}{11} = \frac{a \sqrt{5} - b}{c}$ (Given)

$∴$ $a = 17,$ $b = 24,$ $c = 11$

$∴$    $a + b + c = 17 + 24 + 11 = 52$

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