If the value of frac 3 cos 36 ° +5 sin 18 ° 5 cos 36 ° -3 sin 18 ° is frac a √(5)-b c , where a , b , c are natural numbers and gcd ( a , c )=1 , then a + b + c is equal to :

Mathematics · Trigonometry Ratios and Identities

JEE Main 2024 — 8 April, Shift 2 — Question 9

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 $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are natural numbers and $\operatorname{gcd}(\mathrm{a}, \mathrm{c})=1$ , then

$\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to :

Option A:
50
Option B:
40
Option C:
52
Correct
Option D:
54

Answer: C

Step-by-step solution

\frac{3\cos36^\circ + 5\sin18^\circ}{5\cos36^\circ - 3\sin18^\circ}

Step 1: Use exact trigonometric values

\cos36^\circ = \frac{\sqrt5+1}{4}, \qquad \sin18^\circ = \frac{\sqrt5-1}{4}

Step 2: Substitute the values

Numerator:

3\cos36^\circ + 5\sin18^\circ = \frac{3(\sqrt5+1)+5(\sqrt5-1)}{4}

= \frac{8\sqrt5-2}{4}

Denominator:

5\cos36^\circ - 3\sin18^\circ = \frac{5(\sqrt5+1)-3(\sqrt5-1)}{4}

= \frac{2\sqrt5+8}{4}

Step 3: Form the ratio

\frac{(8\sqrt5-2)/4}{(2\sqrt5+8)/4} = \frac{8\sqrt5-2}{2\sqrt5+8}

Factor out $2$ :

= \frac{4\sqrt5-1}{\sqrt5+4}

Step 4: Rationalize the denominator

\frac{4\sqrt5-1}{\sqrt5+4} \cdot \frac{4-\sqrt5}{4-\sqrt5} = \frac{(4\sqrt5-1)(4-\sqrt5)}{16-5}

Numerator:

(4\sqrt5-1)(4-\sqrt5) = 17\sqrt5 - 24

Denominator:

11

Final Value:

\frac{3\cos36^\circ + 5\sin18^\circ}{5\cos36^\circ - 3\sin18^\circ} = \frac{17\sqrt5 - 24}{11}

Comparing with $\frac{a\sqrt5 - b}{c}$ :

a=17,\quad b=24,\quad c=11

a+b+c = 17+24+11 = \boxed{52}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Trigonometric Ratios of Multiples and Submultiples of angles