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 :
Solutions · 1
Official solution
To find the value of 5cos36∘−3sin18∘3cos36∘+5sin18∘ in the form ca5−b, we need to simplify the given expression. Let's start by using some fundamental trigonometric identities.
We know that:
cos36∘=45+1sin18∘=45−1
First, substitute these values into the expression:
5⋅45+1−3⋅45−13⋅45+1+5⋅45−1
Simplify the numerator and the denominator:
Numerator: 3⋅45+1+5⋅45−1=43(5+1)+5(5−1)=435+3+55−5=485−2=25−21
Denominator: 5⋅45+1−3⋅45−1=45(5+1)−3(5−1)=455+5−35+3=425+8=25+4
Now combine the simplified numerator and denominator:
25+425−21=5+4(25−21)⋅2=5+445−1
Rationalizing the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator:
(5+4)(5−4)(45−1)(5−4)
The denominator simplifies to:
52−42=5−16=−11
The numerator simplifies to:
(45−1)(5−4)=(45⋅5−4⋅45−1⋅5+1⋅4)=(20−165−5+4)=24−175
Combining them, we get:
−1124−175=11175−24
Thus, a=17,b=24,c=11.
Therefore, a+b+c=17+24+11=52.
So the correct option is:
Option B: 52