If 3 + cot 80 ^ o cot 20 ^ o cot 80 ^ o + cot 20 ^ o = tan(π)/(k), then the value of k is (where, (π)/(k) is an acute angle)

Correct answer: 3.6 3sin 80 ^ o sin 20 ^ o + cos 80 ^ o cos 20 ^ o sin 20 ^ o cos 80 ^ o + cos 20 ^ o sin 80 ^ o = 2sin 20 ^ o sin 80 ^ o + (cos 80 ^ o cos 20 ^ o + sin 80 ^ o 20) ( 20 ^ o + 80 ^ o ) = cos 60 ^ o - cos 100 ^ o + ( 80 ^ o - 20 ^ o ) sin 100 ^ o = 2cos 60 ^ o - cos 100 ^ o sin 100 ^ o = 1 - cos 100 ^ o sin 100 ^ o = 2si n ^ 2 5 0 ^ ∘ 2sin 50 ^ o cos 50 ^ o = tan 50 ^ o = tan(5π)/(18) = ((π)/((18)/(5))) ⇒ k = (18)/(5) = 3.6

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

If 3 + cot 80 ^ o cot 20 ^ o cot 80 ^ o + cot 20 ^ o = tan(π)/(k),...

\dfrac{3 + cot {80}^{o}cot {20}^{o}}{cot {80}^{o} + cot {20}^{o}} =

tan\dfrac{\pi}{k},

then the value of

k

is (where,

\dfrac{\pi}{k}

is an acute angle)

Correct answer

3.6

Step-by-step solution

\dfrac{3sin {80}^{o}sin {20}^{o} + cos {80}^{o}cos {20}^{o}}{sin {20}^{o}cos {80}^{o} + cos {20}^{o}sin {80}^{o}}

= \dfrac{2sin {20}^{o}sin {80}^{o} + \left(cos {80}^{o}cos {20}^{o} + sin {80}^{o}\sin 20\right)}{\sin \left({20}^{o} + {80}^{o}\right)}

= \dfrac{cos {60}^{o} - cos {100}^{o} + \cos \left({80}^{o} - {20}^{o}\right)}{sin {100}^{o}}

= \dfrac{2cos {60}^{o} - cos {100}^{o}}{sin {100}^{o}} = \dfrac{1 - cos {100}^{o}}{sin {100}^{o}}

= \dfrac{2si{n}^{2}5{0}^{∘}}{2sin {50}^{o}cos {50}^{o}} = tan {50}^{o}

= tan\dfrac{5\pi}{18} = \tan \left(\dfrac{\pi}{\dfrac{18}{5}}\right)

⇒ k = \dfrac{18}{5} = 3.6

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