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

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

ta n ^ 6 2 0 ^ o - 33ta n ^ 4 2 0 ^ o + 27ta n ^ 2 2 0 ^ o + 4 =

tan620o−33tan420o+27tan220o+4=ta{n}^{6}2{0}^{o} - 33ta{n}^{4}2{0}^{o} + 27ta{n}^{2}2{0}^{o} + 4 =

Correct answer

7

Step-by-step solution

Let, tan 20o=k{20}^{o} = k
then, tan60o=3tan20o−tan320o1−3tan220otan {60}^{o} = \dfrac{3tan {20}^{o} - ta{n}^{3}2{0}^{o}}{1 - 3ta{n}^{2}{20}^{o}}
3=3k−k31−3k2\sqrt{3} = \dfrac{3k - {k}^{3}}{1 - 3{k}^{2}}
(3(1−3k2))2=(3k−k3)2{\left(\sqrt{3}\left(1 - 3{k}^{2}\right)\right)}^{2} = {\left(3k - {k}^{3}\right)}^{2}
3(1+9k4−6k2)=9k2+k6−6k43\left(1 + 9{k}^{4} - 6{k}^{2}\right) = 9{k}^{2} + {k}^{6} - 6{k}^{4}
3+27k4−18k2=9k2+k6−6k43 + 27{k}^{4} - 18{k}^{2} = 9{k}^{2} + {k}^{6} - 6{k}^{4}
k6−33k4+27k2−3=0{k}^{6} - 33{k}^{4} + 27{k}^{2} - 3 = 0
k6−33k4+27k2+4=7{k}^{6} - 33{k}^{4} + 27{k}^{2} + 4 = 7
Hence, tan620o−33tan420o+27tan220o+4=7ta{n}^{6}2{0}^{o} - 33ta{n}^{4}2{0}^{o} + 27ta{n}^{2}2{0}^{o} + 4 = 7
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