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JEE Main 20236 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2023 · 6 April · Evening Shift · Mathematics · Trigonometric Ratios & Identities

The value of tan 9 ^ o - tan 27 ^ o - tan 63 ^ o + tan 81 ^ o is...

The value of tan9o−tan27o−tan63o+tan81otan{9}^{o} - tan{27}^{o} - tan{63}^{o} + tan{81}^{o} is _____.

Correct answer

0

Step-by-step solution

Given,

tan9o−tan27o−tan63o+tan81otan{9}^{{}^{o}} - tan{27}^{{}^{o}} - tan{63}^{{}^{o}} + tan{81}^{{}^{o}}

=(cot81°+tan81o)−(tan27°+cot27o)= \left(cot{81}^{°} + tan{81}^{{}^{o}}\right) - \left(tan{27}^{°} + cot{27}^{{}^{o}}\right)

=(tan9°+cot9°)−(tan27°+cot27°)= \left(tan{9}^{°} + cot{9}^{°}\right) - \left(tan{27}^{°} + cot{27}^{°}\right)

=(sin9°cos9°+cos9°sin9°)−(sin27°cos27°+cos27°sin27°)= \left(\dfrac{sin{9}^{°}}{cos{9}^{°}} + \dfrac{cos{9}^{°}}{sin{9}^{°}}\right) - \left(\dfrac{sin{27}^{°}}{cos{27}^{°}} + \dfrac{cos{27}^{°}}{sin{27}^{°}}\right)

=2sin18o−2sin54o= \dfrac{2}{sin{18}^{o}} - \dfrac{2}{sin{54}^{o}}

=(2×45−1−2×45+1)= \left(\dfrac{2 \times 4}{\sqrt{5} - 1} - \dfrac{2 \times 4}{\sqrt{5} + 1}\right)

=8(15−1−15+1)= 8\left(\dfrac{1}{\sqrt{5} - 1} - \dfrac{1}{\sqrt{5} + 1}\right)

=8(5+1−(5−1)(5−1)(5+1))= 8\left(\dfrac{\sqrt{5} + 1 - \left(\sqrt{5} - 1\right)}{\left(\sqrt{5} - 1\right)\left(\sqrt{5} + 1\right)}\right)

=8(25−1)=4= 8\left(\dfrac{2}{5 - 1}\right) = 4

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