JEE Main Previous Year Trigonometry Questions and Solutions - Trigonometric Identities

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Topic Question Set

Q 11 :
The value of $\tan 9 ° - \tan 27 ° - \tan 63 ° + \tan 81 °$ is ________ . [2023]

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(4)

$We have, \tan 9 ° - \tan 27 ° - \tan 63 ° + \tan 81 °$

$= \tan 9 ° - \tan 27 ° - \tan (90 ° - 27 °) + \tan (90 ° - 9 °)$

$= (\tan 9 ° + \cot 9 °) - (\tan 27 ° + \cot 27 °)$

$= \frac{2 (\sin^{2} 9 ° + \cos^{2} 9 °)}{2 (\sin 9 ° \cos 9 °)} - \frac{2 (\sin^{2} 27 ° + \cos^{2} 27 °)}{2 (\sin 27 ° \cos 27 °)}$

$= \frac{2}{\sin 18 °} - \frac{2}{\sin 54 °}$

$= \frac{8}{\sqrt{5} - 1} - \frac{8}{\sqrt{5} + 1}$

$= \frac{16}{4} = 4$

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