# Prove That: Tan 4 π − Cos 3 π 2 − Sin 5 π 6 Cos 2 π 3 = 1 4 - Mathematics

Prove that:
$\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}$

#### Solution

$4\pi = 720^\circ, \frac{3\pi}{2} = 270^\circ, \frac{5\pi}{6} = 150^\circ, \frac{2\pi}{3} = 120^\circ$
LHS = $\tan\left( 720^\circ \right) - \cos\left( 270^\circ \right) - \sin\left( 150^\circ \right) \cos\left( 120^\circ \right)$
$= \tan\left( 90^\circ \times 8 + 0^\circ \right) - \cos\left( 90^\circ \times 3 + 0^\circ \right) - \sin\left( 90^\circ \times 1 + 60^\circ \right) \cos\left( 90^\circ \times 1 + 30^\circ \right)$
$= \tan\left( 0^\circ \right) - \sin\left( 0^\circ \right) - \cos\left( 60^\circ \right) \left[ - \sin\left( 30^\circ \right) \right]$
$= \tan\left( 0^\circ \right) - \sin\left( 0^\circ \right) + \cos\left( 60^\circ \right) \sin\left( 30^\circ \right)$
$= 0 - 0 + \frac{1}{2} \times \frac{1}{2}$
$= \frac{1}{4}$
= RHS
Hence proved.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 9.1 | Page 40