Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Prove That: Tan 11 π 3 − 2 Sin 4 π 6 − 3 4 C O S E C 2 π 4 + 4 Cos 2 17 π 6 = 3 − 4 √ 3 2 - Mathematics

Prove that: $\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4} {cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3 - 4\sqrt{3}}{2}$

#### Solution

LHS = $\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4}cose c^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6}$
$= \tan\left( \frac{11\pi}{3} \right) - 2\sin\left( \frac{4\pi}{6} \right) - \frac{3}{4} \left[ cosec\left( \frac{\pi}{4} \right) \right]^2 + 4 \left[ \cos\left( \frac{17\pi}{6} \right) \right]^2$
$= \tan\left( \frac{11}{3} \times 180^\circ \right) - 2\sin\left( \frac{4}{6} \times 180^\circ \right) - \frac{3}{4} \left[ cosec\left( \frac{180^\circ}{4} \right) \right]^2 + 4 \left[ \cos\left( \frac{17 \times 180^\circ}{6} \right) \right]^2$
$= \tan \left( 660^\circ \right) - 2\sin \left( 120^\circ \right) - \frac{3}{4} \left[ cosec\left( 45^\circ \right) \right]^2 + 4 \left[ \cos \left( 510^\circ \right) \right]^2$
$= \tan \left( 660^\circ \right) - 2\sin \left( 120^\circ \right) - \frac{3}{4} \left[ cosec\left( 45^\circ \right) \right]^2 + 4 \left[ \cos \left( 510^\circ \right) \right]^2$
$= \tan \left( 90^\circ \times 7 + 30^\circ \right) - 2\sin \left( 90^\circ \times 1 + 30^\circ \right) - \frac{3}{4} \left[ cosec\left( 45^\circ \right) \right]^2 + 4 \left[ \cos\left( 90^\circ \times 5 + 60^\circ \right) \right]^2$
$= \left[ - \cot \left( 30^\circ \right) \right] - \left[ 2\cos \left( 30^\circ \right) \right] - \frac{3}{4} \left[ cosec \left( 45^\circ \right) \right]^2 + 4 \left[ - \sin\left( 60^\circ \right) \right]^2$
$= - \cot \left( 30^\circ \right)-2\cos\left( 30^\circ \right) - \frac{3}{4} \left[ cosec\left( 45^\circ \right) \right]^2 + 4 \left[ \sin \left( 60^\circ \right) \right]^2$
$= - \sqrt{3}-\frac{2\sqrt{3}}{2} - \frac{3}{4} \left[ \sqrt{2} \right]^2 + 4 \left[ \frac{\sqrt{3}}{2} \right]^2$
$= - \sqrt{3} - \sqrt{3} - \frac{3}{2} + 3$
$= \frac{3 - 4\sqrt{3}}{2}$
= RHS
Hence proved .

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 2.6 | Page 39