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Prove That: 3 Sin π 6 Sec π 3 − 4 Sin 5 π 6 Cot π 4 = 1

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Question

Prove that:

\[3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4} = 1\]

 

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Solution

 LHS = \[3\sin\frac{\pi}{6}sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}cot\frac{\pi}{4}\]
\[ = 3\sin\left( \frac{180^\circ}{6} \right)\sec\left( \frac{180^\circ}{3} \right) - 4\sin\left( \frac{5 \times 180^\circ}{6} \right)\cot\left( \frac{180^\circ}{4} \right)\]
\[ = 3\sin\left( 30^\circ \right)\sec\left( 60^\circ \right) - 4\sin\left( 150^\circ \right)\cot\left( 45^\circ \right)\]
\[ = 3\sin\left( 30^\circ \right)\sec\left( 60^\circ \right) - 4\sin\left( 90^\circ \times 1 + 60^\circ \right)\cot\left( 45^\circ \right)\]
\[ = 3\sin \left( 30^\circ \right)\sec \left( 60^\circ \right) - 4\cos \left( 60^\circ \right)\cot \left( 45^\circ \right)\]
\[ = 3 \times \frac{1}{2} \times 2 - 4 \times \frac{1}{2} \times 1\]
\[ = 3 - 2\]
\[ = 1\]
 = RHS
Hence proved .

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Chapter 5: Trigonometric Functions - Exercise 5.3 [Page 39]

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R.D. Sharma Mathematics [English] Class 11
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 2.7 | Page 39

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