sin230º + 4 cot245º – sec260º = ?

Foundation

Mathematics Foundation

Introduction to Trigonometry

Question

sin²30º + 4 cot²45º – sec²60º = ?

$\frac{1}{4}$

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$\sin^{2} 30 º + 4 \cot^{2} 45 º - \sec^{2} 60 º = {(\frac{1}{2})}^{2} + 4 \times {(1)}^{2} - 2^{2} = \frac{1}{4} + 4 - 4 = \frac{1}{4}$