Question

tan76/cot14+sec58/cosec38-sin35•sec55-8sin square30

Answer 1

Follow d image.... The answer is - 1

Answer 2

◀ HERE'S ⏬ YOUR ANSWER▶

$\frac{tan76}{cot14} + \frac{sec58}{cosec32} - sin35.sec55 - 8 {sin}^{2} 30 \\ = \frac{tan(90 - 14)}{cot14} + \frac{sec(90 - 32)}{cosec32} - sin(90 - 55).sec55 - 8. ({ \frac{1}{2} })^{2} \\ = \frac{cot14}{cot14} + \frac{cosec32}{cosec32} - cos55.sec55 - 8. \frac{1}{4} \\ = 1 + 1 - 1 - 2 \\ = - 1$

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