Math, asked by san91, 1 year ago

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

Answers

Answered by Swarup1998
7
◀ 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

⏏HOPE THIS ⬆ HELPS YOU⏏
Similar questions