Math, asked by san91, 1 year ago

2cosec67/sec23-tan70/cot20-cos theta+tan38•tan52

Answers

Answered by Swarup1998
7
◀ HERE'S ⏬ YOUR ANSWER▶

2 \frac{cosec67}{sec23}  -  \frac{tan70}{cot20}  + tan38.tan52 \\  = 2 \frac{sec23}{sec23}  -  \frac{cot20}{cot20}  + cot52.tan52 \\  = 2 - 1 + 1 \\  = 2

⏏ HOPE THIS ⬆ HELPS YOU⏏
Answered by abhikumarcrick
2

Step-by-step explanation:

2cosec67°/sec(90°-67°)-tan70°/cot(90°-70°)-1+tan38°tan(90°-38°)

cos°=1

2cosec67°/cosec67°-tan70°/tan70°-1+tan38°cot38°

2-1-1+1=1

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