Math, asked by Vikram201, 1 year ago

2sin68/cos22-2cot15/5tan75-3tan45tan20tan40tan50tan70/5

Answers

Answered by shreyamahajan
6
Here this is your answer hope it helps plz mark as brainliest ☺☺
2sin68 cos22 − 2tan(90−15) 5cot15 − 3tan45tan20tan40tan50tan70 5( sin 2 70+ sin 2 20) = 2sin(90−22) cos22 − 2tan(90−15) 5cot15 − 3tan45tan20tan40tan(90−40)tan(90−20) 5( sin 2 70+ sin 2 (90−70)) usingsin(90−A)=cosA, tan(90−A)=cotA = 2cos22 cos22 − 2cot15 5cot15 − 3(1)tan20tan40cot40cot20 5( sin 2 70+ cos 2 70) nowusingtanA×cotA=1,alsosin2A +cos2A=1 =2− 2 5 − 3 5 = 10−2−3 5 =1
Similar questions