answer fast please...........
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Answered by
9
let take that thetha = A
2cos²A+(2/1+cot²A)
= 2cos²A+(2/cosec²A)
= 2cos²A+(2×1/cosec²A)
= 2cos²A+2sin²A
= 2(cos²A+sin²A)
= 2(1)
= 2
2cos²A+(2/1+cot²A)
= 2cos²A+(2/cosec²A)
= 2cos²A+(2×1/cosec²A)
= 2cos²A+2sin²A
= 2(cos²A+sin²A)
= 2(1)
= 2
vquantumv:
absolutely correct
Answered by
5
Hey !!!
2cos²¢ + 2 /1 + cot²¢
2cos²¢ + 2/cosec²¢ ✓[1 + cot²¢ = cosec²¢ ]
2cos²¢ + 2 sin²¢ [1 /cosec²¢ = sin²¢]
2 cos²¢ + 2 sin²¢
2(cos²¢ + sin²¢ )
2 (1) ✓[sin²¢ + cos²¢ = 1]
= 2 Answer ...
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Hope it helps you !!
@Rajukumar111
2cos²¢ + 2 /1 + cot²¢
2cos²¢ + 2/cosec²¢ ✓[1 + cot²¢ = cosec²¢ ]
2cos²¢ + 2 sin²¢ [1 /cosec²¢ = sin²¢]
2 cos²¢ + 2 sin²¢
2(cos²¢ + sin²¢ )
2 (1) ✓[sin²¢ + cos²¢ = 1]
= 2 Answer ...
************************************
Hope it helps you !!
@Rajukumar111
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