If sin^2A=cos^3A, prove that, cot^6A-cot^2A=1
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Answered by
19
HELLO DEAR,
=> cos³A = sin²A
on squaring both side
we get,
cos^6A = sin⁴A
[multiply both side by sin²A]
=> sin²Acos^6A = sin⁴Asin²A
=> cos^6/sin^6A = 1/sin²A
=> cot^6A = cosec²A
=> cot^6A = 1 + cot²A
=> cot^6A - cot²A = 1
I HOPE ITS HELP YOU DEAR,
THANKS
=> cos³A = sin²A
on squaring both side
we get,
cos^6A = sin⁴A
[multiply both side by sin²A]
=> sin²Acos^6A = sin⁴Asin²A
=> cos^6/sin^6A = 1/sin²A
=> cot^6A = cosec²A
=> cot^6A = 1 + cot²A
=> cot^6A - cot²A = 1
I HOPE ITS HELP YOU DEAR,
THANKS
Sandy111201:
it is given that sin^2A=cos^3A
r.hs
yea
maine Wahi proff Kiya
we need to prove that cot^6A-cot^2A =1
oka ji
plzz try again. I need help
are done
check
Thankzz a lot
Answered by
5
- sin^2A=cos^3A
- sin^2A×sin^2A=cos^3A×cos^2A
- sin^4A×sin^2A=cos^6A×sin^2A
- cos^6A/sin^6A=1/sin^2A
- cot^6A=cosecA
- cot^6A=1+cot^2A
- cot^6A-cot^2A=1..............PROVED
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