cos2A/secA+sin2A/cosecA=cosA
Anonymous:
Can u tell me which book question
my teacher have given it
I think something wrong
Which class r u
1pu
1pu mean
first pu
Mine 10 class
Answers
Answered by
5
lhs = cos 2A/secA+sin2A/cosecA
= cos2A cosA+sin 2A sinA
( since 1/secA = cosA, 1/cosec A = sinA)
= cos(2A-A)
= cosA
= rhs
= cos2A cosA+sin 2A sinA
( since 1/secA = cosA, 1/cosec A = sinA)
= cos(2A-A)
= cosA
= rhs
ur welcome
bro its cos2A/secA+sin2A/cosecA=cosA
yes i did for that problem only
don't u understand it
srry no ones more will u send
it was typing error i corrected it check once
send once more
go through the soution ,i already corrected it
Answered by
6
cos(2A)/sec(A) + sin(2A)/cosec(A)
= cos(2A)cos(A) + sin(2A)(sin(A)----- 1/sec(x) = cos(x) and 1/cosec(x) = sin(x)
= cos(2A-A) ------ cos(x)cos(y) -+sin(x)sin(y) = cos(x-y)
= cos(A) I hope it helped u
= cos(2A)cos(A) + sin(2A)(sin(A)----- 1/sec(x) = cos(x) and 1/cosec(x) = sin(x)
= cos(2A-A) ------ cos(x)cos(y) -+sin(x)sin(y) = cos(x-y)
= cos(A) I hope it helped u
I hope it helped u
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