Can anyone plz solve my que its urgent for me plz✌✌✌✌
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jharoshan06466:
hii
Hlw.
r u a student
Yes whyn
mtlb
Yes why!!!!
nothing
Ohk.
Answers
Answered by
2
Answer:
L.H.S = square root of (sec^2 A+cosec^2 A)
= square root of (1/cos^2A + 1/sin^2A)
= square root of ((sin^2A + cos^2A)/cos^2A * sin^2A)
I taken LCMof cos^2A & sin^2A. which is cos^2A * sin^2A
now
The solution of ur question is listed below:-
L.H.S = square root of (1/ cos^2A * sin^2A)
because sin^2A + cos^2A =1
so L.H.S = (1/ cosA * sinA) After removed the square root.
L.H.S = (sin^2A + cos^2A/ cosA * sinA)
I have written here (1 = sin^2A + cos^2A)
now, L.H.S = (sin^2A/cosA * sinA) + (cos^2A/cosA * sinA)
= (sinA/cosA) + (cosA/sinA)
= (tanA + cotA) = R.H.S Proved.
mark it the braijleist
i can solve it with. in just 5 steps
Thx.
there is an easy way then this
Why u reciprocal it sis .
Ohk .
what happened
Can u plz tell me that why u had done reciprocal there.
Answered by
1
here's your answer mate
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exam is near dude
Yes.
math kaisa gya
Nyc.
ok
veerai
plss go another question to chat
Means.
yrr baar baar notification aa rha hai
Ohk.^_^
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