√sec square a+ cosec square a = tan a+cot a
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Answered by
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The solution of ur question is listed below:-
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 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.
HOPE IT HELPS YOU......
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 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.
HOPE IT HELPS YOU......
abd12:
what did you do after now
Answered by
0
Answer:
madar chod kuch ve ata hai maths Kya hota hai
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