Math, asked by abd12, 1 year ago

√sec square a+ cosec square a = tan a+cot a

Answers

Answered by abcnehaalosppud
6
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......

abd12: what did you do after now
abd12: i didnt understand anything answwe correctly with expla.
Answered by warofgod647
0

Answer:

madar chod kuch ve ata hai maths Kya hota hai

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