sinA+cosecA=2 then sin^A+cosec^A
Anonymous:
Dear, mark my ans as brainlist
thnks
Answers
Answered by
2
Heya
______________________________
Sin A + Cosec A = 2
SQUARING BOTH SIDE'S
=>
( Sin A + Cosec A )² = 4
=>
Sin² A + Cosec ² A + 2Sin A × (1/Sin A ) = 4
=>
Sin² A + Cosec² A = 4 -2
=>
Sin² A + Cosec² A = 2
______________________________
Sin A + Cosec A = 2
SQUARING BOTH SIDE'S
=>
( Sin A + Cosec A )² = 4
=>
Sin² A + Cosec ² A + 2Sin A × (1/Sin A ) = 4
=>
Sin² A + Cosec² A = 4 -2
=>
Sin² A + Cosec² A = 2
tq so much
Answered by
1
SinA+CosecA =2
$quaring both side we get,
(SinA+CosecA)^2=4
Sin^2A+Cosec^2A+2SinACosA=4
Now,
You should know that SinA=1/CosecA
So,
Sin^2A+Cosec^2A+2=4
Sin^2A+Cosec^2A=2
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