1-cos^2A÷1+sinA=sinA (1-is seperate)
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Hey friend (^_^)
Here is your answer :
1-[cos²A/(1+sinA)]=sinA
LHS :
=1-[cos²A/(1+sinA)]
[USING cos²A+sin²A=1]
=1-[(1-sin²A) /(1+sinA)]
[USING a²-b²=(a+b)(a-b) ]
=1-[(1+sinA)×(1-sinA)/(1+sinA)]
=1-(1-sinA)
=1-1+sinA
=sinA
=RHS
HENCE PROVED
HOPE THIS HELPS YOU
Here is your answer :
1-[cos²A/(1+sinA)]=sinA
LHS :
=1-[cos²A/(1+sinA)]
[USING cos²A+sin²A=1]
=1-[(1-sin²A) /(1+sinA)]
[USING a²-b²=(a+b)(a-b) ]
=1-[(1+sinA)×(1-sinA)/(1+sinA)]
=1-(1-sinA)
=1-1+sinA
=sinA
=RHS
HENCE PROVED
HOPE THIS HELPS YOU
ADAMSHARIEFF:
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