1÷sinA+cosA+1÷sinA-cosA=2sinA÷1-2cos^2A
Answers
Answered by
30
Hey friend (^_^)
Here is your answer
LHS :
=(1/sinA+cosA)+(1/sinA-cosA)
TAKING LCM
=(sinA-cosA+sinA+cosA)/(sin²A-cos²A)
[Using sin²A=1-cos²A]
=2sinA/(1-cos²A-cos²A)
=2sinA/(1-2cos²A)
=RHS
HENCE PROVED
HOPE THIS HELPS YOU
Here is your answer
LHS :
=(1/sinA+cosA)+(1/sinA-cosA)
TAKING LCM
=(sinA-cosA+sinA+cosA)/(sin²A-cos²A)
[Using sin²A=1-cos²A]
=2sinA/(1-cos²A-cos²A)
=2sinA/(1-2cos²A)
=RHS
HENCE PROVED
HOPE THIS HELPS YOU
Answered by
4
Answer:Hey friend (^_^)
Here is your answer
LHS :
=(1/sinA+cosA)+(1/sinA-cosA)
TAKING LCM
=(sinA-cosA+sinA+cosA)/(sin²A-cos²A)
[Using sin²A=1-cos²A]
=2sinA/(1-cos²A-cos²A)
=2sinA/(1-2cos²A)
=RHS
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