(sinA+cosA÷sinA-cosA)+(sinA-cosA÷sinA+cosA)=2÷1-2cos^2A
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Answered by
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Hey friend (^_^)
Here is your answer
LHS :
=[(sinA+cosA/sinA-cosA) +(sinA-cosA)/(sinA+cosA)]
TAKING LCM
=[(sinA+cosA)²+(sinA-cosA)²]/(sin²A-cos²A)
USING IDENTITIES :
•(a+b)²=a²+b²+2ab
•(a-b)²=a²+b²-2ab
•sin²A=1-cos²A
=(sin²A+cos²A+2sinAcosA+sin²A+cos²A-2sinAcosA)/(1-cos²A-cos²A)
=(1+0+1)/(1-2cos²A)
=2/(1-2cos²A)
=RHS
HENCE PROVED
HOPE THIS HELPS YOU
Here is your answer
LHS :
=[(sinA+cosA/sinA-cosA) +(sinA-cosA)/(sinA+cosA)]
TAKING LCM
=[(sinA+cosA)²+(sinA-cosA)²]/(sin²A-cos²A)
USING IDENTITIES :
•(a+b)²=a²+b²+2ab
•(a-b)²=a²+b²-2ab
•sin²A=1-cos²A
=(sin²A+cos²A+2sinAcosA+sin²A+cos²A-2sinAcosA)/(1-cos²A-cos²A)
=(1+0+1)/(1-2cos²A)
=2/(1-2cos²A)
=RHS
HENCE PROVED
HOPE THIS HELPS YOU
ADAMSHARIEFF:
thanks
ur welcome (^_^)
Answered by
1
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