Proove that ( sinA + cosA + 1 ) × ( sin A - 1 + cosA ) × secAcotA = 2
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in this question not 2 possible plz can you check the question.
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akshitakarn:
It's only 2 not 2 secA
hey .sistet can you chek my answerd ..
Answered by
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Given {sinA/1+cosA }+{1+CosA /sinA}=2cosecA
Take LHS
={SinA/1+cosA}+{1+cosA/sinA}
solve….this
It becomes
=SinA. SinA+(1+cosA)(1+cosA)/sinA(1+cosA)
=sin^2A+(1+cosA)^2 / sinA(1+cosA)
=sin^2A+cos^2A+1+2cosA /sinA(1+cosA)
we know identity ,
Sin^2A+Cos^2A=1
It becomes,
=1+1+2cosA/sinA(1+CosA)
=2(1+cosA)/sinA(1+cosA)
=2/sinA
= 2CosecA
Take LHS
={SinA/1+cosA}+{1+cosA/sinA}
solve….this
It becomes
=SinA. SinA+(1+cosA)(1+cosA)/sinA(1+cosA)
=sin^2A+(1+cosA)^2 / sinA(1+cosA)
=sin^2A+cos^2A+1+2cosA /sinA(1+cosA)
we know identity ,
Sin^2A+Cos^2A=1
It becomes,
=1+1+2cosA/sinA(1+CosA)
=2(1+cosA)/sinA(1+cosA)
=2/sinA
= 2CosecA
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