pls answer fast....... i need the answer.... question no. 3?
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Answered by
5
HELLO DEAR ,
LHS:- (sinA +cosA) * (tanA + cotA)
= (sinA + cosA)* (sinA/cosA + cosA/sinA)
= (sinA + cosA ) * (sin2A + cos2A / sinAcosA)
= sinA + cos A * 1/sinAcosA
= sinA+cosA/sinAcosA
= sinA/sinAcosA + cosA/sinAcosA
= 1/cosA +1/sinA
= SecA + CosecA = RHS
Hence. proved.
I HOPE ITS HELP YOU DEAR :-)
LHS:- (sinA +cosA) * (tanA + cotA)
= (sinA + cosA)* (sinA/cosA + cosA/sinA)
= (sinA + cosA ) * (sin2A + cos2A / sinAcosA)
= sinA + cos A * 1/sinAcosA
= sinA+cosA/sinAcosA
= sinA/sinAcosA + cosA/sinAcosA
= 1/cosA +1/sinA
= SecA + CosecA = RHS
Hence. proved.
I HOPE ITS HELP YOU DEAR :-)
christen:
so the lhs side of the question was wrong, right?
Answered by
3
L.H.S =(1-sin A÷1+sin A× 1-sin A÷1-sin A)^1/2
⇒((1- sin A)²÷1-sin² A ))^1/2
⇒((1- sin A)²÷cos² A))^1/2
⇒1-sin A÷ cos A
⇒ l.h.s= 1÷cos A - sin A÷cos A
⇒L.H.S=sec A - tan A = R.H.S
Hence proved.
I hope that helps.
⇒((1- sin A)²÷1-sin² A ))^1/2
⇒((1- sin A)²÷cos² A))^1/2
⇒1-sin A÷ cos A
⇒ l.h.s= 1÷cos A - sin A÷cos A
⇒L.H.S=sec A - tan A = R.H.S
Hence proved.
I hope that helps.
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