Math, asked by christen, 1 year ago

pls answer fast....... i need the answer.... question no. 3?

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Answered by rohitkumargupta
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 :-)

christen: so the lhs side of the question was wrong, right?
Answered by Wimpybrain
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.

Wimpybrain: oops I answered the 4th question
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