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Please can someone solve this quickly? I am preparing for my 10th board exam so please help.(cosA - sinA)(1 + tanA) / 2cos²A - 1 = secA

Answer»

Step-by-step EXPLANATION:

L.H.S=(cosA - sinA)(1 + TANA) / 2cos²A - 1

        =(cosA- sinA)(1+sinA/cosA)/2Cos^2A-1

         [because tanA=sinA/Cos A]

       cosA(1+SinA/cosA)-sinA(1+sinA/cosA)/2Cos^2A-1

       MULTIPLYING inside the brackets,

           cosA+SinA-sinA-sin^2A/cosA/2Cos^2A-1

          =cosA-sin^2A/cosA/2Cos^2A-1

          =cos^2A-sin^2A/cosA/2Cos^2A-1

          =cos^2A-(1-Cos^2A)/cosA/2cos^2A-1

          [because sin^2A+cos^2A=1 so sin^2A=1-Cos^2A

          =cos^2A-1+Cos^2A/cosA/2Cos^2A-1

        =2Cos^2A-1/cosA(2Cos^2A-1)

        2Cos^2A GETS cancelled

      =1/CosA=SecA=R.H.S

i am not sure you understand.....u could write down and try it like this.

I HOPE u undrstand and plss markbrainliest if usefull.


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