Question

(sec A+ tan A)(1- sin A)=cos A
prove​

Answer 1

Answer:

Step-by-step explanation:

(sec A+tan A)(1-sinA)

=(1/cos A + sin A/cos A)(1-sinA)

=[(1+sinA)/cosA](1-sinA)

=(1+sinA)(1-sinA)/cosA

=(1-sin^2 A)/cosA

=(cos^2 A)/cos A

=cos A

Hence proved ....

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Answer 2

Answer:

Step-by-step explanation:

LHS:(Sec A + Tan a)(1 - Sin A)

        =(1/cos A + sin A/cos A)(1-sinA)     {sec A = 1/cos A , Tan A = sin A/cos A }

        As the denominator is common

         =[(1+sinA)/cos A] (1-sinA)  

multiplying (1- Sin A) we get,

         =(1+sinA)(1-sinA)/cos A

{(a+ b)(a - b) = a^2 - b^2}

          =(1-sin^2 A)/cos A  

{ 1 - Sin A = Cos A then, 1 - Sin^2 A = Cos^2 A}

          =(cos^2 A)/cos A  

           =cos A = RHS

Hence, proved.