Question

sin A+ cosec A/ sec A= cos A​

Answer 1

Firstly square on both the sides ,

              (sinA+cosecA)^{2} +(cosA+secA)^{2} = 7+tan^{2} +cot^{2}

here starts,

       LHS:

            =sin^{2}A + cosec^{2} A + 2 sinA cosecA + cos^{2} + sec^{2} + 2 cosA secA.

           = 1+2+2+1+cot^{2} +1+tan^{2}

           =7+tan^{2} + cot^{2}[/tex].

LHS=RHS

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