Question

(Sin A + cosec A)^2 + ( cos A+ sec A)^2= 5+ sec^2.cosec^2A

Answer 1

Step-by-step explanation:

$(sinA+cosecA)²+(cosA+secA)² \\ sin²A+2sinA \: cosecA+cosec²A + cos²A+2cosA \: secA+sec²A \\ sin²A+2+cosec²A+cos²A+2+sec²A \\ 1+2+2+cosec²A+sec²A \\ 5+sec²A+cosec²A\\5+1/cos²A+1/sin2A\\5+sin²A+cos²A/sin²Acos²A\\5+sec²Acosec²A$

Answer 2

Answer:

Step-by-step explanation:

LHS = (sinA + cosecA)² + (cosA + secA)²

       = sin²A + cosec²A + 2 + cos²A + sec²A + 2

       = 4 + (sin²A + cos²A) + (cosec²A + sec²A)

       = 4 + 1 + (1/sin²A + 1/cos²A)

       = 5 + [(sin²A + cos²A)/ (sin²A cos²A)

       = 5 + (1/sin²A cos²A)

       = 5 + sec²A cosec²A

       = RHS