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Question

Prove the following identity :
( sin A + sec A )2 + ( cosA + cosecA ) = ( 1 + secA • cosecA )


Solution

(sinA + secA)² + (cosA + cosecA)² 

= (sinA + 1/cosA)² + (cosA + 1/sinA)² 

= sin²A + 2sinA/cosA + 1/cos²A + cos²A + 2cosA/sinA + 1/sin²A 

= (sin²A + cos²A) + (2sinA/cosA + 2cosA/sinA) + (1/cos²A + 1/sin²A) 

= 1 + 2(sin²A+cos²A)/(sinAcosA) + (sin²A+cos²A)/(sin²Acos²A) 

= 1 + 2/(sinAcosA) + 1/(sin²Acos²A) 

= (1 + 1/(sinAcosA))² 

= (1 + secA cosecA)²
You forgot to put the square on the RHS

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