Math, asked by arif30, 1 year ago

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

Answers

Answered by sarginya

(sinA+cosA)(1/cosA+1/sinA)
(sinA+CosA)(sinA+cosA)/sinA×cosA
(sinA+cosA)²/sinA×cosA
sin²A+cos²A+2sinAcosA/sinA×cosA
(1+2sinAcosA)/sinA×cosA
secAcosecA+2

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