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1/sec-1 + 1/sec+1 = 2cotcosec
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1/sec-1 + 1/sec+1 = 2cotcosec

User Bharat Gulati
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1 Answer

5 votes

Solution :

LHS = 1/(secA -1) + 1/(secA+1)

= [secA+1+secA-1]/[(secA-1)(secA+1)]

= (2secA)/[sec²A-1]

=(2secA)/tan²A

= (2/cosA)/(sin²A/cos²A)

= ( 2cosA)/sin²A

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

= 2cosecAcotA

= RHS

••••

User Frederic Conrotte
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