1+cosA+sinA/1+cosA-sinA=1+sinA/cosA
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prabinbehera1:
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we have RHS= (1+sinA)/cosA
=> secA+tanA
LHS= (1+cosA+sinA)/(1+cosA-sinA)
=>( secA+tanA+1)/(secA-tan+1) [ divided numerator and denominator by cosA]
=> secA+tanA+(sec²A-tan²A)/(secA-tan+1) [since sec²A-tan²A=1]
=> secA+tanA+(secA-tanA)(secA+tanA)/(secA-tan+1)
=> secA+tanA[(secA-tanA+1)/(secA-tan+1) [ taken secA+tanA as common]
=> secA+tanA
LHS=RHS
=> secA+tanA
LHS= (1+cosA+sinA)/(1+cosA-sinA)
=>( secA+tanA+1)/(secA-tan+1) [ divided numerator and denominator by cosA]
=> secA+tanA+(sec²A-tan²A)/(secA-tan+1) [since sec²A-tan²A=1]
=> secA+tanA+(secA-tanA)(secA+tanA)/(secA-tan+1)
=> secA+tanA[(secA-tanA+1)/(secA-tan+1) [ taken secA+tanA as common]
=> secA+tanA
LHS=RHS
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