prove that secA(1-sinA) (secA+tanA)
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Step-by-step explanation:
SinA/secA+tanA-1+cosA/cosecA+cotA-1
=sinA/(1/cosA+sinA/cosA-1)+cosA/(1/sinA+cosA/sinA-1)
=sinA/{(1+sinA-cosA)/cosA}+cosA/{(1+cosA-sinA)/sinA}
=sinAcosA/(1+sinA-cosA)+sinAcosA/(1+cosA-sinA)
=sinAcosA[(1+cosA-sinA+1+sinA-cosA)/(1+sinA-cosA)(1+cosA-sinA)]
=2sinAcosA/(1+sinA-cosA+cosA+sinAcosA-cos²A-sinA-sin²A+sinAcosA)
=2sinAcosA/{1+2sinAcosA-(sin²A+cos²A)}
=2sinAcosA/(1+2sinAcosA-1)
=2sinAcosA/2sinAcosA
=1 (Proved)
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