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tanA+sinA=m, tanA-sinA=n. Then prove that, √mn=tanA.sinA

tanA+sinA=m, tanA-sinA=n.
Then prove that, √mn=tanA.sinA

Grade:12th pass

1 Answers

Niranjan Pandey of Cambridge
34 Points
2 years ago
A/q    mn= (tanA+sinA)(tanA-sinA)
              =tan2A-sinAtanA+sinAtanA-sin2A
              =tan2A-sin2A
              =(tanA+sinA)(tanA-sinA)
              =(sinA/cosA + sinA)(sinA/cosA – sinA)
              ={(sinA+sinAcosA)/cosA}{(sinA-sinAcosA)/cosA}
              ={sinA(1+ cosA)/cosA}{sinA(1- cosA)/cosA}
              ={sin2A(1-cosA2)}/cos2A
              ={sin2A(sin2A)}/cos2A
   √mn   =sinAsinA/cosA
              =sinA*sinA/cosA
              =sinA*tanA
              =tanAsinA

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