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If `sec theta + tan theta = m, ` show that ` ((m^(2) -1))/((m^(2) +1)) = sin theta . `

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We have
`(m^(2) -1)=(sec theta + tan theta)^(2) -1 `
` = sec^(2) theta + tan^(2)theta + 2sec theta tan theta -1 `
` = (sec^(2)theta -1) + tan^(2)theta + 2 sec theta tan theta `
`= 2 tan^(2)theta + 2sec theta tan theta " " [ because sec^(2)theta -1 = tan^(2)theta] `
` = 2 tan theta(tan theta + sec theta). " "...(i)`
` (m^(2) +1) = (sec theta + tan theta)^(2) +1 `
` = sec^(2) theta + tan^(2)theta + 2 sec theta tan theta +1 `
` = (1+ tan^(2) theta)+ sec^(2)theta + 2sec theta tan theta `
`= 2 sec^(2) theta + 2 sec theta tan theta " " [ because 1+ tan^(2)theta = sec^(2)theta ] `
`=2 sec theta (sec theta + tan theta ). " "...(ii) `
From (i) and (ii), we get
`((m^(2) -1))/((m^(2) +1)) = (tan theta)/(sec theta) = ((sin theta)/(cos theta ) xx cos theta) = sin theta . `
Hence, ` ((m^(2) -1))/((m^(2) +1)) = sin theta . `

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