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