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| 1. |
If tanA + secA =l prove that secA= l^2+1/2l |
| Answer» tanA + secA =l\xa0or secA+tanA= I ---------------------(1)We know\xa0that\xa0{tex}\\sec^2\\mathrm A-\\tan^2\\mathrm A=\\mathrm 1-------(2){/tex}Dividing (2) by (1) we get{tex}\\begin{array}{l}\\frac{\\sec^2\\mathrm A-\\tan^2\\mathrm A}{\\mathrm{secA}+\\mathrm{tanA}}=\\frac1{\\mathrm I}\\\\\\mathrm{secA}-\\mathrm{tanA}=\\frac1{\\mathrm I}-----------(3)\\end{array}{/tex}Adding (1) and (3) we get{tex}\\begin{array}{l}2\\mathrm{secA}=\\mathrm I+\\frac1{\\mathrm I}\\\\2\\mathrm{secA}=\\frac{\\mathrm I^2+1}{\\mathrm I}\\\\\\mathrm{secA}=\\frac{\\mathrm I^2+1}{2\\mathrm I}\\\\\\mathrm{Hence}\\;\\mathrm{proved}\\end{array}{/tex} | |