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Ram Kushwah 6 years, 4 months ago
tanA + secA =l
or secA+tanA= I ---------------------(1)
We know that
{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}
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Janu Likki 6 years, 4 months ago
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