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If tanA + secA =l prove …

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If tanA + secA =l prove that secA= l^2+1/2l

Posted by Ritish Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

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}

1Thank You

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