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Prove the following identity: $\frac{1}{(\sec A + \tan A )}= \sec A - \tan A$

I searched for an answer on the net but found it really confusing. Can anyone help me with this with less number of steps?

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    $\begingroup$ $$(\sec A-\tan A)(\sec A+\tan A)=?$$ $\endgroup$ Aug 28, 2016 at 9:06
  • $\begingroup$ @lab bhattacharjee (sec A - tan A ) (sec A + tan A) = 1 $\endgroup$ Aug 28, 2016 at 9:09
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    $\begingroup$ So, if $ab=1, a=?$ $\endgroup$ Aug 28, 2016 at 9:11

1 Answer 1

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$$\color{red}{\sec^2 A-\tan^2 A} = \frac{1-\sin^2 A}{\cos^2 A} = \color{red}{1}$$ hence by dividing both sides by $\sec A+\tan A$...

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