Question
$(\sec \theta + \tan \theta ) . (\sec \theta - \tan \theta ) =$ ?

Answer

$(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)$
$=\sec ^2 \theta-\tan ^2 \theta \ldots \ldots\left[\because(a+b)(a-b)=a^2-b^2\right]$
$=1 \ldots \ldots\left[\begin{array}{l}\because 1+\tan ^2 \theta=\sec ^2 \theta \\ \therefore \sec ^2 \theta-\tan ^2 \theta=1\end{array}\right]$

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