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If $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma )$$ = \tan \alpha \tan \beta \tan \gamma $, then $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )$$(\sec \gamma - \tan \gamma ) = $
$\cot \alpha \cot \beta \cot \gamma $
$\tan \alpha \tan \beta \tan \gamma $
$\cot \alpha + \cot \beta + \cot \gamma $
$\tan \alpha + \tan \beta + \tan \gamma $
Solution
(a)Given : $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma )$
$ = \tan \alpha \tan \beta \tan \gamma $ …(i)
Let$x = (\sec \alpha – \tan \alpha )(\sec \beta – \tan \beta )(\sec \gamma – \tan \gamma )$ …(ii)
Multiply both equations, (i) and (ii), we get
$({\sec ^2}\alpha – {\tan ^2}\alpha )({\sec ^2}\beta – {\tan ^2}\beta )({\sec ^2}\gamma – {\tan ^2}\gamma )$
$ = x.(\tan \alpha \tan \beta \tan \gamma )$
$ \Rightarrow x = \frac{1}{{\tan \alpha \tan \beta \tan \gamma }}$$\therefore x = \cot \alpha \cot \beta \cot \gamma $
