જો (sec alpha + tan alpha )(sec beta + tan be | vedclass

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Question

(a) Given : $(\sec \alpha + 	an \alpha )(\sec \beta + 	an \beta )(\sec \gamma + 	an \gamma )$

$ = 	an \alpha 	an \beta 	an \gamma $ ...$(i)$

Vedclass · Accepted Answer

$\cot \alpha \cot \beta \cot \gamma $

Vedclass · Answer

(a) Given : $(\sec \alpha + 	an \alpha )(\sec \beta + 	an \beta )(\sec \gamma + 	an \gamma )$

$ = 	an \alpha 	an \beta 	an \gamma $ ...$(i)$

Let $ x = (\sec \alpha - 	an \alpha )(\sec \beta - 	an \beta )(\sec \gamma - 	an \gamma )$ ...$(ii)$

Multiply both equations, $(i)$ and $(ii)$, we get

$({\sec ^2}\alpha - {	an ^2}\alpha )({\sec ^2}\beta - {	an ^2}\beta )({\sec ^2}\gamma - {	an ^2}\gamma )$

$ = x.(	an \alpha 	an \beta 	an \gamma )$

$ \Rightarrow x = \frac{1}{{	an \alpha 	an \beta 	an \gamma }}$

$	herefore x = \cot \alpha \cot \beta \cot \gamma $

જો $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma )$

$ = \tan \alpha \tan \beta \tan \gamma $, તો $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )$$(\sec \gamma - \tan \gamma ) = $

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જો $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma )$ $ = \tan \alpha \tan \beta \tan \gamma $, તો $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )$$(\sec \gamma - \tan \gamma ) = $