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

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Question

(A) આપેલ છે: $(\sec \alpha + 	an \alpha )(\sec \beta + 	an \beta )(\sec \gamma + 	an \gamma ) = 	an \alpha 	an \beta 	an \gamma$ ... $(i)$ધારો ક

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) આપેલ છે: $(\sec \alpha + 	an \alpha )(\sec \beta + 	an \beta )(\sec \gamma + 	an \gamma ) = 	an \alpha 	an \beta 	an \gamma$ ... $(i)$ધારો કે $x = (\sec \alpha - 	an \alpha )(\sec \beta - 	an \beta )(\sec \gamma - 	an \gamma )$ ... $(ii)$સમીકરણ $(i)$ અને $(ii)$ નો ગુણાકાર કરતા,આપણને મળે છે:$(\sec^2 \alpha - 	an^2 \alpha )(\sec^2 \beta - 	an^2 \beta )(\sec^2 \gamma - 	an^2 \gamma ) = x \cdot (	an \alpha 	an \beta 	an \gamma )$કોઈપણ ખૂણા $	heta$ માટે $\sec^2 	heta - 	an^2 	heta = 1$ હોવાથી,ડાબી બાજુ $(1)(1)(1) = 1$ થશે.તેથી,$1 = x \cdot (	an \alpha 	an \beta 	an \gamma )$.$x = \frac{1}{	an \alpha 	an \beta 	an \gamma } = \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 $ હોય,તો $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )(\sec \gamma - \tan \gamma ) = $

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