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Question: If \((\sec\alpha + \tan\alpha)(\sec\beta + \tan\beta)(\sec\gamma + \tan\gamma) = \tan\alpha\tan\beta...

If (sec⁡α+tan⁡α)(sec⁡β+tan⁡β)(sec⁡γ+tan⁡γ)=tan⁡αtan⁡βtan⁡γ(\sec\alpha + \tan\alpha)(\sec\beta + \tan\beta)(\sec\gamma + \tan\gamma) = \tan\alpha\tan\beta\tan\gamma, then (sec⁡α−tan⁡α)(sec⁡β−tan⁡β)(sec⁡γ−tan⁡γ)=(\sec\alpha - \tan\alpha)(\sec\beta - \tan\beta)(\sec\gamma - \tan\gamma) =

A

cot⁡αcot⁡βcot⁡γ\cot\alpha\cot\beta\cot\gamma

B

tan⁡αtan⁡βtan⁡γ\tan\alpha\tan\beta\tan\gamma

C

cot⁡α+cot⁡β+cot⁡γ\cot\alpha + \cot\beta + \cot\gamma

D

tan⁡α+tan⁡β+tan⁡γ\tan\alpha + \tan\beta + \tan\gamma

Answer

cot⁡αcot⁡βcot⁡γ\cot\alpha\cot\beta\cot\gamma

Explanation

Solution

Given : (sec⁡α+tan⁡α)(sec⁡β+tan⁡β)(sec⁡γ+tan⁡γ)(\sec\alpha + \tan\alpha)(\sec\beta + \tan\beta)(\sec\gamma + \tan\gamma)

=tan⁡αtan⁡βtan⁡γ= \tan\alpha\tan\beta\tan\gamma ...(i)

Letx=(sec⁡α−tan⁡α)(sec⁡β−tan⁡β)(sec⁡γ−tan⁡γ)x = (\sec\alpha - \tan\alpha)(\sec\beta - \tan\beta)(\sec\gamma - \tan\gamma) ...(ii)

Multiply both equations, (i) and (ii), we get

(sec⁡2α−tan⁡2α)(sec⁡2β−tan⁡2β)(sec⁡2γ−tan⁡2γ)(\sec^{2}\alpha - \tan^{2}\alpha)(\sec^{2}\beta - \tan^{2}\beta)(\sec^{2}\gamma - \tan^{2}\gamma)

=x.(tan⁡αtan⁡βtan⁡γ)= x.(\tan\alpha\tan\beta\tan\gamma)

⇒x=1tan⁡αtan⁡βtan⁡γ\Rightarrow x = \frac{1}{\tan\alpha\tan\beta\tan\gamma} ∴x=cot⁡αcot⁡βcot⁡γ\therefore x = \cot\alpha\cot\beta\cot\gamma