Question 87

If (sec α + tan α) (sec β + tan β ) (sec γ + tan γ) = tan α tan β tan γ, then (sec α - tan α)(sec β - tan β ) (sec γ - tan γ) = ?

Solution

Expression : (sec α + tan α) (sec β + tan β ) (sec γ + tan γ) = tan α tan β tan γ -----------(i)

Using, $$sec^2 x - tan^2 x = 1$$

=> $$sec^2 \alpha - tan^2 \alpha = (sec \alpha + tan \alpha) (sec \alpha - tan \alpha) = 1$$ ---------(ii)

Similarly, $$(sec \beta + tan \beta) (sec \beta - tan \beta) = 1$$ -----------(iii)

$$(sec \gamma + tan \gamma) (sec \gamma - tan \gamma) = 1$$ -----------(iv)

Multiplying Eqns (ii),(iii)&(iv), we get :

=> $$[(sec \alpha + tan \alpha) (sec \beta + tan \beta) (sec \gamma + tan \gamma)]$$ $$\times [(sec \alpha - tan \alpha) (sec \beta - tan \beta) (sec \gamma - tan \gamma)] = 1$$

Using (i), => $$[tan \alpha.tan \beta.tan \gamma]$$ $$\times[(sec \alpha - tan \alpha) (sec \beta - tan \beta) (sec \gamma - tan \gamma)] = 1$$

=> $$(sec \alpha - tan \alpha) (sec \beta - tan \beta) (sec \gamma - tan \gamma)$$ = $$\frac{1}{tan \alpha.tan \beta.tan \gamma}$$

=> $$(sec \alpha - tan \alpha) (sec \beta - tan \beta) (sec \gamma - tan \gamma)$$ = $$cot \alpha.cot \beta.cot \gamma$$