Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

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) =

1) cot α cot β cot γ

2) tan α tan β tan γ

3) cot α + cot β + cot γ

4) tan α + tan β + tan γ

Answer: (1) cot α cot β cot γ

Solution:

Given,

(sec α + tan α)(sec β + tan β)(sec γ + tan γ) = tan α tan β tan γ….(i)

Using the identity sec2x – tan2x = 1,

(sec α + tan α)(sec α – tan α) = 1

(sec α – tan α) = 1/(sec α + tan α)….(ii)

(sec β + tan β)(sec β – tan β) = 1

(sec β – tan β) = 1/(sec β + tan β)….(iii)

(sec γ + tan γ)(sec γ – tan γ) = 1

(sec γ – tan γ) = 1/(sec γ + tan γ)….(iv)

Multiplying (ii), (iii) and (iv),

(sec α – tan α)(sec β – tan β)(sec γ – tan γ) = [1/(sec α + tan α)] [1/(sec β + tan β)][1/(sec γ + tan γ)]

= 1/[(sec α + tan α)(sec β + tan β)(sec γ + tan γ)]

= 1/(tan α tan β tan γ) {from (i)}

= cot α cot β cot γ

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