If 0 λ μ δ 2, then secλ+secμ+secδcosecλ+cosecμ+cosecδ lies between Solution:- 3 secλ secλ + secμ - Maths - Trigonometric Functions

Question

If 0<λ<μ<δ<π2, then
secλ+secμ+secδcosecλ+cosecμ+cosecδ
lies between

Solution:-
3 secλ < secλ + secμ + sec∂ <
3sec∂

sir please tell me whyy it becomes 3seclambda

Brijendra Pal · Accepted Answer

Dear Student, From 0<λ<μ<δ<π2we can judge that minimum value can be λ and maximum value can be taken as δ So minimum value caseμ=δ=λ= secλ+secλ+secλcosecλ+cosecλ+cosecλ= 2secλ+1sinλcosλ+2cosecλ=2secλ+cosecλ+22sinλcosλBut it is not coming to be 3 secλ So kindly recheck the question as some sign must come between   secλ+secλ+secλ and cosecλ+cosecλ+cosecλ

If 0<λ<μ<δ<π2, then secλ+secμ+secδcosecλ+cosecμ+cosecδ lies between Solution:- 3 secλ < secλ + secμ + sec∂ < 3sec∂ sir please tell me whyy it becomes 3seclambda

If 0<λ<μ<δ<π2, then secλ+secμ+secδcosecλ+cosecμ+cosecδ lies between

Solution:-
3 secλ < secλ + secμ + sec∂ < 3sec∂

sir please tell me whyy it becomes 3seclambda