Question

Prove that sec^2A + cosec^2A can never be less than 2.

Answer 1

see in attachment:-

either see the solution given below

(secx)^2+(cosec)^2=2+((tanx)^2+(cotx)^2)

Using A.M is greater than or equal to G.M we get

((tanx)^2+(cotx)^2)/2 is greater than or equal to ((tanx)^2(cotx)^2)^1/2 i.e 1.

(secx)^2+(cosecx)^2 is greater than or equal to 2+2 i.e 4.

Hence minimum value of (secx)^2+(cosecx)^2 is 4.
means greater then 2

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i m taking x in place of A