Question

The minimum value of (sinx+cosecx)^2 + (cosx+secx)^2

Answer 1

Sin^2 x + 2 + cosec^2 x + cos^2 x + 2 + sec^2 x

5 + cosec^2 x + sec^2 x

Answer 2

\[\sin^2x+2+\frac{ 1 }{ \sin^2x }+\cos^2x+2+\frac{ 1 }{ \cos^2x }=1+4+\frac{ 1 }{ (\sin^2x)(\cos^2x) }\]
\[1+4+1=6\]

Answer 3

minimum 6

Answer 4

nope...the answer shuld be 9

Answer 5

yep its 9 only,,
@Jonask just missed this:
1/sin^2x cos^2x = 4/(sin2x)^2 => minima is 4
hence total = 1 + 4 +4 = 9

Answer 6

Hw Did u Do that @shubhamsrg

Answer 7

\[\frac{ 1 }{ (\frac{ 1 }{ 2 }\sin2x)^2 }=\frac{ 4 }{ \sin^22x }\]

Answer 8

min \[\frac{ 4 }{ 1 }=4\]
instead of 1+4+1
1+4+4

Answer 9

anyways, you may also go for
AM>= GM

(sinx+cosecx)^2 + (cosx+secx)^2 >= 2 (sinx+cosecx) (cosx+secx)
consider RHS

2 (sinx cosx + sinx sec x + cosecx cosx + cosecx sec x)
2( sinx cos x+ 1/sinx cosx + tanx + cot x)
sin 2x + 4/sin 2x + 2tanx + 2/tanx

x=pi/4 gives a maximum = 1 + 4 + 2+ 2 = 9
which is a minima for LHS..