NT P 2 M 36

Ya Even I felt

But this is given as an integer type where max is 9

& ans given is 3

(Though it can never be 3

o
---- )
o

=\sqrt{\frac{(1+sin^{2}\alpha )(1+cos^{2}\alpha) }{sin^{2}\alpha cos^{2}\alpha }}
=\sqrt{\frac{sin^{2}\alpha cos^{2}\alpha +2}{sin^{2}\alpha cos^{2}\alpha }}
=\sqrt{\frac{4sin^{2}\alpha cos^{2}\alpha +2}{4sin^{2}\alpha cos^{2}\alpha }}
= \sqrt{}{\frac{{2sin^{2}2\alpha+2 }}{sin^{2}2\alpha }}
= \sqrt{2+\frac{2}{sin^{2}2\alpha }}
=let y=\sqrt{2+\frac{2}{sin^{2}2\alpha }}
= dy/dx=\frac{1}{2\sqrt{2+\frac{2}{sin^{2}2\alpha }}}\left(2\left(\frac{-1}{sin^{2}2\alpha } \right) \right).2sin2\alpha cos2\alpha
= dy/dx= - \frac{2cos2\alpha }{\sqrt{2sin^{2}2\alpha +2}}

now for MAXIMA or MINIMA , dy/dx = 0 ,\Rightarrow - \frac{2cos2\alpha }{\sqrt{2sin^{2}2\alpha +2}}=0
\Rightarrow cos2\alpha =0\Rightarrow 2\alpha =\pi /2\Rightarrow \alpha =\pi /4
Now substituting in the above equation the value of \alpha we get minimum value = 3
please correct me if i'm rong. i didn't check value of second differentiation for value of alpha is minimum or max. Got the value and just substituted.

@MKM: maxima is asked

@uttara: he solved for minima

ANSWER SHOULD BE VERY CLOSE TO INFINITY AS

y = √2 + 2 / sin ² 2α as corretcly stated by Manmay ,

α cannot take the value 0 as α belongs to open interval of ( 0 , pi / 2 ) , but still it will take such a small value that the sin part is approximately 0.

Hence y ≈ ∞

Ya Asish & Gallardo I agree with ure explanations

Manmay's method is wrong as explained by Asish he solved for minima

But any way s thanks for contributing all 3 of u

MAXIMA IS NOT DEFINED IN THIS CASE