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Question Number 59220 by Aditya789 last updated on 06/May/19

find the minimum value of

y = \frac{b^{2}}{\sin^{2} x} + \frac{a^{2}}{\cos^{2} x}

Answered by MJS last updated on 06/May/19

\frac{d y}{d x} = - \frac{2 b^{2} \cos x}{\sin^{3} x} + \frac{2 a^{2} \sin x}{\cos^{3} x} = 0

\sin x = t

\cos x = \sqrt{1 - t^{2}}

2 \frac{(a^{2} - b^{2}) t^{4} + 2 b^{2} t^{2} - b^{2}}{t^{3} {(1 - t^{2})}^{\frac{3}{2}}} = 0

\Rightarrow t = \pm \sqrt{\frac{b}{a + b}} \lor t = \pm \sqrt{\frac{b}{b - a}}

\Rightarrow y = {(a + b)}^{2} \lor y = {(a - b)}^{2}

\Rightarrow the minimal value of y = \min ({(a + b)}^{2}, {(a - b)}^{2})

Commented by MJS last updated on 06/May/19

{you}^{'} re welcome!

Commented by mr W last updated on 06/May/19

y o u a r e r i g h t s i r! t h a n k s!

Answered by tanmay last updated on 06/May/19

a n o t h e r a p p r o a c h

y = b^{2} {c o s e c}^{2} x + a^{2} {s e c}^{2} x

y = a^{2} (1 + {t a n}^{2} x) + b^{2} (1 + {c o t}^{2} x)

y = a^{2} + b^{2} + a^{2} {t a n}^{2} x + b^{2} {c o t}^{2} x

t = t a n x

y = a^{2} + b^{2} + a^{2} t^{2} + \frac{b^{2}}{t^{2}}

y = (a^{2} + b^{2}) + {(a t - \frac{b}{t})}^{2} + 2 \times a t \times \frac{b}{t}

y = a^{2} + b^{2} + 2 a b + {(a t - \frac{b}{t})}^{2}

y_{m i n} = a^{2} + 2 a b + b^{2} = {(a + b)}^{2} [w h e n {(a t - \frac{b}{t})}^{2} = 0]

y_{m i n} = {(a + b)}^{2} w h e n a t - \frac{b}{t} = 0

t^{2} = \frac{b}{a} \to t = \pm \sqrt{\frac{b}{a}}

t a n x = \pm \sqrt{\frac{b}{a}}

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