Question Number 59220 by Aditya789 last updated on 06/May/19
$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{y}=\:\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \mathrm{x}} \\ $$
Answered by MJS last updated on 06/May/19
$$\frac{{dy}}{{dx}}=−\frac{\mathrm{2}{b}^{\mathrm{2}} \mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{3}} \:{x}}+\frac{\mathrm{2}{a}^{\mathrm{2}} \mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{3}} \:{x}}=\mathrm{0} \\ $$$$\mathrm{sin}\:{x}\:={t} \\ $$$$\mathrm{cos}\:{x}\:=\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\mathrm{2}\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}^{\mathrm{4}} +\mathrm{2}{b}^{\mathrm{2}} {t}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{t}^{\mathrm{3}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\mathrm{0} \\ $$$$\Rightarrow\:{t}=\pm\sqrt{\frac{{b}}{{a}+{b}}}\:\vee\:{t}=\pm\sqrt{\frac{{b}}{{b}−{a}}} \\ $$$$\Rightarrow\:{y}=\left({a}+{b}\right)^{\mathrm{2}} \:\vee\:{y}=\left({a}−{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{minimal}\:\mathrm{value}\:\mathrm{of}\:{y}=\mathrm{min}\left(\left({a}+{b}\right)^{\mathrm{2}} ,\:\left({a}−{b}\right)^{\mathrm{2}} \right) \\ $$
Commented by MJS last updated on 06/May/19
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}! \\ $$
Commented by mr W last updated on 06/May/19
$${you}\:{are}\:{right}\:{sir}!\:{thanks}! \\ $$
Answered by tanmay last updated on 06/May/19
![another approach y=b^2 cosec^2 x+a^2 sec^2 x y=a^2 (1+tan^2 x)+b^2 (1+cot^2 x) y=a^2 +b^2 +a^2 tan^2 x+b^2 cot^2 x t=tanx y=a^2 +b^2 +a^2 t^2 +(b^2 /t^2 ) y=(a^2 +b^2 )+(at−(b/t))^2 +2×at×(b/t) y=a^2 +b^2 +2ab+(at−(b/t))^2 y_(min) =a^2 +2ab+b^2 =(a+b)^2 [when (at−(b/t))^2 =0] y_(min) =(a+b)^2 when at−(b/t)=0 t^2 =(b/a)→t=±(√(b/a)) tanx=±(√(b/a))](https://www.tinkutara.com/question/Q59241.png)
$${another}\:{approach} \\ $$$${y}={b}^{\mathrm{2}} {cosec}^{\mathrm{2}} {x}+{a}^{\mathrm{2}} {sec}^{\mathrm{2}} {x} \\ $$$${y}={a}^{\mathrm{2}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)+{b}^{\mathrm{2}} \left(\mathrm{1}+{cot}^{\mathrm{2}} {x}\right) \\ $$$${y}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {cot}^{\mathrm{2}} {x} \\ $$$${t}={tanx} \\ $$$${y}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{{t}^{\mathrm{2}} } \\ $$$${y}=\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+\left({at}−\frac{{b}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}×{at}×\frac{{b}}{{t}} \\ $$$${y}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}+\left({at}−\frac{{b}}{{t}}\right)^{\mathrm{2}} \\ $$$${y}_{{min}} ={a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} \:\:\:\left[{when}\:\left({at}−\frac{{b}}{{t}}\right)^{\mathrm{2}} =\mathrm{0}\right] \\ $$$${y}_{{min}} =\left({a}+{b}\right)^{\mathrm{2}} \:\:\:{when}\:{at}−\frac{{b}}{{t}}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} =\frac{{b}}{{a}}\rightarrow{t}=\pm\sqrt{\frac{{b}}{{a}}}\: \\ $$$${tanx}=\pm\sqrt{\frac{{b}}{{a}}}\: \\ $$