Question Number 62347 by rajesh4661kumar@gamil.com last updated on 20/Jun/19 | ||
Answered by Kunal12588 last updated on 20/Jun/19 | ||
$$\frac{{csc}\:\theta\:−\:{sin}\:\theta}{{sec}\:\theta\:−\:{cos}\:\theta}\:=\:\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} } \\ $$$$\Rightarrow{cot}\:\theta\:\frac{\mathrm{1}−{sin}^{\mathrm{2}} \theta}{\mathrm{1}−{cos}^{\mathrm{2}} \theta}=\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} } \\ $$$$\Rightarrow{cot}\:\theta=\frac{{a}}{{b}} \\ $$$$\Rightarrow{cot}^{\mathrm{2}} \theta=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{1}+{cot}^{\mathrm{2}} \theta=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{csc}^{\mathrm{2}} \theta=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\:\frac{{csc}^{\mathrm{2}} \theta−\mathrm{1}}{{csc}\:\theta}={a}^{\mathrm{3}} \\ $$$$\Rightarrow\frac{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}{\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{b}}}={a}^{\mathrm{3}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={a}^{\mathrm{3}} {b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{1}={ab}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{1} \\ $$ | ||
Commented by peter frank last updated on 20/Jun/19 | ||
$${nice} \\ $$ | ||
Answered by Kunal12588 last updated on 20/Jun/19 | ||
$${ANOTHER}\:{WAY} \\ $$$$\mathrm{cosec}\:\theta\:−\:\mathrm{sin}\:\theta\:=\:{a}^{\mathrm{3}} \\ $$$$\underset{\mathrm{tan}^{−\mathrm{1}} \:\:{t}\:=\:\theta} {\Rightarrow}\:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}−\frac{{t}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}={a}^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{1}+{t}^{\mathrm{2}} −{t}^{\mathrm{2}} ={a}^{\mathrm{3}} {t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{3}} {t}}=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{sec}\:\theta\:−\:\mathrm{cos}\:\theta\:={b}^{\mathrm{3}} \\ $$$$\underset{\mathrm{tan}^{−\mathrm{1}} \:{t}\:=\:\theta} {\Rightarrow}\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}={b}^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{1}={b}^{\mathrm{3}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{t}^{\mathrm{2}} }{{b}^{\mathrm{3}} }=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} {t}}=\frac{{t}^{\mathrm{2}} }{{b}^{\mathrm{3}} } \\ $$$$\Rightarrow{t}=\frac{{b}}{{a}} \\ $$$${putting}\:{the}\:{value}\:{of}\:\:{t} \\ $$$$\frac{\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}{{b}^{\mathrm{3}} }=\sqrt{\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{a}} \\ $$$$\Rightarrow{ab}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{1} \\ $$ | ||