Question Number 29413 by math solver last updated on 08/Feb/18
Commented by math solver last updated on 08/Feb/18
$${q}.\mathrm{3}\:? \\ $$
Answered by beh.i83417@gmail.com last updated on 09/Feb/18
$${x}^{\mathrm{2}} −\mathrm{2}{x}.{cos}\alpha+{cos}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \alpha=\mathrm{0} \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{y}.{cos}\beta+{cos}^{\mathrm{2}} \beta+{sin}^{\mathrm{2}} \beta=\mathrm{0} \\ $$$$\left({x}−{cos}\alpha\right)^{\mathrm{2}} ={i}^{\mathrm{2}} {sin}^{\mathrm{2}} \alpha\Rightarrow{x}={cos}\alpha\pm{i}.{sin}\alpha \\ $$$$\left({y}−{cos}\beta\right)^{\mathrm{2}} ={i}^{\mathrm{2}} {sin}^{\mathrm{2}} \beta\Rightarrow{y}={cos}\beta\pm{i}.{sin}\beta \\ $$$$\left(\pm{ve}\right): \\ $$$$\begin{cases}{{x}={cos}\alpha+{i}.{sin}\alpha}\\{\frac{\mathrm{1}}{{x}}={cos}\alpha−{i}.{sin}\alpha}\end{cases}\Rightarrow\begin{cases}{{cos}\alpha=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{{x}}\right)}\\{{sin}\alpha=\frac{\mathrm{1}}{\mathrm{2}{i}}\left({x}−\frac{\mathrm{1}}{{x}}\right)}\end{cases} \\ $$$${cos}\left(\alpha+\beta\right)={cos}\alpha.{cos}\beta−{sin}\alpha.{sin}\beta \\ $$$${cos}\alpha.{cos}\beta=\frac{\mathrm{1}}{\mathrm{4}}.\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)}{{xy}} \\ $$$${sin}\alpha.{sin}\beta=\frac{−\mathrm{1}}{\mathrm{4}}.\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}{{xy}} \\ $$$$\mathrm{4}{cos}\left(\alpha+\beta\right)=\frac{{x}^{\mathrm{2}} .{y}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}+{x}^{\mathrm{2}} .{y}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{1}}{{xy}} \\ $$$$\Rightarrow\mathrm{2}{cos}\left(\alpha+\beta\right)=\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{1}}{{xy}}={xy}+\frac{\mathrm{1}}{{xy}}\:.\blacksquare \\ $$
Commented by math solver last updated on 09/Feb/18
$${sir},{although}\:{your}\:{method}\:{is}\:{right} \\ $$$${but}\:{ans}.\:{given}\:{in}\:{book}\:{is}\:{option}\left({c}\right). \\ $$
Commented by beh.i83417@gmail.com last updated on 09/Feb/18
$${you}\:{are}\:{right}\:{sir}.{it}\:{is}\:{now}\:{corrected}. \\ $$
Commented by math solver last updated on 09/Feb/18
$${thank}\:{you}\:{sir}. \\ $$