Question Number 59811 by ANTARES VY last updated on 15/May/19
$$\left(\frac{\boldsymbol{\mathrm{tg}}^{\mathrm{2}} \left(\mathrm{590}°\right)}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\mathrm{320}°\right)}+\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{111}°\right)}{\boldsymbol{\mathrm{cos}}\left(\mathrm{159}°\right)}\right)\left(\frac{\boldsymbol{\mathrm{cos}}\left(\mathrm{279}°\right)}{\boldsymbol{\mathrm{sin}}\left(\mathrm{549}°\right)}+\frac{\boldsymbol{\mathrm{ctg}}\left(\mathrm{950}°\right)}{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\mathrm{400}°\right)}\right) \\ $$$$\boldsymbol{\mathrm{simplify}}. \\ $$
Commented by Kunal12588 last updated on 15/May/19
$${sir}\:{is}\:{it} \\ $$$$\left(\frac{\boldsymbol{\mathrm{tg}}^{\mathrm{2}} \left(\mathrm{590}°\right)}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\mathrm{320}°\right)}+\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{111}°\right)}{\boldsymbol{\mathrm{cos}}\left(\mathrm{159}°\right)}\right)\left(\frac{\boldsymbol{\mathrm{cos}}\left(\mathrm{279}°\right)}{\boldsymbol{\mathrm{sin}}\left(\mathrm{549}°\right)}+\frac{\boldsymbol{\mathrm{ctg}}^{\mathrm{2}} \left(\mathrm{950}°\right)}{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\mathrm{400}°\right)}\right)? \\ $$
Commented by ANTARES VY last updated on 15/May/19
$$\boldsymbol{\mathrm{yes}} \\ $$
Answered by Kunal12588 last updated on 15/May/19
$${tan}\left(\mathrm{590}°\right)={tan}\left(\mathrm{360}°+\mathrm{180}°+\mathrm{50}°\right)={tan}\left(\mathrm{50}°\right)={cot}\left(\mathrm{40}°\right) \\ $$$${cos}\left(\mathrm{320}°\right)={cos}\left(\mathrm{360}°−\mathrm{40}°\right)={cos}\left(\mathrm{40}°\right) \\ $$$${sin}\left(\mathrm{111}°\right)={sin}\left(\mathrm{90}°+\mathrm{21}°\right)={cos}\left(\mathrm{21}°\right) \\ $$$${cos}\left(\mathrm{159}°\right)={cos}\left(\mathrm{180}°−\mathrm{21}°\right)=−{cos}\left(\mathrm{21}°\right) \\ $$$${cos}\left(\mathrm{279}°\right)={cos}\left(\mathrm{270}°+\mathrm{9}°\right)={sin}\left(\mathrm{9}°\right) \\ $$$${sin}\left(\mathrm{549}°\right)={sin}\left(\mathrm{360}°+\mathrm{180}°+\mathrm{9}°\right)=−{sin}\left(\mathrm{9}°\right) \\ $$$${cot}\left(\mathrm{950}°\right)={cot}\left(\mathrm{1080}°−\mathrm{180}°+\mathrm{50}°\right)={cot}\left(\mathrm{50}°\right)={tan}\left(\mathrm{40}°\right) \\ $$$${sin}\left(\mathrm{400}°\right)={sin}\left(\mathrm{360}°+\mathrm{40}°\right)={sin}\left(\mathrm{40}°\right) \\ $$$$\left(\frac{\boldsymbol{\mathrm{tg}}^{\mathrm{2}} \left(\mathrm{590}°\right)}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\mathrm{320}°\right)}+\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{111}°\right)}{\boldsymbol{\mathrm{cos}}\left(\mathrm{159}°\right)}\right)\left(\frac{\boldsymbol{\mathrm{cos}}\left(\mathrm{279}°\right)}{\boldsymbol{\mathrm{sin}}\left(\mathrm{549}°\right)}+\frac{\boldsymbol{\mathrm{ctg}}^{\mathrm{2}} \left(\mathrm{950}°\right)}{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\mathrm{400}°\right)}\right) \\ $$$$=\left(\frac{{cot}^{\mathrm{2}} \left(\mathrm{40}°\right)}{{cos}^{\mathrm{2}} \left(\mathrm{40}°\right)}+\frac{{cos}\left(\mathrm{21}°\right)}{−{cos}\left(\mathrm{21}°\right)}\right)\left(\frac{{sin}\left(\mathrm{9}°\right)}{−{sin}\left(\mathrm{9}°\right)}+\frac{{tan}^{\mathrm{2}} \left(\mathrm{40}°\right)}{{sin}^{\mathrm{2}} \left(\mathrm{40}°\right)}\right) \\ $$$$=\left(\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\mathrm{40}°\right)}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\mathrm{40}°\right)}−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\mathrm{40}°\right){cos}^{\mathrm{2}} \left(\mathrm{40}°\right)}−\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\mathrm{40}°\right)}−\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\mathrm{40}°\right)}+\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\mathrm{40}°\right){cos}^{\mathrm{2}} \left(\mathrm{40}°\right)}−\frac{{sin}^{\mathrm{2}} \mathrm{40}°+{cos}^{\mathrm{2}} \mathrm{40}°}{{sin}^{\mathrm{2}} \left(\mathrm{40}°\right){cos}^{\mathrm{2}} \left(\mathrm{40}°\right)}+\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\mathrm{40}°\right){cos}^{\mathrm{2}} \left(\mathrm{40}°\right)}−\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\mathrm{40}°\right){cos}^{\mathrm{2}} \left(\mathrm{40}°\right)}+\mathrm{1} \\ $$$$=\mathrm{1} \\ $$$$ \\ $$