Question Number 151959 by john_santu last updated on 24/Aug/21
Answered by iloveisrael last updated on 24/Aug/21
$$\left(\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} +\left(\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{1}\right)\left(\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}−\left(\mathrm{sin}\:{x}\mathrm{cos}\:{x}\right)^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{3}\left(\mathrm{sin}\:{x}\mathrm{cos}\:{x}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{sin}\:{x}\mathrm{cos}\:{x}\right)^{\mathrm{2}} =\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{6}} {x}}+\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{6}} {x}}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\left(\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }\:=\:\mathrm{16}. \\ $$