Question Number 72931 by Maclaurin Stickker last updated on 05/Nov/19
$${Find}\:{the}\:{numeric}\:{value}\:{of} \\ $$$$\frac{\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} \left(\alpha\right)}{\mathrm{cos}\:^{\mathrm{2}} \left(\alpha\right)−\mathrm{tan}\:^{\mathrm{2}} \left(\alpha\right)} \\ $$
Answered by ajfour last updated on 05/Nov/19
$${say}\:\alpha=\mathrm{0}° \\ $$$${numeric}\:{value}\:=\:\frac{\mathrm{2}}{\mathrm{1}}\:=\:\mathrm{2}\: \\ $$$${say}\:\alpha=\mathrm{45}° \\ $$$${numeric}\:{value}\:=\:\frac{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}}=\:−\mathrm{3}\:. \\ $$
Commented by Maclaurin Stickker last updated on 05/Nov/19
$${Can}\:{you}\:{solve}\:{by}\:{factoring} \\ $$$${the}\:{expression}? \\ $$
Commented by ajfour last updated on 05/Nov/19
$${perhaps}\:{someone}\:{else}\:{can}.. \\ $$
Answered by MJS last updated on 05/Nov/19
$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{general}\:\mathrm{numeric}\:\mathrm{value}. \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{express}\:\mathrm{using} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{only}\:\mathrm{sin}\:\alpha\:={s} \\ $$$$\frac{{s}^{\mathrm{4}} −\mathrm{3}{s}^{\mathrm{2}} +\mathrm{2}}{{s}^{\mathrm{4}} −\mathrm{3}{s}^{\mathrm{2}} +\mathrm{1}}=\frac{\left({s}−\mathrm{1}\right)\left({s}+\mathrm{1}\right)\left({s}^{\mathrm{2}} −\mathrm{2}\right)}{\left({s}^{\mathrm{2}} −{s}−\mathrm{1}\right)\left({s}^{\mathrm{2}} +{s}−\mathrm{1}\right)} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{only}\:\mathrm{cos}\:\alpha\:={c} \\ $$$$\frac{{c}^{\mathrm{2}} \left({c}^{\mathrm{2}} +\mathrm{1}\right)}{{c}^{\mathrm{4}} +{c}^{\mathrm{2}} −\mathrm{1}} \\ $$$$ \\ $$$$\left(\mathrm{3}\right)\:\mathrm{only}\:\mathrm{tan}\:\alpha\:={t} \\ $$$$−\frac{{t}^{\mathrm{2}} +\mathrm{2}}{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} −\mathrm{1}} \\ $$