Question Number 168605 by cortano1 last updated on 14/Apr/22
$$\:\:\:\:\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{sin}\:^{\mathrm{2}} \mathrm{25}°−\mathrm{cos}\:^{\mathrm{2}} \mathrm{15}°}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{sin}\:^{\mathrm{2}} \mathrm{25}°−\mathrm{sin}\:^{\mathrm{2}} \mathrm{15}°}=? \\ $$
Commented by som(math1967) last updated on 14/Apr/22
$${tan}\mathrm{15}×{cot}\mathrm{10}\:? \\ $$
Commented by cortano1 last updated on 14/Apr/22
$${how}\:{sir}? \\ $$
Commented by som(math1967) last updated on 14/Apr/22
$${let}\:\frac{{cos}^{\mathrm{2}} \mathrm{10}+{sin}^{\mathrm{2}} \mathrm{25}−{cos}^{\mathrm{2}} \mathrm{15}}{{sin}^{\mathrm{2}} \mathrm{10}+{sin}^{\mathrm{2}} \mathrm{25}−{sin}^{\mathrm{2}} \mathrm{15}}=\boldsymbol{{p}} \\ $$$$\:\therefore\frac{{cos}^{\mathrm{2}} \mathrm{10}+{sin}^{\mathrm{2}} \mathrm{25}−{cos}^{\mathrm{2}} \mathrm{15}+{sin}^{\mathrm{2}} \mathrm{10}+{sin}^{\mathrm{2}} \mathrm{25}−{sin}^{\mathrm{2}} \mathrm{15}}{\boldsymbol{{cos}}^{\mathrm{2}} \mathrm{10}+\boldsymbol{{sin}}^{\mathrm{2}} \mathrm{25}−\boldsymbol{{cos}}^{\mathrm{2}} \mathrm{15}−{sin}^{\mathrm{2}} \mathrm{10}−{sin}^{\mathrm{2}} \mathrm{25}+{sin}^{\mathrm{2}} \mathrm{15}} \\ $$$$=\frac{\boldsymbol{{p}}+\mathrm{1}}{\boldsymbol{{p}}−\mathrm{1}}\bigstar \\ $$$$\frac{\mathrm{1}+\mathrm{2}\boldsymbol{{sin}}^{\mathrm{2}} \mathrm{25}−\mathrm{1}}{\boldsymbol{{cos}}\mathrm{2}.\mathrm{10}−\boldsymbol{{cos}}\mathrm{2}.\mathrm{15}} \\ $$$$\Rightarrow\frac{\mathrm{2}\boldsymbol{{sin}}^{\mathrm{2}} \mathrm{25}}{\mathrm{2}\boldsymbol{{sin}}\mathrm{25}\boldsymbol{{sin}}\mathrm{5}}=\frac{\boldsymbol{{p}}+\mathrm{1}}{\boldsymbol{{p}}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\boldsymbol{{sin}}\mathrm{25}}{\boldsymbol{{sin}}\mathrm{5}}=\frac{\boldsymbol{{p}}+\mathrm{1}}{\boldsymbol{{p}}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\boldsymbol{{sin}}\mathrm{25}+\boldsymbol{{sin}}\mathrm{5}}{\boldsymbol{{sin}}\mathrm{25}−\boldsymbol{{sin}}\mathrm{5}}=\frac{\boldsymbol{{p}}+\mathrm{1}+\boldsymbol{{p}}−\mathrm{1}}{\boldsymbol{{p}}+\mathrm{1}−\boldsymbol{{p}}+\mathrm{1}}\bigstar \\ $$$$\Rightarrow\frac{\mathrm{2}\boldsymbol{{sin}}\mathrm{15}\boldsymbol{{cos}}\mathrm{10}}{\mathrm{2}\boldsymbol{{cos}}\mathrm{15}\boldsymbol{{sin}}\mathrm{10}}=\frac{\mathrm{2}\boldsymbol{{p}}}{\mathrm{2}} \\ $$$$\therefore\boldsymbol{{p}}=\boldsymbol{{tan}}\mathrm{15}\boldsymbol{{cot}}\mathrm{10} \\ $$$$\bigstar\:\:\boldsymbol{{using}}\:\boldsymbol{{componendo}}\:\boldsymbol{{dividendo}} \\ $$
Commented by cortano1 last updated on 14/Apr/22
$${yes}.\:{great} \\ $$