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Question Number 168605 by cortano1 last updated on 14/Apr/22

((cos^2 10°+sin^2 25°−cos^2 15°)/(sin^2 10°+sin^2 25°−sin^2 15°))=?

$$\:\:\:\:\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{sin}\:^{\mathrm{2}} \mathrm{25}°−\mathrm{cos}\:^{\mathrm{2}} \mathrm{15}°}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{sin}\:^{\mathrm{2}} \mathrm{25}°−\mathrm{sin}\:^{\mathrm{2}} \mathrm{15}°}=? \\ $$

Commented by som(math1967) last updated on 14/Apr/22

tan15×cot10 ?

$${tan}\mathrm{15}×{cot}\mathrm{10}\:? \\ $$

Commented by cortano1 last updated on 14/Apr/22

how sir?

$${how}\:{sir}? \\ $$

Commented by som(math1967) last updated on 14/Apr/22

let ((cos^2 10+sin^2 25−cos^2 15)/(sin^2 10+sin^2 25−sin^2 15))=p ∴((cos^2 10+sin^2 25−cos^2 15+sin^2 10+sin^2 25−sin^2 15)/(cos^2 10+sin^2 25−cos^2 15−sin^2 10−sin^2 25+sin^2 15)) =((p+1)/(p−1))★ ((1+2sin^2 25−1)/(cos2.10−cos2.15)) ⇒((2sin^2 25)/(2sin25sin5))=((p+1)/(p−1)) ⇒((sin25)/(sin5))=((p+1)/(p−1)) ⇒((sin25+sin5)/(sin25−sin5))=((p+1+p−1)/(p+1−p+1))★ ⇒((2sin15cos10)/(2cos15sin10))=((2p)/2) ∴p=tan15cot10 ★ using componendo dividendo

$${let}\:\frac{{cos}^{\mathrm{2}} \mathrm{10}+{sin}^{\mathrm{2}} \mathrm{25}−{cos}^{\mathrm{2}} \mathrm{15}}{{sin}^{\mathrm{2}} \mathrm{10}+{sin}^{\mathrm{2}} \mathrm{25}−{sin}^{\mathrm{2}} \mathrm{15}}=\boldsymbol{{p}} \\ $$$$\:\therefore\frac{{cos}^{\mathrm{2}} \mathrm{10}+{sin}^{\mathrm{2}} \mathrm{25}−{cos}^{\mathrm{2}} \mathrm{15}+{sin}^{\mathrm{2}} \mathrm{10}+{sin}^{\mathrm{2}} \mathrm{25}−{sin}^{\mathrm{2}} \mathrm{15}}{\boldsymbol{{cos}}^{\mathrm{2}} \mathrm{10}+\boldsymbol{{sin}}^{\mathrm{2}} \mathrm{25}−\boldsymbol{{cos}}^{\mathrm{2}} \mathrm{15}−{sin}^{\mathrm{2}} \mathrm{10}−{sin}^{\mathrm{2}} \mathrm{25}+{sin}^{\mathrm{2}} \mathrm{15}} \\ $$$$=\frac{\boldsymbol{{p}}+\mathrm{1}}{\boldsymbol{{p}}−\mathrm{1}}\bigstar \\ $$$$\frac{\mathrm{1}+\mathrm{2}\boldsymbol{{sin}}^{\mathrm{2}} \mathrm{25}−\mathrm{1}}{\boldsymbol{{cos}}\mathrm{2}.\mathrm{10}−\boldsymbol{{cos}}\mathrm{2}.\mathrm{15}} \\ $$$$\Rightarrow\frac{\mathrm{2}\boldsymbol{{sin}}^{\mathrm{2}} \mathrm{25}}{\mathrm{2}\boldsymbol{{sin}}\mathrm{25}\boldsymbol{{sin}}\mathrm{5}}=\frac{\boldsymbol{{p}}+\mathrm{1}}{\boldsymbol{{p}}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\boldsymbol{{sin}}\mathrm{25}}{\boldsymbol{{sin}}\mathrm{5}}=\frac{\boldsymbol{{p}}+\mathrm{1}}{\boldsymbol{{p}}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\boldsymbol{{sin}}\mathrm{25}+\boldsymbol{{sin}}\mathrm{5}}{\boldsymbol{{sin}}\mathrm{25}−\boldsymbol{{sin}}\mathrm{5}}=\frac{\boldsymbol{{p}}+\mathrm{1}+\boldsymbol{{p}}−\mathrm{1}}{\boldsymbol{{p}}+\mathrm{1}−\boldsymbol{{p}}+\mathrm{1}}\bigstar \\ $$$$\Rightarrow\frac{\mathrm{2}\boldsymbol{{sin}}\mathrm{15}\boldsymbol{{cos}}\mathrm{10}}{\mathrm{2}\boldsymbol{{cos}}\mathrm{15}\boldsymbol{{sin}}\mathrm{10}}=\frac{\mathrm{2}\boldsymbol{{p}}}{\mathrm{2}} \\ $$$$\therefore\boldsymbol{{p}}=\boldsymbol{{tan}}\mathrm{15}\boldsymbol{{cot}}\mathrm{10} \\ $$$$\bigstar\:\:\boldsymbol{{using}}\:\boldsymbol{{componendo}}\:\boldsymbol{{dividendo}} \\ $$

Commented by cortano1 last updated on 14/Apr/22

yes. great

$${yes}.\:{great} \\ $$

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