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

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

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

$$tan15\times cot10?$$

Commented by cortano1 last updated on 14/Apr/22

$$howsir?$$

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

$$let\frac{{cos}^{2}10+{sin}^{2}25-{cos}^{2}15}{{sin}^{2}10+{sin}^{2}25-{sin}^{2}15}=\mathit{p}$$$$\therefore \frac{{cos}^{2}10+{sin}^{2}25-{cos}^{2}15+{sin}^{2}10+{sin}^{2}25-{sin}^{2}15}{{\mathit{c}\mathit{o}\mathit{s}}^{2}10+{\mathit{s}\mathit{i}\mathit{n}}^{2}25-{\mathit{c}\mathit{o}\mathit{s}}^{2}15-{sin}^{2}10-{sin}^{2}25+{sin}^{2}15}$$$$=\frac{\mathit{p}+1}{\mathit{p}-1}★$$$$\frac{1+2{\mathit{s}\mathit{i}\mathit{n}}^{2}25-1}{\mathit{c}\mathit{o}\mathit{s}2.10-\mathit{c}\mathit{o}\mathit{s}2.15}$$$$\Rightarrow \frac{2{\mathit{s}\mathit{i}\mathit{n}}^{2}25}{2\mathit{s}\mathit{i}\mathit{n}25\mathit{s}\mathit{i}\mathit{n}5}=\frac{\mathit{p}+1}{\mathit{p}-1}$$$$\Rightarrow \frac{\mathit{s}\mathit{i}\mathit{n}25}{\mathit{s}\mathit{i}\mathit{n}5}=\frac{\mathit{p}+1}{\mathit{p}-1}$$$$\Rightarrow \frac{\mathit{s}\mathit{i}\mathit{n}25+\mathit{s}\mathit{i}\mathit{n}5}{\mathit{s}\mathit{i}\mathit{n}25-\mathit{s}\mathit{i}\mathit{n}5}=\frac{\mathit{p}+1+\mathit{p}-1}{\mathit{p}+1-\mathit{p}+1}★$$$$\Rightarrow \frac{2\mathit{s}\mathit{i}\mathit{n}15\mathit{c}\mathit{o}\mathit{s}10}{2\mathit{c}\mathit{o}\mathit{s}15\mathit{s}\mathit{i}\mathit{n}10}=\frac{2\mathit{p}}{2}$$$$\therefore \mathit{p}=\mathit{t}\mathit{a}\mathit{n}15\mathit{c}\mathit{o}\mathit{t}10$$$$★\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{c}\mathit{o}\mathit{m}\mathit{p}\mathit{o}\mathit{n}\mathit{e}\mathit{n}\mathit{d}\mathit{o}\mathit{d}\mathit{i}\mathit{v}\mathit{i}\mathit{d}\mathit{e}\mathit{n}\mathit{d}\mathit{o}$$

Commented by cortano1 last updated on 14/Apr/22

$$yes.great$$

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