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Question Number 134446 by physicstutes last updated on 03/Mar/21

$$\mathrm{prove}\mathrm{that}$$$$\frac{{\cos }^{4}x+{\sin }^{4}x}{{\cos }^{4}x-{\sin }^{4}x}=\frac{\cos 2A+\sec 2A}{2}$$

Answered by Ar Brandon last updated on 03/Mar/21

$${\cos }^4\mathrm{x}={\left(\frac{{\mathrm{e}}^{\mathrm{ix}}+{\mathrm{e}}^{-\mathrm{ix}}}{2}\right)}^4=\frac{1}{16}({\mathrm{e}}^{4\mathrm{ix}}+{4\mathrm{e}}^{2\mathrm{ix}}+6+{4\mathrm{e}}^{-2\mathrm{ix}}+{\mathrm{e}}^{-4\mathrm{ix}})$$$$=\frac{1}{16}(2\cos 4\mathrm{x}+8\cos 2\mathrm{x}+6)$$$${\sin }^4\mathrm{x}={\left(\frac{{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}}{2\mathrm{i}}\right)}^4=\frac{1}{16}({\mathrm{e}}^{4\mathrm{ix}}-{4\mathrm{e}}^{2\mathrm{ix}}+6-{4\mathrm{e}}^{-2\mathrm{ix}}+{\mathrm{e}}^{-4\mathrm{ix}})$$$$=\frac{1}{16}(2\cos 4\mathrm{x}-8\cos 2\mathrm{x}+6)$$$$\frac{{\cos }^4\mathrm{x}+{\sin }^4\mathrm{x}}{{\cos }^4\mathrm{x}-{\sin }^4\mathrm{x}}=\frac{4\cos 4\mathrm{x}+12}{16\cos 2\mathrm{x}}=\frac{{8\cos }^{2}2\mathrm{x}+8}{16\cos 2\mathrm{x}}$$$$=\frac{\cos 2\mathrm{x}+\sec 2\mathrm{x}}{2}$$

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