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Question Number 143348 by gsk2684 last updated on 13/Jun/21

$$\mathrm{if}\frac{\cos {}^4\mathrm{x}}{\cos {}^2\mathrm{y}}+\frac{\sin {}^4\mathrm{x}}{\sin {}^2\mathrm{y}}=1\mathrm{then}$$$$\mathrm{find}\frac{\cos {}^4\mathrm{y}}{\cos {}^2\mathrm{x}}+\frac{\sin {}^4\mathrm{y}}{\sin {}^2\mathrm{x}}=?$$

Answered by som(math1967) last updated on 13/Jun/21

$$\frac{{cos}^{4}x}{{cos}^{2}y}+\frac{{sin}^{4}x}{{sin}^{2}y}=1$$$${cos}^4{xsin}^{2}y+{sin}^4{xcos}^{2}y={sin}^2{ycos}^{2}y$$$${(1-{sin}^{2}x)}^2{sin}^{2}y+{sin}^{4}x(1-{sin}^{2}y)={sin}^2{ycos}^{2}y$$$$\cancel{{sin}^{2}y}-2{sin}^2{xsin}^{2}y+\cancel{{sin}^4{xsin}^{2}y}$$$$+{sin}^{4}x-\cancel{{sin}^4{xsin}^{2}y}=\cancel{{sin}^{2}y}-{sin}^{4}y$$$${sin}^{4}x-2{sin}^2{xsin}^{2}y+{sin}^{4}y=0$$$${({sin}^{2}x-{sin}^{2}y)}^2=0$$$$\therefore {sin}^{2}x={sin}^{2}y$$$$1-{cos}^{2}x=1-{cos}^{2}y$$$$\therefore {cos}^{2}x={cos}^{2}y$$$$$$$$=\frac{{cos}^{4}y}{{cos}^{2}x}+\frac{{sin}^{4}y}{{sin}^{2}x}$$$$=\frac{{cos}^{4}y}{{cos}^{2}y}+\frac{{sin}^{4}y}{{sin}^{2}y}$$$$={cos}^{2}y+{sin}^{2}y=1ans$$

Commented by gsk2684 last updated on 13/Jun/21

$$\mathcal{T}\mathrm{hank}\mathrm{you}\mathrm{sir}$$

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