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Question Number 116891 by bemath last updated on 07/Oct/20

$$\mathrm{Proof}\mathrm{that}\frac{4(\cos {}^4\left(a\right)+\sin {}^4\left(a\right))}{\cos {}^4\left(a\right)-\sin {}^4\left(a\right)}=(3+\cos \left(4a\right))\sec \left(2a\right)$$

Answered by john santu last updated on 07/Oct/20

$$\Rightarrow \frac{4\{{(\sin {}^{2}a+\cos {}^{2}a)}^2-2\sin {}^{2}a\cos {}^{2}a\}}{\cos {}^{2}a-\sin {}^{2}a}=$$$$\frac{4(1-\frac{1}{2}\sin {}^{2}2a)}{\cos 2a}=\frac{4-2\sin {}^{2}2a}{\cos 2a}$$$$=\frac{4-2(\frac{1}{2}-\frac{1}{2}\cos 4a)}{\cos 2a}=(3+\cos 4a)\sec 2a$$$$proved$$

Answered by Dwaipayan Shikari last updated on 07/Oct/20

$$\frac{4({cos}^{4}a+{sin}^{4}a)}{({cos}^{2}a+{sin}^{2}a)({cos}^{2}a-{sin}^{2}a)}$$$$=\frac{4({({cos}^{2}a+{sin}^{2}a)}^2-2{sin}^2{acos}^{2}a)}{cos2a}$$$$=\frac{4(1-\frac{1}{2}{sin}^{2}2a)}{cos2a}=\frac{4-2{sin}^{2}2a}{cos2a}=\frac{4-1+cos4a}{cos2a}=(3+cos4a)sec2a$$

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