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Question Number 129073 by amns last updated on 12/Jan/21

$$\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t}:$$$${sec}^4\theta -{sec}^2\theta ={tan}^4\theta +{tan}^2\theta$$$$Pleasehelpmequickly!$$

Commented by benjo_mathlover last updated on 12/Jan/21

$$\sec {}^4\theta -\sec {}^2\theta =\sec {}^2\theta (\sec {}^2\theta -1)$$$$=(1+\tan {}^2\theta )\left(\tan {}^2\theta \right)$$$$=\tan {}^2\theta +\tan {}^4\theta$$

Answered by MJS_new last updated on 12/Jan/21

$$\frac{1}{{\cos }^{4}x}-\frac{1}{{\cos }^{2}x}=$$$$$$$$=\frac{1}{c^4}-\frac{1}{c^2}=\frac{1-c^2}{c^4}=\frac{s^2}{c^4}=\frac{s^2\times 1}{c^4}=\frac{s^2(s^2+c^2)}{c^4}=$$$$=\frac{s^4+s^2{c}^2}{c^4}=\frac{s^4}{c^4}+\frac{s^2}{c^2}=$$$$={\tan }^{4}x+{\mathrm{ran}}^{2}x$$

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