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Question Number 114988 by ZiYangLee last updated on 22/Sep/20

$$\mathrm{Prove}{\tan }^{2}x={\sin }^{2}x+{\sec }^{2}x$$

Answered by Olaf last updated on 22/Sep/20

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{false}\mathrm{sir}.$$$${\tan }^{2}x\ne {\sin }^{2}x+{\sec }^{2}x$$

Commented by ZiYangLee last updated on 22/Sep/20

$$\mathrm{Ok}\mathrm{sir}\mathrm{tks}$$

Answered by Olaf last updated on 22/Sep/20

$$\mathrm{But}{\tan }^{2}x=\frac{{\sin }^{2}x}{{\cos }^{2}x}={\sin }^{2}x{\left(\frac{1}{\cos x}\right)}^2={\sin }^{2}x\times {\sec }^{2}x$$$${\mathrm{It}}^{\prime }\mathrm{s}\times \mathrm{not}+$$$$$$

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