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Question Number 190877 by 073 last updated on 13/Apr/23

Commented by 073 last updated on 13/Apr/23

$$\mathrm{solution}\mathrm{please}?$$

Commented by Frix last updated on 13/Apr/23

$${\mathrm{There}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{solution}\mathrm{because}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}$$$$\mathrm{equation}.\mathrm{We}\mathrm{can}\mathrm{transform}\mathrm{it}.$$$$\sin (\alpha +\beta )+\sin (\alpha -\beta )=2\sin \alpha \cos \beta$$$$\tan (\alpha +\beta )+\tan (\alpha -\beta )=\frac{2\sin \alpha \cos \alpha }{{\cos }^2\alpha -{\sin }^2\beta }$$$$\Rightarrow \mathrm{Answer}\mathrm{is}\frac{({\cos }^{2}3-{\sin }^{2}x)\cos x}{\cos 3}$$

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