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