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Question Number 18693 by Arnab Maiti last updated on 27/Jul/17
prove that 2tan^(−1) [tan(α/2)tan((π/4)−(β/2))] =tan^(−1) (((sinα cosβ)/(cosα +sinβ)))
$$\mathrm{prove}\:\mathrm{that}\:\mathrm{2tan}^{−\mathrm{1}} \left[\mathrm{tan}\frac{\alpha}{\mathrm{2}}\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)\right] \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\alpha\:\mathrm{cos}\beta}{\mathrm{cos}\alpha\:+\mathrm{sin}\beta}\right) \\ $$
Answered by 951172235v last updated on 01/Feb/19
tan (Θ/(2 )) =tan (α/2) tan ((Λ^− /4) −(β/2)) =((tan (α/2)(1−tan (β/2)))/((1+tan (β/2)))) tan Θ = ((2tan (Θ/2))/(1−tan^2 (Θ/2))) = ((2tan (α/2)(1−tan^2 (β/2)))/((1−tan (β/2))^2 −tan^2 (α/2)(1+tan (β/2))^2 )) = ((2tan (α/2)(1−tan^2 (β/2)))/((1−tan^2 (α/2))(1+tan^2 (β/2)) + 2tan (β/2)(1+tan^2 (α/2)))) Dividing by (1+tan^2 (α/2))(1+tan^2 (β/2)) tan Θ = ((sin α cos β)/(cos α+sin β)) Θ= tan^(−1) (((sin αcos β)/(cos α+sin β)))
$$\mathrm{tan}\:\frac{\Theta}{\mathrm{2}\:}\:=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\left(\frac{\overset{−} {\Lambda}}{\mathrm{4}}\:−\frac{\beta}{\mathrm{2}}\right)\:=\frac{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\left(\mathrm{1}−\mathrm{tan}\:\frac{\beta}{\mathrm{2}}\right)}{\left(\mathrm{1}+\mathrm{tan}\:\frac{\beta}{\mathrm{2}}\right)} \\ $$$$\mathrm{tan}\:\Theta\:=\:\frac{\mathrm{2tan}\:\frac{\Theta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\Theta}{\mathrm{2}}}\:=\:\frac{\mathrm{2tan}\:\frac{\alpha}{\mathrm{2}}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\beta}{\mathrm{2}}\right)}{\left(\mathrm{1}−\mathrm{tan}\:\frac{\beta}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{tan}\:^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}\left(\mathrm{1}+\mathrm{tan}\:\frac{\beta}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2tan}\:\frac{\alpha}{\mathrm{2}}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\beta}{\mathrm{2}}\right)}{\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}\right)\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \frac{\beta}{\mathrm{2}}\right)\:+\:\mathrm{2tan}\:\frac{\beta}{\mathrm{2}}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}\right)} \\ $$$$\mathrm{Dividing}\:\mathrm{by}\:\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}\right)\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \frac{\beta}{\mathrm{2}}\right) \\ $$$$\mathrm{tan}\:\Theta\:=\:\frac{\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta}{\mathrm{cos}\:\alpha+\mathrm{sin}\:\beta} \\ $$$$\Theta=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\alpha\mathrm{cos}\:\beta}{\mathrm{cos}\:\alpha+\mathrm{sin}\:\beta}\right) \\ $$

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