Question Number 75303 by vishalbhardwaj last updated on 09/Dec/19
$$\mathrm{If}\:{cos}\theta\:=\:\frac{{cos}\alpha−\mathrm{cos}\beta}{\mathrm{1}−{cos}\alpha\:{cos}\beta}\:,\:\mathrm{prove}\:\mathrm{that}\:\mathrm{one} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\frac{\theta}{\mathrm{2}}\:\mathrm{is}\:{tan}\frac{\alpha}{\mathrm{2}}\:.\:{tan}\frac{\beta}{\mathrm{2}}\:?? \\ $$
Answered by mind is power last updated on 09/Dec/19
$$\frac{\mathrm{1}−\mathrm{cos}\left(\theta\right)}{\mathrm{1}+\mathrm{cos}\left(\theta\right)}=\mathrm{tg}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\left(\theta\right)}{\mathrm{1}+\mathrm{cos}\left(\theta\right)}=\frac{\mathrm{1}−\mathrm{cos}\left(\alpha\right)\mathrm{cos}\left(\beta\right)−\mathrm{cos}\left(\alpha\right)+\mathrm{cos}\left(\beta\right)}{\mathrm{1}−\mathrm{cos}\left(\alpha\right)\mathrm{cos}\left(\beta\right)+\mathrm{cos}\left(\alpha\right)−\mathrm{cos}\left(\beta\right)}=\frac{\left(\mathrm{1}+\mathrm{cos}\left(\beta\right)\right)\left(\mathrm{1}−\mathrm{cos}\left(\alpha\right)\right)}{\left(\mathrm{1}+\mathrm{cos}\left(\alpha\right)\right)\left(\mathrm{1}−\mathrm{cos}\left(\beta\right)\right.} \\ $$$$=\frac{\mathrm{tg}^{\mathrm{2}} \left(\alpha\right)}{\mathrm{tg}^{\mathrm{2}} \left(\beta\right)}\Rightarrow\mathrm{tg}\left(\frac{\theta}{\mathrm{2}}\right)=\underset{−} {+}\frac{\mathrm{tg}\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{tg}\left(\frac{\beta}{\mathrm{2}}\right)} \\ $$
Commented by vishalbhardwaj last updated on 09/Dec/19
$$\mathrm{what}\:\mathrm{is}\:\mathrm{tg} \\ $$
Commented by JDamian last updated on 09/Dec/19
$${tg}\:\:{is}\:\:{tan} \\ $$
Commented by mind is power last updated on 09/Dec/19
$$\mathrm{tg}=\mathrm{tan}\: \\ $$
Commented by peter frank last updated on 10/Dec/19
$${please}\:{help}\:{sir}\:{mind}\:{is}\:{power}\:\:{and}\:{Damian}\:\:{am}\:{stucking} \\ $$$${Qn}\:\mathrm{75302} \\ $$