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

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