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Question Number 158335 by mr W last updated on 02/Nov/21

if α,β,γ are the angles of a triangle, find (1/(tan 𝛂 tan 𝛃))+(1/(tan 𝛃 tan 𝛄))+(1/(tan 𝛄 tan 𝛂))=?

$${if}\:\alpha,\beta,\gamma\:{are}\:{the}\:{angles}\:{of}\:{a}\:{triangle}, \\ $$$${find}\: \\ $$$$\frac{\mathrm{1}}{\boldsymbol{\mathrm{tan}}\:\boldsymbol{\alpha}\:\boldsymbol{\mathrm{tan}}\:\boldsymbol{\beta}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{tan}}\:\boldsymbol{\beta}\:\boldsymbol{\mathrm{tan}}\:\boldsymbol{\gamma}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{tan}}\:\boldsymbol{\gamma}\:\boldsymbol{\mathrm{tan}}\:\boldsymbol{\alpha}}=? \\ $$

Answered by puissant last updated on 03/Nov/21

α + β + γ = π → α + β = π−γ ; ⇒ tan(α+β)=tan(π−γ) ⇒ ((tanα+tanβ)/(1−tanαtanβ)) = −tanγ ⇒ tanα+tanβ = −tanγ+tanαtanβtanγ ⇒ tanα+tanβ+tanγ=tanαtanβtanγ (1/(tanαtanβ))+(1/(tanβtanγ))+(1/(tanαtanγ)) =((tanγ)/(tanαtanβtanγ))+((tanα)/(tanαtanβtanγ))+((tanβ)/(tanαtanβtanγ)) =((tanα+tanβ+tanγ)/(tanαtanβtanγ)) = 1 ...........Le puissant............

$$\:\:\:\:\:\:\:\:\:\:\:\alpha\:+\:\beta\:+\:\gamma\:=\:\pi\:\rightarrow\:\alpha\:+\:\beta\:=\:\pi−\gamma\:; \\ $$$$ \\ $$$$\Rightarrow\:{tan}\left(\alpha+\beta\right)={tan}\left(\pi−\gamma\right)\:\Rightarrow\:\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}\:=\:−{tan}\gamma \\ $$$$ \\ $$$$\Rightarrow\:{tan}\alpha+{tan}\beta\:=\:−{tan}\gamma+{tan}\alpha{tan}\beta{tan}\gamma \\ $$$$ \\ $$$$\Rightarrow\:{tan}\alpha+{tan}\beta+{tan}\gamma={tan}\alpha{tan}\beta{tan}\gamma \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{tan}\alpha{tan}\beta}+\frac{\mathrm{1}}{{tan}\beta{tan}\gamma}+\frac{\mathrm{1}}{{tan}\alpha{tan}\gamma}\: \\ $$$$=\frac{{tan}\gamma}{{tan}\alpha{tan}\beta{tan}\gamma}+\frac{{tan}\alpha}{{tan}\alpha{tan}\beta{tan}\gamma}+\frac{{tan}\beta}{{tan}\alpha{tan}\beta{tan}\gamma} \\ $$$$=\frac{{tan}\alpha+{tan}\beta+{tan}\gamma}{{tan}\alpha{tan}\beta{tan}\gamma}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...........\mathscr{L}{e}\:{puissant}............ \\ $$

Commented by mr W last updated on 03/Nov/21

thanks!

$${thanks}! \\ $$

Commented by puissant last updated on 03/Nov/21

you′re welcome sir

$${you}'{re}\:{welcome}\:{sir} \\ $$

Answered by ajfour last updated on 03/Nov/21

Commented by ajfour last updated on 03/Nov/21

required=((pq)/r^2 )+(((1−((pq)/r^2 )))/(((p/q)+(q/r))))((p/r)+(q/r)) = 1 .

$${required}=\frac{{pq}}{{r}^{\mathrm{2}} }+\frac{\left(\mathrm{1}−\frac{{pq}}{{r}^{\mathrm{2}} }\right)}{\left(\frac{{p}}{{q}}+\frac{{q}}{{r}}\right)}\left(\frac{{p}}{{r}}+\frac{{q}}{{r}}\right) \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{1}\:. \\ $$

Commented by mr W last updated on 03/Nov/21

great!

$${great}! \\ $$

Commented by ajfour last updated on 03/Nov/21

yeah, but not worth a like..

Commented by mr W last updated on 03/Nov/21

not many people can come to this method, i think.

$${not}\:{many}\:{people}\:{can}\:{come}\:{to}\:{this} \\ $$$${method},\:{i}\:{think}. \\ $$

Commented by otchereabdullai@gmail.com last updated on 07/Nov/21

nice one!

$$\mathrm{nice}\:\mathrm{one}! \\ $$

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