Question Number 164426 by cortano1 last updated on 17/Jan/22
$$\:\:{In}\:\Delta{ABC}\:{if}\:\begin{cases}{\mathrm{cot}\:{A}.\mathrm{cot}\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{cot}\:{B}.\mathrm{cot}\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{18}}}\end{cases} \\ $$$$\:{then}\:\mathrm{tan}\:\:{C}\:=\:? \\ $$
Commented by bobhans last updated on 17/Jan/22
$$\:\mathrm{let}\:\mathrm{tan}\:\mathrm{C}=\mathrm{x}>\mathrm{0}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{2}}{\mathrm{x}}+\frac{\mathrm{18}}{\mathrm{x}}+\mathrm{x}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:.\frac{\mathrm{18}}{\mathrm{x}}.\:\mathrm{x} \\ $$$$\:\Rightarrow\:\mathrm{20}\:+\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{36}\Rightarrow\mathrm{x}=\mathrm{tan}\:\mathrm{C}\:=\:\mathrm{4}\:\checkmark \\ $$
Answered by som(math1967) last updated on 17/Jan/22
$${A}+{B}+{C}=\pi \\ $$$${C}=\pi−\left({A}+{B}\right) \\ $$$${cotC}=−{cot}\left({A}+{B}\right) \\ $$$${cotC}=\:−\frac{{cotAcotB}−\mathrm{1}}{{cotB}+{cotA}} \\ $$$${cotBcotC}+{cotAcotC}=−{cotAcotB}+\mathrm{11} \\ $$$$\Rightarrow{cotAcotB}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{18}−\mathrm{1}−\mathrm{9}}{\mathrm{18}}=\frac{\mathrm{8}}{\mathrm{18}}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$${cot}^{\mathrm{2}} {Acot}^{\mathrm{2}} {Bcot}^{\mathrm{2}} {C}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{18}}×\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{81}} \\ $$$${cotAcotBcotC}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${tanC}=\frac{\mathrm{1}}{{cotC}}=\frac{{cotAcotB}}{{cotAcotBcotC}}=\frac{\mathrm{4}}{\mathrm{9}}×\frac{\mathrm{9}}{\mathrm{1}} \\ $$$${tanC}=\mathrm{4} \\ $$
Commented by cortano1 last updated on 17/Jan/22
$${only}\:\:\mathrm{4}\:{sir} \\ $$
Commented by som(math1967) last updated on 17/Jan/22
$${yes}\:{tanC}>\mathrm{0} \\ $$