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Question Number 81944 by M±th+et£s last updated on 16/Feb/20
show that   cot(40°)−cot(50°)=2tan(10°)  cos(70°) cos(50^° ) cos(10^° )=((√3)/8)
$${show}\:{that}\: \\ $$$${cot}\left(\mathrm{40}°\right)−{cot}\left(\mathrm{50}°\right)=\mathrm{2}{tan}\left(\mathrm{10}°\right) \\ $$$${cos}\left(\mathrm{70}°\right)\:{cos}\left(\mathrm{50}^{°} \right)\:{cos}\left(\mathrm{10}^{°} \right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$
Answered by TANMAY PANACEA last updated on 16/Feb/20
tan10^o =tan(50^o −40^o )                =((tan50−tan40)/(1+tan50.tan40))                =(((1/(cot50))−(1/(cot40)))/(1+1))   [tan50×tan40=tan50×cot50=1]           2tan10=((cot40−co50)/(cot50cot40))     [cot50×cot40=1]  so cot4o−cot50=2tan10 proved
$${tan}\mathrm{10}^{{o}} ={tan}\left(\mathrm{50}^{{o}} −\mathrm{40}^{{o}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{tan}\mathrm{50}−{tan}\mathrm{40}}{\mathrm{1}+{tan}\mathrm{50}.{tan}\mathrm{40}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{1}}{{cot}\mathrm{50}}−\frac{\mathrm{1}}{{cot}\mathrm{40}}}{\mathrm{1}+\mathrm{1}}\:\:\:\left[{tan}\mathrm{50}×{tan}\mathrm{40}={tan}\mathrm{50}×{cot}\mathrm{50}=\mathrm{1}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}{tan}\mathrm{10}=\frac{{cot}\mathrm{40}−{co}\mathrm{50}}{{cot}\mathrm{50}{cot}\mathrm{40}}\:\:\:\:\:\left[{cot}\mathrm{50}×{cot}\mathrm{40}=\mathrm{1}\right] \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{cot}}\mathrm{4}\boldsymbol{{o}}−\boldsymbol{{cot}}\mathrm{50}=\mathrm{2}\boldsymbol{{tan}}\mathrm{10}\:\boldsymbol{{proved}} \\ $$
Answered by TANMAY PANACEA last updated on 16/Feb/20
cos(60+10)cos(60−10)cos10  =(cos60cos10−sin60sin10)(cos60cos10+sin60sin10)cos10  =[((1/(2 ))cos10)^2 −(((√3)/2)sin10)^2 ]×cos10  =((cos^2 10−3sin^2 10)/4)×cos10  =((cos^3 10−3(1−cos^2 10)cos10)/4)  =((4cos^3 10−3cos10)/4)  =((cos(3×10))/4)=((√3)/(2×4))=((√3)/8)
$${cos}\left(\mathrm{60}+\mathrm{10}\right){cos}\left(\mathrm{60}−\mathrm{10}\right){cos}\mathrm{10} \\ $$$$=\left({cos}\mathrm{60}{cos}\mathrm{10}−{sin}\mathrm{60}{sin}\mathrm{10}\right)\left({cos}\mathrm{60}{cos}\mathrm{10}+{sin}\mathrm{60}{sin}\mathrm{10}\right){cos}\mathrm{10} \\ $$$$=\left[\left(\frac{\mathrm{1}}{\mathrm{2}\:}{cos}\mathrm{10}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\mathrm{10}\right)^{\mathrm{2}} \right]×{cos}\mathrm{10} \\ $$$$=\frac{{cos}^{\mathrm{2}} \mathrm{10}−\mathrm{3}{sin}^{\mathrm{2}} \mathrm{10}}{\mathrm{4}}×{cos}\mathrm{10} \\ $$$$=\frac{{cos}^{\mathrm{3}} \mathrm{10}−\mathrm{3}\left(\mathrm{1}−{cos}^{\mathrm{2}} \mathrm{10}\right){cos}\mathrm{10}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{4}{cos}^{\mathrm{3}} \mathrm{10}−\mathrm{3}{cos}\mathrm{10}}{\mathrm{4}} \\ $$$$=\frac{{cos}\left(\mathrm{3}×\mathrm{10}\right)}{\mathrm{4}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}×\mathrm{4}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$
Commented by M±th+et£s last updated on 16/Feb/20
god bless you sir
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 16/Feb/20
god bless to all of us
$${god}\:{bless}\:{to}\:{all}\:{of}\:{us} \\ $$

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