Question-204250

Question Number 204250 by Noorzai last updated on 10/Feb/24

Answered by AST last updated on 10/Feb/24

tan(9)+tan(81)−(tan27+tan63) =((sin9)/(cos9))+((sin81)/(cos81))−(((sin27)/(cos27))+((sin63)/(cos63))) =((2sin(9+81)=2)/(2cos9cos81))−((2sin(27+63)=2)/(2cos(27)cos(63))) =(2/(2sin(9)cos(9)))−(2/(2sin(27)cos(27)))=(2/(sin(18)))−(2/(sin(54))) Let θ=18°⇒sin(5θ)=16sin^5 (θ)−20sin^3 (θ)+5sin(θ) ⇒sin(18)=(((√5)−1)/4);sin(54)=sin(3θ)=(((√5)+1)/4) ⇒?=((8((√5)+1))/4)−((8((√5)−1))/4)=4

$${tan}\left(\mathrm{9}\right)+{tan}\left(\mathrm{81}\right)−\left({tan}\mathrm{27}+{tan}\mathrm{63}\right) \\ $$$$=\frac{{sin}\mathrm{9}}{{cos}\mathrm{9}}+\frac{{sin}\mathrm{81}}{{cos}\mathrm{81}}−\left(\frac{{sin}\mathrm{27}}{{cos}\mathrm{27}}+\frac{{sin}\mathrm{63}}{{cos}\mathrm{63}}\right) \\ $$$$=\frac{\mathrm{2}{sin}\left(\mathrm{9}+\mathrm{81}\right)=\mathrm{2}}{\mathrm{2}{cos}\mathrm{9}{cos}\mathrm{81}}−\frac{\mathrm{2}{sin}\left(\mathrm{27}+\mathrm{63}\right)=\mathrm{2}}{\mathrm{2}{cos}\left(\mathrm{27}\right){cos}\left(\mathrm{63}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{9}\right){cos}\left(\mathrm{9}\right)}−\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{27}\right){cos}\left(\mathrm{27}\right)}=\frac{\mathrm{2}}{{sin}\left(\mathrm{18}\right)}−\frac{\mathrm{2}}{{sin}\left(\mathrm{54}\right)} \\ $$$${Let}\:\theta=\mathrm{18}°\Rightarrow{sin}\left(\mathrm{5}\theta\right)=\mathrm{16}{sin}^{\mathrm{5}} \left(\theta\right)−\mathrm{20}{sin}^{\mathrm{3}} \left(\theta\right)+\mathrm{5}{sin}\left(\theta\right) \\ $$$$\Rightarrow{sin}\left(\mathrm{18}\right)=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}};{sin}\left(\mathrm{54}\right)={sin}\left(\mathrm{3}\theta\right)=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow?=\frac{\mathrm{8}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{4}}−\frac{\mathrm{8}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{4}}=\mathrm{4} \\ $$

Answered by universe last updated on 10/Feb/24

tan(9)+tan(81)−(tan27+tan63) =((sin9)/(cos9))+((sin81)/(cos81))−(((sin27)/(cos27))+((sin63)/(cos63))) =((2sin(9+81)=2)/(2cos9cos81))−((2sin(27+63))/(2cos(27)cos(63))) =(2/(2sin(9)cos(9)))−(2/(2sin(27)cos(27)))=(2/(sin(18)))−(2/(sin(54))) 2[((sin54−sin18 )/(sin54 sin18 ))] = 4((cos36 sin18 )/(sin54 sin18 )) =4

$${tan}\left(\mathrm{9}\right)+{tan}\left(\mathrm{81}\right)−\left({tan}\mathrm{27}+{tan}\mathrm{63}\right) \\ $$$$=\frac{{sin}\mathrm{9}}{{cos}\mathrm{9}}+\frac{{sin}\mathrm{81}}{{cos}\mathrm{81}}−\left(\frac{{sin}\mathrm{27}}{{cos}\mathrm{27}}+\frac{{sin}\mathrm{63}}{{cos}\mathrm{63}}\right) \\ $$$$=\frac{\mathrm{2}{sin}\left(\mathrm{9}+\mathrm{81}\right)=\mathrm{2}}{\mathrm{2}{cos}\mathrm{9}{cos}\mathrm{81}}−\frac{\mathrm{2}{sin}\left(\mathrm{27}+\mathrm{63}\right)}{\mathrm{2}{cos}\left(\mathrm{27}\right){cos}\left(\mathrm{63}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{9}\right){cos}\left(\mathrm{9}\right)}−\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{27}\right){cos}\left(\mathrm{27}\right)}=\frac{\mathrm{2}}{{sin}\left(\mathrm{18}\right)}−\frac{\mathrm{2}}{{sin}\left(\mathrm{54}\right)} \\ $$$$\:\mathrm{2}\left[\frac{\mathrm{sin54}−\mathrm{sin18}\:}{\mathrm{sin54}\:\mathrm{sin18}\:}\right]\:=\:\mathrm{4}\frac{\mathrm{cos36}\:\mathrm{sin18}\:\:}{\mathrm{sin54}\:\mathrm{sin18}\:\:}\:=\mathrm{4} \\ $$

Answered by cortano12 last updated on 10/Feb/24

⇒ tan 9°+cot 9°−(tan 27°+cot 27°) = (2/(sin 18°)) − (2/(sin 54°)) = ((4(sin 54°−sin 18°))/(2sin 54° sin 18°)) = ((8cos 36° sin 18°)/(2 cos 36° sin 18°)) = determinant ((4))

$$\:\:\Rightarrow\:\mathrm{tan}\:\mathrm{9}°+\mathrm{cot}\:\mathrm{9}°−\left(\mathrm{tan}\:\mathrm{27}°+\mathrm{cot}\:\mathrm{27}°\right) \\ $$$$\:\:=\:\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{18}°}\:−\:\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{54}°} \\ $$$$\:\:\:=\:\frac{\mathrm{4}\left(\mathrm{sin}\:\mathrm{54}°−\mathrm{sin}\:\mathrm{18}°\right)}{\mathrm{2sin}\:\mathrm{54}°\:\mathrm{sin}\:\mathrm{18}°} \\ $$$$\:\:=\:\frac{\mathrm{8cos}\:\mathrm{36}°\:\mathrm{sin}\:\mathrm{18}°}{\mathrm{2}\:\mathrm{cos}\:\mathrm{36}°\:\mathrm{sin}\:\mathrm{18}°} \\ $$$$\:\:=\:\begin{array}{|c|}{\mathrm{4}}\\\hline\end{array} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply