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

Answered by deleteduser1 last updated on 10/Feb/24

$$tan\left(9\right)+tan\left(81\right)-(tan27+tan63)$$$$=\frac{sin9}{cos9}+\frac{sin81}{cos81}-(\frac{sin27}{cos27}+\frac{sin63}{cos63})$$$$=\frac{2sin(9+81)=2}{2cos9cos81}-\frac{2sin(27+63)=2}{2cos\left(27\right)cos\left(63\right)}$$$$=\frac{2}{2sin\left(9\right)cos\left(9\right)}-\frac{2}{2sin\left(27\right)cos\left(27\right)}=\frac{2}{sin\left(18\right)}-\frac{2}{sin\left(54\right)}$$$$Let\theta =18°\Rightarrow sin\left(5\theta \right)=16{sin}^5\left(\theta \right)-20{sin}^3\left(\theta \right)+5sin\left(\theta \right)$$$$\Rightarrow sin\left(18\right)=\frac{\sqrt{5}-1}{4};sin\left(54\right)=sin\left(3\theta \right)=\frac{\sqrt{5}+1}{4}$$$$\Rightarrow ?=\frac{8(\sqrt{5}+1)}{4}-\frac{8(\sqrt{5}-1)}{4}=4$$

Answered by universe last updated on 10/Feb/24

$$tan\left(9\right)+tan\left(81\right)-(tan27+tan63)$$$$=\frac{sin9}{cos9}+\frac{sin81}{cos81}-(\frac{sin27}{cos27}+\frac{sin63}{cos63})$$$$=\frac{2sin(9+81)=2}{2cos9cos81}-\frac{2sin(27+63)}{2cos\left(27\right)cos\left(63\right)}$$$$=\frac{2}{2sin\left(9\right)cos\left(9\right)}-\frac{2}{2sin\left(27\right)cos\left(27\right)}=\frac{2}{sin\left(18\right)}-\frac{2}{sin\left(54\right)}$$$$2\left[\frac{\sin 54-\sin 18}{\sin 54\sin 18}\right]=4\frac{\cos 36\sin 18}{\sin 54\sin 18}=4$$

Answered by cortano12 last updated on 10/Feb/24

$$\Rightarrow \tan 9°+\cot 9°-(\tan 27°+\cot 27°)$$$$=\frac{2}{\sin 18°}-\frac{2}{\sin 54°}$$$$=\frac{4(\sin 54°-\sin 18°)}{2\sin 54°\sin 18°}$$$$=\frac{8\cos 36°\sin 18°}{2\cos 36°\sin 18°}$$$$=\overline{)\begin{array}{c}4\end{array}}$$

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