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Question Number 177542 by peter frank last updated on 06/Oct/22

Answered by Frix last updated on 07/Oct/22

$$9°=x$$$$27°=3x$$$$\tan (90°-x)=\frac{1}{\tan x}$$$$81°=90°-9°$$$$63°=90°-27°$$$$\tan x=\frac{\sin x}{\cos x}$$$$\tan 3x=\frac{{\tan }^{3}x-3\tan x}{{3\tan }^{2}x-1}=\frac{({\sin }^{2}x-{3\cos }^{2}x)\sin x}{({3\sin }^{2}x-{\cos }^{2}x)\cos x}$$$$\tan x+\frac{1}{\tan x}-(\tan 3x+\frac{1}{\tan 3x})=$$$$=\frac{2({\sin }^{2}x-2\cos x\sin x-{\cos }^{2}x)({\sin }^{2}x+2\cos x\sin x-{\cos }^{2}x)}{\cos x\sin x({\sin }^{2}x-{3\cos }^{2}x)({3\sin }^{2}x-{\cos }^{2}x)}$$$${\sin }^{2}x=\frac{1-\cos 2x}{2}$$$${\cos }^{2}x=\frac{1+\cos 2x}{2}$$$$\cos x\sin x=\frac{\sin 2x}{2}$$$$\mathrm{now}\mathrm{we}\mathrm{have}$$$$\frac{4({\cos }^{2}2x-{\sin }^{2}2x)}{({4\cos }^{2}2x-1)\sin 2x}=\frac{4({2\cos }^{2}2x-1)}{({4\cos }^{2}2x-1)\sin 2x}=$$$$=\frac{4\cos 4x}{\sin 6x}$$$$\mathrm{but}x=9°$$$$\frac{4\cos 36°}{\sin 54°}$$$$\mathrm{but}\sin (90°-x)=\cos x$$$$\frac{4\cos 36°}{\cos 36°}=4$$

Commented by peter frank last updated on 07/Oct/22

$$\mathrm{thank}\mathrm{you}$$

Answered by Ar Brandon last updated on 07/Oct/22

$$T=\tan 9°-\tan 27°-\tan 63°+\tan 81°$$$$=(\tan 9°+\frac{1}{\tan 9°})-(\tan 27°+\frac{1}{\tan 27°})$$$$=\frac{{\tan }^{2}9°+1}{\tan 9°}-\frac{{\tan }^{2}27°+1}{\tan 27°}=\frac{{\sec }^{2}9°}{\tan 9°}-\frac{{\sec }^{2}27°}{\tan 27°}$$$$=\frac{1}{\sin 9°\cos 9°}-\frac{1}{\sin 27°\cos 27°}=\frac{2}{\sin 18°}-\frac{2}{\sin 54°}$$$$=\frac{2(\sin 54°-\sin 18°)}{\sin 18°\sin 54°}=\frac{4\sin 18°\cos 36°}{\sin 18°\sin 54°}$$$$=\frac{4\sin 18°\sin 54°}{\sin 18°\sin 54°}=4$$

Commented by peter frank last updated on 07/Oct/22

$$\mathrm{thank}\mathrm{you}$$

Answered by Frix last updated on 07/Oct/22

$$\mathrm{or}:$$$$\tan 9°+\tan 81°=\tan 9°+\frac{1}{\tan 9°}=$$$$[\frac{\mathrm{s}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{s}}=\frac{{\mathrm{s}}^2+{\mathrm{c}}^2}{\mathrm{cs}}=\frac{1}{\mathrm{cs}}]$$$$=\frac{1}{\cos 9°\sin 9°}$$$$\mathrm{similar}$$$$-(\tan 27°+\tan 63°)=-\frac{2}{\sin 54°}=-\frac{2}{\sin \left(6\times 9°\right)}=$$$$[\sin 6x=2\cos x\sin x(3-{16\cos }^{2}x{\sin }^{2}x)$$$$=\frac{1}{\cos 9°\sin 9°({16\cos }^{2}9°{\sin }^{2}9°-3)}$$$$\frac{1}{\cos 9°\sin 9°}+\frac{1}{\cos 9°\sin 9°({16\cos }^{2}9°{\sin }^{2}9°-3)}=$$$$=\frac{2(1-{8\cos }^{2}9°{\sin }^{2}9°)}{\cos 9°\sin 9°(3-{16\cos }^{2}9°{\sin }^{2}9°)}=$$$$=\frac{4\cos 36°}{\sin 54°}=4$$

Commented by peter frank last updated on 07/Oct/22

$$\mathrm{thank}\mathrm{you}$$

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