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Question Number 194767 by horsebrand11 last updated on 15/Jul/23

$$\tan \theta =2$$$$\frac{8\sin \theta +5\cos \theta }{\sin {}^3\theta +\cos {}^3\theta +\cos \theta }=?$$

Answered by dimentri last updated on 15/Jul/23

$$\sec {}^{2}x=5,\tan {}^{2}x=4$$$$\equiv \frac{8\tan x\sec {}^{2}x+5\sec {}^{2}x}{\tan {}^{3}x+1+{\sec }^{2}x}$$$$=\frac{105}{14}=\frac{15}{2}$$

Answered by Frix last updated on 15/Jul/23

$$\sin \theta =\frac{\tan \theta }{\sqrt{1+{\tan }^2\theta }}\wedge \cos \theta =\frac{1}{\sqrt{1+{\tan }^2\theta }}$$$$\sin \theta =\frac{2}{\sqrt{5}}\wedge \cos \theta =\frac{1}{\sqrt{5}}$$$$\frac{\frac{16}{\sqrt{5}}+\frac{5}{\sqrt{5}}}{\frac{8}{5\sqrt{5}}+\frac{1}{5\sqrt{5}}+\frac{1}{\sqrt{5}}}=\frac{21}{\frac{9}{5}+1}=\frac{105}{14}=\frac{15}{2}$$

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