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Question Number 190527 by Rupesh123 last updated on 04/Apr/23

Commented by Frix last updated on 05/Apr/23

$$\pm \frac{5}{6}$$

Answered by cortano12 last updated on 05/Apr/23

$$\Rightarrow \cos \theta -\sin \theta =\frac{\sqrt{13}}{12}\sin 2\theta$$$$\mathrm{let}\sin 2\theta =\mathrm{y}$$$$\Rightarrow 1-\mathrm{y}=\frac{13}{144}{\mathrm{y}}^2$$$$\Rightarrow {13\mathrm{y}}^2+144\mathrm{y}-144=0$$$$\Rightarrow \mathrm{y}=\frac{12}{13}\Rightarrow \sin 2\theta =\frac{12}{13}$$$$\Rightarrow \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{12}{5}$$$$\Rightarrow 5+5\tan \theta =12-12\tan \theta$$$$\Rightarrow 12-12\tan {}^2\theta =10\tan \theta$$$$\Rightarrow 6\tan {}^2\theta +5\tan \theta -6=0$$$$\Rightarrow \{\begin{array}{l}\tan \theta =\frac{2}{3}\\ \tan \theta =-\frac{3}{2}\end{array}$$

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