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Question Number 120468 by help last updated on 31/Oct/20

Commented by bemath last updated on 01/Nov/20

$$\cot (\theta +30°)\cot (\theta -30°)=$$$$\frac{1}{\tan (\theta +30°)\tan (\theta -30°)}=$$$$\frac{1}{(\frac{\tan \theta +\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}\tan \theta }{3}}.\frac{\tan \theta -\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}\tan \theta }{3}})}=$$$$\frac{1-\frac{1}{3}\tan {}^2\theta }{\tan {}^2\theta -\frac{1}{3}}=\frac{3-\tan {}^2\theta }{3\tan {}^2\theta -1}:\left[\frac{\tan {}^2\theta }{\tan {}^2\theta }\right]$$$$=\frac{3\cot {}^2\theta -1}{3-\cot {}^2\theta }$$$$$$

Commented by bemath last updated on 01/Nov/20

$$\left(\mathrm{b}\right)-3\cot (\theta +30°)\cot (\theta -30°)=\mathrm{cosec}{}^2\theta$$$$\Rightarrow -3\left(\frac{3\cot {}^2\theta -1}{3-\cot {}^2\theta }\right)=1+\cot {}^2\theta$$$$\Rightarrow 3-9\cot {}^2\theta =(3-\cot {}^2\theta )(1+\cot {}^2\theta )$$$$\mathrm{let}\cot {}^2\theta =\mathrm{s}$$$$\Rightarrow 3-9\mathrm{s}=3+2\mathrm{s}-{\mathrm{s}}^2$$$$\Rightarrow {\mathrm{s}}^2-11\mathrm{s}=0\to \{\begin{array}{l}\mathrm{s}=0\left(\mathrm{rejected}\right)\\ \mathrm{s}=11\end{array}$$$$\Rightarrow \cot \theta =\pm \sqrt{11}$$$$\Rightarrow \tan \theta =\pm \frac{1}{\sqrt{11}}\Rightarrow \theta =\pm 16.77°+\mathrm{n}.180°$$

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