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Question Number 116359 by bemath last updated on 03/Oct/20

$$\mathrm{If}0<\theta <\frac{\pi }{4}\mathrm{such}\mathrm{that}\mathrm{cosec}\theta -\sec \theta =\frac{\sqrt{13}}{6}$$ $$\mathrm{then}\cot \theta -\tan \theta \mathrm{equals}\mathrm{to}\mathrm{\_}\mathrm{\_}$$

Answered by bobhans last updated on 03/Oct/20

$$\Rightarrow \mathrm{let}\sin \theta \cos \theta =\mathrm{r}.\mathrm{Then}{(\frac{1}{\sin \theta }-\frac{1}{\cos \theta })}^2=\frac{13}{36}$$ $$\Rightarrow \frac{{(\cos \theta -\sin \theta )}^2}{{\left(\sin \theta \cos \theta \right)}^2}=\frac{13}{36}$$ $$\Rightarrow \frac{1-2\mathrm{r}}{{\mathrm{r}}^2}=\frac{13}{36};{13\mathrm{r}}^2+72\mathrm{r}-36=0$$ $$\Rightarrow (13\mathrm{r}-6)(\mathrm{r}+6)=0\to \{\begin{array}{l}\mathrm{r}=-6\left(\mathrm{rejected}\right)\\ \mathrm{r}=\frac{6}{13}\end{array}$$ $$\mathrm{So}\mathrm{we}\mathrm{have}\sin \theta \cos \theta =\frac{6}{13}\mathrm{and}\sin 2\theta =\frac{12}{13}$$ $$\mathrm{so}\cot \theta -\tan \theta =\frac{2\cos 2\theta }{\sin 2\theta }=\frac{2\times \left(\frac{5}{13}\right)}{\left(\frac{12}{13}\right)}=\frac{5}{6}$$ $$$$ $$$$

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