Basic trigonometry formula and double angle formula

If $\displaystyle 0<\theta <\frac{\pi }{4}$ is such that $\displaystyle \operatorname{cosec}\theta -\sec \theta =\frac{{\sqrt{{13}}}}{6}$ , then find $\displaystyle \left( {\cot \theta -\tan \theta } \right)$ .

Solution

$\displaystyle \operatorname{cosec}\theta -\sec \theta =\frac{{\sqrt{{13}}}}{6}\ ,\ \ \ 0<\theta <\frac{\pi }{4}$

$\displaystyle \ \ \frac{1}{{\sin \theta }}-\frac{1}{{\cos \theta }}=\frac{{\sqrt{{13}}}}{6}$

$\displaystyle \ \ \ \frac{{\cos \theta -\sin \theta }}{{\sin \theta \cos \theta }}=\frac{{\sqrt{{13}}}}{6}$

$\displaystyle \frac{{{{{\cos }}^{2}}\theta -2\sin \theta \cos \theta +{{{\sin }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta {{{\cos }}^{2}}\theta }}=\frac{{13}}{{36}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{1-\sin 2\theta }}{{\frac{1}{4}{{{\sin }}^{2}}2\theta }}=\frac{{13}}{{36}}$ 

$\displaystyle \ 13{{\sin }^{2}}2\theta +144\sin 2\theta -144=0$ 

$\displaystyle \ (13\sin 2\theta -12)(\sin 2\theta +12)=0$

$\displaystyle 13\sin 2\theta -12=0\ \ (or)\ \ \sin 2\theta +12=0$

$\displaystyle \sin 2\theta =\frac{{12}}{{13}}\ \ \ (or)\ \ \sin 2\theta =-12\ (reject)$

$\displaystyle \therefore \ \sin 2\theta =\frac{{12}}{{13}}\ $

$\displaystyle \cot \theta -\tan \theta =\frac{{\cos \theta }}{{\sin \theta }}-\frac{{\sin \theta }}{{\cos \theta }}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}{{\sin \theta \cos \theta }}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\cos 2\theta }}{{\sin 2\theta }}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\cot 2\theta$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\times \frac{5}{{12}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{5}{6}\ \ \ $