Trigonometry and Quadratic equation

 If $ \displaystyle \sec \theta +\cos ec\ \theta =2$ , find exact value of $ \displaystyle \sin \theta $.

  Solution

$ \displaystyle \sec \theta +\cos ec\ \theta =2$

$ \displaystyle \frac{1}{{\cos \theta }}\ +\ \frac{1}{{\sin \theta }}\ \ =\ 2$

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

$ \displaystyle \sin \theta +\cos \theta =\sin 2\theta \ \ ......(1)$

$ \displaystyle {{(\sin \theta +\cos \theta )}^{2}}={{\sin }^{2}}2\theta $

$ \displaystyle {{\sin }^{2}}\theta +2\sin \theta \cos \theta +{{\cos }^{2}}\theta ={{\sin }^{2}}2\theta $

$ \displaystyle 1+\sin 2\theta ={{\sin }^{2}}2\theta $

$ \displaystyle {{\sin }^{2}}2\theta -\sin 2\theta -1=0$

$ \displaystyle \sin 2\theta =\frac{{1-\sqrt{5}}}{2}\ \ [\because -1\le {{\sin }^{2}}2\theta \le 1]$

$ \displaystyle 2\sin \theta \cos \theta =\frac{{1-\sqrt{5}}}{2}$

$ \displaystyle \cos \theta =\frac{{1-\sqrt{5}}}{{4\sin \theta }}$

$ \displaystyle \sin \theta +\frac{{1-\sqrt{5}}}{{4\sin \theta }}=\frac{{1-\sqrt{5}}}{2}\ \ \ \ \ [\ By\ (1)\ ]$

$ \displaystyle 4{{\sin }^{2}}\theta +1-\sqrt{5}=(2-2\sqrt{5})\sin \theta $

$ \displaystyle 4{{\sin }^{2}}\theta -(2-2\sqrt{5})\sin \theta +1-\sqrt{5}=0$

$ \displaystyle \sin \theta =\frac{{2-2\sqrt{5}\pm \sqrt{{{{{(2-2\sqrt{5})}}^{2}}-16(1-\sqrt{5})}}}}{8}$

$ \displaystyle \ \ \ \ \ \ \ =\frac{{2-2\sqrt{5}\pm \sqrt{{8+8\sqrt{5}}}}}{8}$

$ \displaystyle \ \ \ \ \ \ \ =\frac{{2-2\sqrt{5}\pm 2\sqrt{{2+2\sqrt{5}}}}}{8}$

$ \displaystyle \ \ \ \ \ \ \ =\frac{{1-\sqrt{5}\pm \sqrt{{2+2\sqrt{5}}}}}{4}$