What is the value of $$\frac{{\left[ {1 - \tan \left( {90 - \theta } \right) + \sec \left( {90 - \theta } \right)} \right]}}{{\left[ {\tan \left( {90 - \theta } \right) - \sec \left( {90 - \theta } \right) + 1} \right]}}?$$

Solution (By Examveda Team)

\[\begin{array}{l} \frac{{\left[ {1 - \tan \left( {90 - \theta } \right) + \sec \left( {90 - \theta } \right)} \right]}}{{\left[ {\tan \left( {90 - \theta } \right) - \sec \left( {90 - \theta } \right) + 1} \right]}}\\ \Rightarrow \frac{{\left[ {1 - \cot \theta + {\rm{cosec}}\,\theta } \right]}}{{\left[ {\cot \theta + {\rm{cosec}}\,\theta + 1} \right]}}\\ \Rightarrow \frac{{\left[ {1 - \frac{{\cos \theta }}{{\sin \theta }} + \frac{1}{{\sin \theta }}} \right]}}{{\left[ {\frac{{\cos \theta }}{{\sin \theta }} + \frac{1}{{\sin \theta }} + 1} \right]}}\\ \Rightarrow \frac{{\left[ {\sin \theta - \cos \theta + 1} \right]}}{{\left[ {\sin \theta + \cos \theta + 1} \right]}}\\ \Rightarrow \frac{{\left( {\sin \theta + 1} \right) - \cos \theta }}{{\left( {\sin \theta + 1} \right) + \cos \theta }}\\ \left[ \begin{array}{l} \therefore \sin \theta = 2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}\\ \cos \theta = 1 - 2{\sin ^2}\frac{\theta }{2}\\ \cos \frac{\theta }{2} = 2{\cos ^2}\frac{\theta }{2} - 1 \end{array} \right]\\ \Rightarrow \frac{{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2} + 1 - 1 + 2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2} + 1 + 2{{\cos }^2}\frac{\theta }{2} - 1}}\\ \Rightarrow \frac{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} + \cos \frac{\theta }{2}} \right)}}{{2\cos \frac{\theta }{2}\left( {\sin \frac{\theta }{2} + \cos \frac{\theta }{2}} \right)}}\\ \Rightarrow \tan \frac{\theta }{2} \end{array}\]