Correct Answer - Option 1 : sec θ
Concept:
\(tan(A+B) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\)
Application:
Given:
\(u = \log \left[ {\tan \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right)} \right]\)
\(u= \log \left[ {\frac{{\tan \frac{\pi }{4} + \tan \frac{\theta }{2}}}{{1 - \tan \frac{x}{4} \cdot \frac{\theta }{2}}}} \right]\)
\(u = \log \left[ {\frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}}} \right]\)
\({e^u} = \frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}}\) ---(1)
\({e^{ - u}} = \frac{{1 - \tan \frac{\theta }{2}}}{{1 + \tan \frac{\theta }{2}}}\) ---(2)
Add equation (1) and (2)
\({e^u} + {e^{ - u}} = \frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}} + \frac{{1 - \tan \frac{\theta }{2}}}{{1 + \tan \frac{\theta }{2}}}\)
\({e^u} + {e^{ - u}} = \frac{{{{\left( {1 + \tan \frac{\theta }{2}} \right)}^2} + {{\left( {1 - \tan \frac{\theta }{2}} \right)}^2}}}{{1 - {{\tan }^2}\theta }}\)
\({e^u} + {e^{ - u}} = \frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}} + \frac{{1 - \tan \frac{\theta }{2}}}{{1 + \tan \frac{\theta }{2}}}\)
\({e^u} + {e^{ - u}} = \frac{{{{\left( {1 + \tan \frac{\theta }{2}} \right)}^2} + {{\left( {1 - \tan \frac{\theta }{2}} \right)}^2}}}{{1 - {{\tan }^2}\theta }}\)
\(= \frac{{1 + {{\tan }^2}\frac{\theta }{2} + 2\tan \frac{\theta }{2} + 1 + {{\tan }^2}\frac{\theta }{2} - 2\tan \frac{\theta }{2}}}{{1 - {{\tan }^2}\theta }}\)
\(\cosh u = \frac{{{{\sec }^2}\frac{\theta }{2}}}{{1 - \frac{{{{\sin }^2}\frac{\theta }{2}}}{{{{\cos }^2}\frac{\theta }{2}}}}} \)
\(= \frac{{{{\sec }^2}\frac{\theta }{2}}}{{{{\cos }^2}\frac{\theta }{2} - {{\sin }^2}\frac{\theta }{2}}} \times {\cos ^2}\frac{\theta }{2}\)
\(\cosh u = \frac{1}{{\cos \theta }} = \sec \theta\)
cosh u = sec θ
