Solution :
In right angled triangle ABC,
\(sec \theta\) = \(AC\over AB\) \(\implies\) \(sec^2 \theta\) = \(AC^2\over AB^2\)
\(tan \theta\) = \(BC\over AB\) \(\implies\) \(tan^2 \theta\) = \(BC^2\over AB^2\)
\(\implies\) 1 + \(tan^2 \theta\) = 1 + \(BC^2\over AB^2\) = \(AB^2 + BC^2\over AB^2\) = \(AC^2\over AB^2\)
[ By Pythagoras theorem, \(AC^2\) = \(BC^2 + AB^2\) ]
\(\implies\) 1 + \(tan^2 \theta\) = \(sec^2 \theta\)
