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