Prove that 1 + \(cot^2 \theta\) = \(cosec^2 \theta\).

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 + \(cot^2 \theta\) = \(cosec^2 \theta\)

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