Prove that 1 + \(tan^2 \theta\) = \(sec^2 \theta\).

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 + \(tan^2 \theta\) = \(sec^2 \theta\)

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