Prove that \(sin^2 \theta\) + \(cos^2 \theta\) = 1.

Solution :

In right angled triangle ABC,

\(sin \theta\) = \(BC\over AC\)  \(\implies\)   \(sin^2 \theta\) = \(BC^2\over AC^2\)

\(cos \theta\) = \(AB\over AC\)  \(\implies\)   \(cos^2 \theta\) = \(AB^2\over AC^2\)

On adding,

\(sin^2 \theta\) + \(cos^2 \theta\) = \(BC^2\over AC^2\) + \(AB^2\over AC^2\)

\(sin^2 \theta\) + \(cos^2 \theta\) = \(BC^2 + AB^2\over AC^2\) = \(AC^2\over AC^2\) = 1

[ By Pythagoras theorem,  \(AC^2\) = \(BC^2 + AB^2\) ]

\(\implies\) \(sin^2 \theta\) + \(cos^2 \theta\) = 1

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