Evaluate: $sin^{2}19^{o}+sin^{2}71^{o}$.

Given: $sin^{2}19^{o}+sin^{2}71^{o}$.

To do: To evaluate the given expression.

Solution:

Given expression:

 $sin^{2}19^{o}+sin^{2}71^{o}$

$=sin^{2}19^{o}+sin^{2}( 90^{o}-19^{o})$

$=sin^{2}19^{o}+cos^{2}19^{o}$

$=1$

Thus, $sin^{2}19^{o}+sin^{2}71^{o}=1$.

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Updated on: 10-Oct-2022

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