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If $Δ A B C$ is a equilateral triangle then the values of $\sin^{2} A + \cos^{2} B + \tan^{2} C$ is

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answer is D.

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Detailed Solution

$\sin^{2} A + \cos^{2} B + \tan^{2} C$

$= \sin^{2} 60 ° + \cos^{2} 60 ° + \tan^{2} 60 °$

$= {(\frac{\sqrt{3}}{2})}^{2} + {(\frac{1}{2})}^{2} + {(\sqrt{3})}^{2}$

$= \frac{3}{4} + \frac{1}{4} + 3$

$= 1 + 3 = 4$

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