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Q. The value of $\sin ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}-\tan ^2 \frac{\pi}{4}$ is equal to ...$A$.... Here, $A$ refers to

Trigonometric Functions

Solution:

$\sin ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}-\tan ^2 \frac{\pi}{4}$
$=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-(1)^2=\frac{1}{4}+\frac{1}{4}-1$
$=\frac{1}{2}-1=-\frac{1}{2}$