Question

Prove that ${\sin }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}-{\tan }^2\frac{\pi }{4}=-\frac{1}{2}$.

Answer 1

Now,

Let $a={\sin }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}-{\tan }^2\frac{\pi }{4}$

We know that, $\sin \frac{\pi }{6}=\frac{1}{2}$ and $\cos \frac{\pi }{6}=\frac{1}{\sqrt{2}},\tan \frac{\pi }{4}=1$

$a=\frac{1}{4}+\frac{1}{4}-1$

$a=\frac{1}{2}-1$

$a=-\frac{1}{2}$.

Hence, proved