Question

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

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $-\frac{3}{2}$

Answer 1

Answer: (B)

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