Question

The value of ${\sin }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}-{\tan }^2\frac{\pi }{4}+{\cot }^2\frac{\pi }{2}$ is equal to

• A. $-\frac{1}{2}$
• B. $1$
• C. $\frac{\sqrt{3}}{2}$
• D. $0$

Answer 1

The correct option is A $-\frac{1}{2}$

${\sin }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}-{\tan }^2\frac{\pi }{4}+{\cot }^2\frac{\pi }{2}$
$={\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2-(1{)}^2+0$
$=-\frac{1}{2}$