Question

The value of $\cos ({270}^o+\theta )\cos ({90}^o-\theta )-\sin ({270}^o-\theta )\cos \theta$ is

• A. $0$
• B. $-1$
• C. $1/2$
• D. $1$

Answer 1

The correct option is D $1$

$cos({270}^{\circ }+\theta )cos(90-\theta )-sin(270-\theta )cos\theta$

$sin\theta .sin\theta -(-cos\theta )cos\theta$

$si{n}^2\theta +co{s}^2\theta$

= $1$

[$\because$ By using trignometry table]