Question

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

• A. $1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

The correct option is A $1$

The value of $\frac{\sin \theta \cos ({90}^o-\theta )\cos \theta }{\sin ({90}^o-\theta )}+\frac{\cos \theta \sin ({90}^o-\theta )\sin \theta }{\cos ({90}^o-\theta )}$ is
$=\frac{\sin \theta \sin \theta \cos \theta }{\cos \theta }+\frac{\cos \theta \cos \theta \sin \theta }{\sin \theta }$
$={\sin }^2\theta +{\cos }^2\theta =1$