Question

The value of
$\frac{\cos \theta }{\sin ({90}^0+\theta )}+\frac{\sin \theta }{\sin ({180}^0+\theta )}+\frac{\cot ({90}^0-\theta )}{\tan \theta }$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$\frac{cos\theta }{sin({90}^0+\theta )}+\frac{sin\theta }{sin({180}^0+\theta )}+\frac{cot({90}^0-\theta )}{tan\theta }$
$=\frac{\cos \theta }{\cos \theta }+\frac{\sin \theta }{-\sin \theta }+\frac{\tan \theta }{\tan \theta }=1-1+1=1$