Question

$[\sin \alpha +\sin ({180}^{\circ }-\alpha )+\sin ({180}^{\circ }+\alpha )]cosec\alpha =$ ______

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

We know that $\sin (n\pi +x)=\sin x$ if n is an even number and $\sin (n\pi +x)=-\sin x$ if $n$ is an odd number

So, $[\sin \alpha +\sin ({180}^0-\alpha )+\sin ({180}^0+\alpha \left)\right]cosec\alpha$
$=[\sin \alpha -(-\sin \alpha )-\sin \alpha ]cosec\alpha =\sin \alpha \times cosec\alpha =1$