If $\sin \left({\theta }_1\right)+\sin \left({\theta }_2\right)+\sin \left({\theta }_3\right)=3$,then $\cos \left({\theta }_1\right)+\cos \left({\theta }_2\right)+\cos \left({\theta }_3\right)=$

The correct option is D

$0$

Explanation for the correct option

Step 1: Given expression

$\sin \left({\theta }_1\right)+\sin \left({\theta }_2\right)+\sin \left({\theta }_3\right)=3$

Step:2 Simplification and calculate the value of $\cos \left({\theta }_1\right)+\cos \left({\theta }_2\right)+\cos \left({\theta }_3\right)$

We know that , the maximum value of $\sin \left(\theta \right)$ is $\left(1\right)$

$\therefore$$\begin{array}{rcl}& \Rightarrow & \sin \left(\theta \right)=1\\ & \Rightarrow & \sin \left(\theta \right)=\sin \left({90}^0\right)\\ & \Rightarrow & \left(\theta \right)={90}^0\\ & & \end{array}$

Then , by the value of $\left(\theta \right)$ from the above statement

$\left({\theta }_1\right)=\left({\theta }_2\right)=\left({\theta }_3\right)={90}^0.....\left(1\right)$

Step 3: Put the value of equation $\left(1\right)$ in $\cos \left({\theta }_1\right)+\cos \left({\theta }_2\right)+\cos \left({\theta }_3\right)$

$\begin{array}{rcl}& \Rightarrow & \cos \left({90}^0\right)+\cos \left({90}^0\right)+\cos \left({90}^0\right)\\ & \Rightarrow & 0+0+0\left(\because valueof\cos \left({90}^0\right)=0\right)\\ & \Rightarrow & 0\end{array}$

Hence, option $\left(\mathit{D}\right)$ is correct.