समाधान {l}{cosθ_{1}+cosθ_{2}+cosθ_{3}=3text{,thenwhatis}sinθ_{1}+}{+sinθ_{3}text{equalto?}}{{llll}{text{(A)}0}&{text{(B)}1}&{text{(C)}2}&{text{(D)}3}}

$$\left. \begin{array} { l } { \cos \theta _ { 1 } + \cos \theta _ { 2 } + \cos \theta _ { 3 } = 3 \text { , then what is } \sin \theta _ { 1 } + } \\ { + \sin \theta _ { 3 } \text { equal to? } } \\ { \left. \begin{array} { l l l l } { \text { (A) } 0 } & { \text { (B) } 1 } & { \text { (C) } 2 } & { \text { (D) } 3 } \end{array} \right. } \end{array} \right.$$

θ_1, θ_2, θ_3 को लागि हल गर्नुहोस्

$$\theta _{1}=ArcCosI(\left(-1\right)CosI(\theta _{2})+\left(-1\right)CosI(\theta _{3})+3)+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }\theta _{2}\in \begin{bmatrix}\left(-\frac{1}{2}\right)\pi +2n_{45}\pi +arcSin(\left(-1\right)CosI(\theta _{3})+2),\left(-1\right)arcSin(\left(-1\right)CosI(\theta _{3})+2)+2n_{45}\pi +\frac{1}{2}\pi \end{bmatrix}\text{, }n_{45}\in \mathrm{Z}\text{, }\theta _{3}=2n_{42}\pi \text{, }n_{42}\in \mathrm{Z}\text{; }\theta _{1}=\left(-1\right)ArcCosI(\left(-1\right)CosI(\theta _{2})+\left(-1\right)CosI(\theta _{3})+3)+2n_{2}\pi \text{, }n_{2}\in \mathrm{Z}\text{, }\theta _{2}\in \begin{bmatrix}\left(-\frac{1}{2}\right)\pi +2n_{45}\pi +arcSin(\left(-1\right)CosI(\theta _{3})+2),\left(-1\right)arcSin(\left(-1\right)CosI(\theta _{3})+2)+2n_{45}\pi +\frac{1}{2}\pi \end{bmatrix}\text{, }n_{45}\in \mathrm{Z}\text{, }\theta _{3}=2n_{42}\pi \text{, }n_{42}\in \mathrm{Z}$$