Given that (sin‌θ−csc‌θ)2+(cos‌θ+secθ)2=5 and θ is in the third quadrant, we need to find (sin‌θ+cos‌θ)3. First, simplify the given expression: (sin‌θ−csc‌θ)2+(cos‌θ+secθ)2=5 By expanding, we have: sin‌2θ+csc2θ−2+cos2θ+sec2θ+2=5 This simplifies to: 1+1+cot2θ+1+tan2θ=5 Thus: tan2θ+cot2θ=2 This implies: (tan‌θ−cot‌θ)2=0 So, tan‌θ=cot‌θ, leading to tan2θ=1. Therefore, tan‌θ=±1. Since θ lies in the third quadrant, where both sine and cosine are negative, we have tan‌θ=1. Thus, θ=‌