Trigonometric Ratios and Identities

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=‌

5π

4

.
Now, find (sin‌θ+cos‌θ)3 :

sin‌θ=−‌

1

√2

,‌‌cos‌θ=−‌

1

√2

Therefore:
(sin‌θ+cos‌θ)=(−‌

1

√2

−‌

1

√2

)=−√2
Thus:
(sin‌θ+cos‌θ)3=(−√2)3=−2√2