We are given:
[
\sin A = \frac{4}{5}, \quad \tan C = 1
]
Step 1: Find (\cos A)
Since (\sin A = \frac{4}{5}), we use:
[
\cos A = \frac{3}{5}
]
(assuming (A) is acute, as is standard in triangle problems)
Step 2: Find (\sin C) and (\cos C)
[
\tan C = 1 \Rightarrow C = 45^\circ
]
So:
[
\sin C = \cos C = \frac{\sqrt{2}}{2}
]
Step 3: Use angle sum
[
B = 180^\circ - (A + C)
]
So:
[
\cos B = -\cos(A + C)
]
Step 4: Compute (\cos(A + C))
[
\cos(A + C) = \cos A \cos C - \sin A \sin C
]
[
= \frac{3}{5}\cdot \frac{\sqrt{2}}{2} - \frac{4}{5}\cdot \frac{\sqrt{2}}{2}
= \frac{\sqrt{2}}{10}(3 - 4)
= -\frac{\sqrt{2}}{10}
]
Step 5: Final answer
[
\cos B = -(-\frac{\sqrt{2}}{10}) = \frac{\sqrt{2}}{10}
]
✅ Final Answer:
[
\boxed{\frac{\sqrt{2}}{10}}
