The sides of a right-angled triangle, right-angled at B, are 6, 8 and 10 units. C is the vertex opposite to the side with length 8 units. What is the value of tan
2
A+cos
2
C?

Given a right-angled triangle △ABC right-angled at B. The sides are 6, 8, and 10 units. Since 10 is the largest side, the hypotenuse AC=10.
It is specified that C is the vertex opposite to the side with length 8 units. Therefore, AB=8 units.
This leaves the remaining side BC=6 units.

For angle A:

Opposite side=BC=6

Adjacent side=AB=8

tanA=
Adjacent
Opposite
​
=
AB
BC
​
=
8
6
​
=
4
3
​

For angle C:

Adjacent side=BC=6

Hypotenuse=AC=10

cosC=
Hypotenuse
Adjacent
​
=
AC
BC
​
=
10
6
​
=
5
3
​

Now, compute tan
2
A+cos
2
C:

tan
2
A+cos
2
C=(
4
3
​
)
2
+(
5
3
​
)
2

tan
2
A+cos
2
C=
16
9
​
+
25
9
​

Taking LCM of 16 and 25, which is 400:

tan
2
A+cos
2
C=
400
9×25+9×16
​

tan
2
A+cos
2
C=
400
225+144
​
=
400
369
​