tan θ = \frac{5}{7}

⇒ perpendicular (p) = 5k and base (b) = 7k

Applying Pythagoras theorem:

hypotenuse (h) = {\sqrt{(5k)^2 + (7k)^2}}

= {\sqrt{25k^2 + 49k^2}}

= √74 k

∴ sin θ = \frac{5k}{√74 k} = \frac{5}{√74 }

and cos θ = \frac{7k}{√74 k} = \frac{7}{√74 }

Now, compute the numerator:

5 sin θ + 7 cos θ = 5 × \frac{5}{\sqrt{74}} + 7 × \frac{7}{\sqrt{74}}

= \frac{25}{\sqrt{74}} + \frac{49}{\sqrt{74}} = \frac{74}{\sqrt{74}} = \sqrt{74}

Similarly, compute the denominator:

7 sin θ + 5 cos θ = 7 × \frac{5}{\sqrt{74}} + 5 × \frac{7}{\sqrt{74}}

= \frac{35}{\sqrt{74}} + \frac{35}{\sqrt{74}} = \frac{70}{\sqrt{74}}

Dividing:

\frac{5 sin θ + 7 cos θ}{7 sin θ + 5 cos θ} = \frac{\sqrt{74}}{\sqrt{74}} = 1