If tan θ = \frac{5}{7}, find the value of \frac{5 sin θ + 7 cos θ}{7 sin θ + 5 cos θ}
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
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