If 3sin θ + 4cos θ = 5 ; then find value of tan θ

3sin θ + 4cos θ = 5

⇒ \(\frac{3}{5}\)sin θ + \(\frac{4}{5}\)cos θ = 1 .....(i)

*(sin²θ + cos²θ = 1)*

Therefore, from (i)

sin θ = \(\frac{3}{5}\) &

cos θ = \(\frac{4}{5}\)

⇒  tan θ = \(\frac{sin θ}{cos θ}\) = \(\frac{\frac{3}{5}}{\frac{4}{5}}\) = \(\frac{3}{4}\)