If cosθ = 3/5, find the value of {sin​θ - 1/tanθ}/2tanθ

Question

If cosθ = \(\frac{3}5\), find the value of \(\cfrac{sinθ-{\frac{1}{tanθ}}}{2tanθ}\)

Answer 1

Given: cosθ = \(\frac{3}5\)

To find: the value of ⇒ \(\cfrac{sinθ-{\frac{1}{tanθ}}}{2tanθ}\)

Solution: We know:

cosθ = \(\frac{Base}{Hypotenuse}\)

By applying Pythagoras theorem, we have(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ BC² = AB² + AC² 

⇒ BC² = 3² + 4² 

⇒ BC² = 9 + 16 

⇒ BC² = 25 

⇒ BC = \(\sqrt{25}\)

⇒ BC = 5

Use:sinθ = \(\frac{perpendicular}{Hypotenuse}\) and tanθ = \(\frac{sinθ }{cosθ }\)

⇒ sinθ = \(\frac{4}5\)

Substitute the known values,