Question

If cos θ=0.6 show that (5sin θ -3tan θ) = 0

Answer 1

Let us consider a right ΔABC right angled at B.
Now, we know that cos θ = 0.6 = `(BC)/(AC) = 3/5`

So, if BC = 3k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:

`Ac^2 = AB^2 + BC^2`
`⟹ AB^2 = AC^2 − BC^2`
`⟹ AB^2 = (5K)^2 − (3K)^2 = 25K^2 − 9K^2`
`⟹ AB^2 = 16K^2`
⟹ 𝐴𝐵 = 4𝑘

Finding out the other T-rations using their definitions, we get:

sin θ = `(AB)/(AC) = (4K)/(5K) = 4/5`

TAN θ = `(AB)/(BC)  = (4K)/(3K) = 4/3`

Substituting the values in the given expression, we get:
5 sin 𝜃 − 3 tan 𝜃

`⇒ 5 (4/5) - 3(4/3)`

= 4-4=0= RHS

i.e., LHS = RHS
Hence, Proved.