If α,−π2<α<π2 is the solution of 4cosθ+5sinθ=1, then the value of tanα is

The given equation can be rewritten in the form of sin(p + α) = sinβ, where p is the phase angle, α is the angle in question, and β is another angle.

First, rewrite the equation as follows:

4cosθ + 5sinθ = 1

This can be written in the form of Rcos(θ - α) = 1, where R is the resultant vector and α is the phase angle.

Rcos(θ - α) = 1 can be rewritten as Rcosθcosα + Rsinθsinα = 1

Comparing this with the original equation, we get:

Rcosα = 4 and Rsinα = 5

Squaring and adding these two equations, we get:

R^2 = (4^2 + 5^2) = 41

So, R = sqrt(41)

Now, divide Rsinα by Rcosα to get tanα:

tanα = Rsinα / Rcosα = 5 / 4

So, the value of tanα is 5/4.