What is d/(dθ)(r(θ))+r(θ)tan(θ)=tan(θ) ?

The answer to d/(dθ)(r(θ))+r(θ)tan(θ)=tan(θ) is r=(tan(θ))/(1+tan(θ)θ);\quad 1+tan(x)x\ne 0

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d/(dθ)(r(θ))+r(θ)tan(θ)=tan(θ)

\frac{d}{d θ} (r (θ)) + r (θ) tan (θ) = tan (θ)

r = \frac{tan ( θ )}{1 + tan ( θ ) θ}; 1 + tan (x) x \neq = 0

Solution steps

\frac{d}{d θ} (r θ) + r θ tan (θ) = tan (θ)

\frac{d}{d θ} (r θ) = r

r + r θ tan (θ) = tan (θ)

r + r θ tan (θ) = tan (θ) : r = \frac{tan ( θ )}{1 + tan ( θ ) θ}; 1 + tan (x) x \neq = 0

r = \frac{tan ( θ )}{1 + tan ( θ ) θ}; 1 + tan (x) x \neq = 0

\frac{d y}{d x} = \frac{1 + x ^{2}}{1 - y ^{2}}, y = 0, x = 0

f^{^{'}} (x) + 2 x f (x) = 0, f (0) = 3

- 2 y \frac{d y}{d x} + x^{2} + 2 x^{4} = 0

\frac{d y}{d x} = \frac{- 0.8}{cos ( y )}, y (0) = \frac{π}{2}

4 y \frac{d y}{d x} = x + x y^{2}

What is d/(dθ)(r(θ))+r(θ)tan(θ)=tan(θ) ?
The answer to d/(dθ)(r(θ))+r(θ)tan(θ)=tan(θ) is r=(tan(θ))/(1+tan(θ)θ);\quad 1+tan(x)x\ne 0