Any help with the following question would be greatly appreciated.

Q: Alan pulls a container with weight of magnitudeWnewtons at a constant speed up a rough plane, with coefficient of friction $\displaystyle \mu$, inclined at an acute angle $\displaystyle \theta$ to the horizontal by means of a light inextensible rope. The rope also makes an angle of $\displaystyle \theta$ to the inclined plane.

(a) Show that the magnitude of the tension of the rope is given by

$\displaystyle \frac{tan\theta + \mu}{1 + \mu tan\theta} W$ newtons.

(b) Determine the range of values of $\displaystyle \theta$ for which the tension in the rope is less than the weight of the container.

(Although it is not stated in the question it is clear that $\displaystyle \theta < 45$ or the string would be vertical.

Attempt:

Resolving Parallel to the plane:

$\displaystyle \mu R + W sin\theta = T cos\theta$

Resolving Perpendicular to the plane:

$\displaystyle W cos\theta - R = T sin\theta$

Squaring and adding gives

$\displaystyle T^2cos^2\theta + T^2sin^2\theta = (\mu R + Wsin\theta)^2 + (Wcos\theta - R)^2$

$\displaystyle T^2 = \mu^2R^2 + 2\mu RWsin\theta + W^2sin^2\theta + W^2cos^2\theta - 2RWcos\theta + R^2$

$\displaystyle T^2 = \mu^2 R^2 + W^2 + 2\mu RWsin\theta - 2RWcos\theta$

And I'm a bit stuck now - I seem to be a long way from the target answer