Any help with the following question would be greatly appreciated.

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

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

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

(b) Determine the range of values of \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 \theta < 45 or the string would be vertical.


Attempt:

Resolving Parallel to the plane:

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

Resolving Perpendicular to the plane:
W cos\theta - R = T sin\theta

Squaring and adding gives

T^2cos^2\theta + T^2sin^2\theta = (\mu R + Wsin\theta)^2 + (Wcos\theta - R)^2
T^2 = \mu^2R^2 + 2\mu RWsin\theta + W^2sin^2\theta + W^2cos^2\theta - 2RWcos\theta + R^2
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