Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ  = 5$$

Find 

$$4\cosθ -3\sinθ$$

Help me guys (PS:I'm not that good in maths)
HINT:

using Brahmagupta–Fibonacci identity

$$(4\sin\theta+3\cos\theta)^2+(4\cos\theta-3\sin\theta)^2=(4^2+3^2)(\sin^2\theta+\cos^2\theta)$$
A very fast solution
Maximum value of $\displaystyle4\sin\theta+3\cos\theta$ is $\displaystyle{5}$. This value is achieved when $\displaystyle\tan\theta=\dfrac{4}{3}$ by differentiating.So $\cos\theta=\dfrac{3}{5}$ and $\sin\theta=\dfrac{4}{5}$.
Hence $4\cos\theta-3\sin\theta=0$.