If $3\sin \alpha +5\cos \alpha =5$, then the value of $(3\cos \alpha -5\sin \alpha {)}^2$ is equal to

Let $3\sin \alpha +5\cos \alpha =5\text{— (Equation 1)}$
Let $3\cos \alpha -5\sin \alpha =k\text{— (Equation 2)}$

Square both equations and add them together:

$$(3\sin \alpha +5\cos \alpha {)}^2+(3\cos \alpha -5\sin \alpha {)}^2=5^2+k^2$$

Expand the left side:

$$(9{\sin }^2\alpha +25{\cos }^2\alpha +30\sin \alpha \cos \alpha )+(9{\cos }^2\alpha +25{\sin }^2\alpha -30\sin \alpha \cos \alpha )=25+k^2$$

The cross-product terms cancel out:

$$9({\sin }^2\alpha +{\cos }^2\alpha )+25({\sin }^2\alpha +{\cos }^2\alpha )=25+k^2$$

Using the fundamental identity ${\sin }^2\alpha +{\cos }^2\alpha =1$:

$$9(1)+25(1)=25+k^2$$

$$9+25=25+k^2$$

$$k^2=9$$

Since we need the value of $(3\cos \alpha -5\sin \alpha {)}^2$, which is exactly $k^2$:

$$\text{Value}=9$$