Prove that type question of Trigonometric Identities 
If $3\sin A + 5\cos A = 5$, then prove that:
     $$5\sin A + 3\cos A = ±3.$$

 A: The question as stated seems to be incorrect. Most likely it is asking for $$\color{blue}{5 \sin A} \color{red}{-} \color{blue}{3 \cos A}.$$
If such is the case then from the first equation if we divide throughout by $\sqrt{34}$, we get
$$\frac{3}{\sqrt{34}} \sin A + \frac{5}{\sqrt{34}} \cos A=\frac{5}{\sqrt{34}}.$$
Let $\sin \alpha=\frac{3}{\sqrt{34}}$, then we get
$$\cos(A-\alpha)=\frac{5}{\sqrt{34}}.$$
But then we also get
$$\cos(A-\alpha)=\frac{5}{\sqrt{34}}=\cos \alpha.$$
This implies that
$$A-\alpha = 2n \pi \pm \alpha, \qquad \text{ for } n \in \mathbb{Z}.$$
Thus 
$$A=2n \pi \qquad \text{ or } \qquad A=2n\pi+2\alpha.$$
Let $\color{blue}{5 \sin A} \color{red}{-} \color{blue}{3 \cos A}=x$. 


*

*If $A=2n \pi$, then $x=-3$.

*If $A=2n\pi+2\alpha$, then $5 \sin A \color{red}{-} 3 \cos A=5 \sin 2\alpha \color{red}{-} 3 \cos 2\alpha$. 


Using the values of $\sin \alpha$ and $\cos \alpha$, we get $x=3$. 
Thus $x = \pm 3$.
NOTE: If the actual problem (with $+$ sign) stands as is then $x=-3$ or $x=\frac{99}{17}$. 
A: HINT. Try squaring the first identity and then use $\sin^2A+\cos^2A=1$
