QUESTION

If $a\cos \theta +b\sin \theta =c$ then prove that $a\sin \theta -b\cos \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}$.

Hint: For solving this problem, we will adopt the method of squaring the trigonometric equation. On squaring both sides will transform our result to obtain the required equation.

As per the given question, we have $a\cos \theta +b\sin \theta =c$. To obtain the final result $a\sin \theta -b\cos \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}$ we will adopt the squaring method.
Let, the given part be equation (1) such that $a\cos \theta +b\sin \theta =c\ldots (1)$
Squaring both the sides of equation (1), we get
\begin{align} & {{\left( a\cos \theta +b\sin \theta \right)}^{2}}={{c}^{2}} \\ & {{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +2ab\cos \theta \sin \theta ={{c}^{2}} \\ \end{align}
Now, expanding each term as, ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \text{ and }{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$
\begin{align} & {{a}^{2}}\left( 1-{{\sin }^{2}}\theta \right)+{{b}^{2}}\left( 1-{{\cos }^{2}}\theta \right)+2ab\cos \theta \sin \theta ={{c}^{2}} \\ & {{a}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}-{{b}^{2}}{{\cos }^{2}}\theta +2ab\cos \theta \sin \theta ={{c}^{2}} \\ \end{align}
Rearranging the terms so that trigonometric functions are on one side, we get
${{a}^{2}}+{{b}^{2}}-{{c}^{2}}={{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -2ab\cos \theta \sin \theta$
By using the identity: ${{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}$, we get
${{a}^{2}}+{{b}^{2}}-{{c}^{2}}={{\left( a\sin \theta -b\cos \theta \right)}^{2}}$
Taking the square root of both the sides, we get
\begin{align} & \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}=\sqrt{{{\left( a\sin \theta -b\cos \theta \right)}^{2}}} \\ & {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=\pm \left( a\sin \theta -b\cos \theta \right) \\ \end{align}
Therefore, the left hand side is equal to the right hand side. So, it is proven.
Hence, we get the same expression which is required in the problem statement.

Note: The key step for solving this problem is the technique of squaring both the sides. Some knowledge of basic identities of trigonometric functions is also required. Students must notice the rearrangement done in the question because without rearranging it will be hard to obtain the required expression.