If$\tan \theta =\dfrac{a}{b}$, show that: $\dfrac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)}$\[\]

Hint: We will try to convert sine and cosine of angle $\theta $ present in the numerator at the left hand side of the equation in terms of tangent of the angle $\theta $ using the formula ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta $ so that we can use the give value $\tan \theta =\dfrac{a}{b}$. We replace $\sin \theta ,\cos \theta $ with the obtained expressions in $a,b$ at the left hand side and simplify till we arrive at the right hand side. \[\]

Complete step-by-step answer:
Now, from the given question we have
\[\tan \theta =\dfrac{a}{b}\ \ \ \ \ ...(a)\]
Now, by using the trigonometric identity which gives the relation between the function that are mentioned in the hint, we get the following
\[\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }\]
Now, this can also be written as the following using the other relations given in the hint as follows
\[\begin{align}
  & \Rightarrow {{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \\
 & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}-1={{\tan }^{2}}\theta \\
 & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}={{\tan }^{2}}\theta +1 \\
\end{align}\]
Let us now substitute the value from the question and as well as from equation (a) in this
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}={{\left( \dfrac{a}{b} \right)}^{2}}+1 \\
 & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}=\left( \dfrac{{{a}^{2}}}{{{b}^{2}}} \right)+1 \\
 & \Rightarrow {{\left( \dfrac{1}{\cos \theta } \right)}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}} \\
\end{align}\]
Now, this can be further written as
\[\begin{align}
  & \Rightarrow \left( \dfrac{1}{\cos \theta } \right)=\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}}} \\
 & \Rightarrow \cos \theta =\sqrt{\dfrac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}} \\
\end{align}\]
Now, using the relation between the sin and cos function, we have
\[\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
Now, this can be used to get the expression which can be written as
\[\begin{align}
  & \Rightarrow {{\left( \sqrt{\dfrac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}} \right)}^{2}}+{{\sin }^{2}}\theta =1 \\
 & \Rightarrow \dfrac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}+{{\sin }^{2}}\theta =1 \\
 & \Rightarrow {{\sin }^{2}}\theta =1-\dfrac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \\
 & \Rightarrow {{\sin }^{2}}\theta =\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}} \\
 & \Rightarrow \sin \theta =\sqrt{\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}} \\
\end{align}\]
Now, from the given expression in the question, on substituting the values, we have
\[\dfrac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\]
Let us first consider the left hand side and calculate its value
\[\begin{align}
  & L.H.S=\dfrac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta } \\
 & L.H.S=\dfrac{a\times \sqrt{\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}}-b\times \sqrt{\dfrac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}}{a\times \sqrt{\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}}+b\times \sqrt{\dfrac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}} \\
 & L.H.S=\dfrac{{{a}^{2}}\sqrt{\dfrac{1}{a+b}}-{{b}^{2}}\sqrt{\dfrac{1}{a+b}}}{{{a}^{2}}\sqrt{\dfrac{1}{a+b}}+{{b}^{2}}\sqrt{\dfrac{1}{a+b}}} \\
 & L.H.S=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \\
\end{align}\]
Thus, the value of right hand side is equal to left hand side
Hence, it is verified that,
\[\dfrac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\]

Note: We note that the angles of the variables at the denominator at any expression cannot be zero, so we have $b\ne 0$. We also note that for $\tan \theta $ to exist $\theta $ cannot be $\dfrac{{\left( {2n + 1} \right)}}{2}\pi $ where $n$ is any integer. We can alternatively solve by dividing $\cos \theta $ at the denominator and numerator at the left hand side expression and then putting $\tan \theta =\dfrac{a}{b}$.