Using basic trigonometry show that:
\[\dfrac{\sec A-\tan A}{\sec A+\tan A}=1-2\sec A\tan A+2{{\tan }^{2}}A\]

Hint: Since, the denominator value is 1 in the RHS try to multiply the numerator and denominator with a term so that denominator becomes unity and then use the trigonometric identity \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\] to simplify it further.

Complete step by step solution:
Now, from the given equation we get,
\[\Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}\]
Now, on multiplying both the numerator and denominator with:
\[\sec A-\tan A\]
\[\begin{align}
  & \Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}\times dfrac{\sec A-\tan A}{\sec A-\tan A} \\
 & \Rightarrow \dfrac{{{\left( \sec A-\tan A \right)}^{2}}}{\left( \sec A-\tan A \right)\left( \sec A+\tan A \right)} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( \sec A-\tan A \right)}^{2}}}{{{\sec }^{2}}A-{{\tan }^{2}}A} \\
 & \Rightarrow {{\left( \sec A-\tan A \right)}^{2}}\text{ }\left[ \because {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \right] \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {{\sec }^{2}}A+{{\tan }^{2}}A-2\sec A\tan A \\
 & \Rightarrow {{\tan }^{2}}A+1+{{\tan }^{2}}A-2\sec A\tan A\text{ }\left[ \because {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \right] \\
\end{align}\]
\[\Rightarrow 1-2\sec A\tan A+2{{\tan }^{2}}A\]
\[\therefore \dfrac{\sec A-\tan A}{\sec A+\tan A}=1-2\sec A\tan A+2{{\tan }^{2}}A\]

Note: It is important to note that multiplying the numerator and denominator with and term such that we can simplify it using the trigonometric identities. We multiplied the numerator and denominator with \[\sec A-\tan A\] here because we then get the result required accordingly.
We can also multiply the numerator and denominator with \[\sec A+\tan A\]. But, then we get 1 in the numerator and the denominator becomes a square term which we again need to simplify by multiplying with suitable terms in order to get the given equation. So, it is better to multiply the other term which gives the result quickly.