Prove the following statement:
\[\dfrac{1}{\sec A-\tan A}=\sec A+\tan A\]

Hint: First of all, consider the LHS of the given equation and convert it in terms of \[\sin \theta \] and \[\cos \theta \] by using \[\sec \theta =\dfrac{1}{\cos \theta }\] and \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Now, simplify the given expression and multiply it by \[\left( \dfrac{1+\sin A}{1+\sin A} \right)\] and use \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] to prove the desired result.
Complete step-by-step answer:
In this question, we have to prove that \[\dfrac{1}{\sec A-\tan A}=\sec A+\tan A\]. Let us consider the LHS of the equation given in the question.
\[E=\dfrac{1}{\sec A-\tan A}\]
We know that \[\sec \theta =\dfrac{1}{\cos \theta }\]. By using this in the above expression, we get,
\[E=\dfrac{1}{\dfrac{1}{\cos A}-\tan A}\]
We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. By using this in the above expression, we get,
\[E=\dfrac{1}{\dfrac{1}{\cos A}-\dfrac{\sin A}{\cos A}}\]
By taking cos A as LCM and simplifying the above expression, we get,
\[E=\dfrac{1}{\dfrac{1-\sin A}{\cos A}}\]
\[E=\dfrac{\cos A}{1-\sin A}\]
By multiplying \[\left( \dfrac{1+\sin A}{1+\sin A} \right)\] in the above expression, we get,
\[E=\left( \dfrac{\cos A}{1-\sin A} \right).\left( \dfrac{1+\sin A}{1+\sin A} \right)\]
We know that \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]. By using this in the above expression and simplifying it, we get,
\[E=\dfrac{\cos A+\cos A\sin A}{{{\left( 1 \right)}^{2}}-{{\left( \sin A \right)}^{2}}}\]
We know that \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] or \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]. By using this in the above expression, we get,
\[E=\dfrac{\cos A+\cos A\sin A}{{{\cos }^{2}}A}\]
By separating the different terms of the above equation, we get,
\[E=\dfrac{\cos A}{{{\cos }^{2}}A}+\dfrac{\cos A\sin A}{{{\cos }^{2}}A}\]
By canceling the like terms of the above equation, we get,
\[E=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]
We know that \[\dfrac{1}{\cos \theta }=\sec \theta \] and \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta \]. By using this in the above expression, we get,
\[E=\sec A+\tan A\]
E = RHS
So, we get, LHS = RHS
Hence proved.
So, we have proved that \[\dfrac{1}{\sec A-\tan A}=\sec A+\tan A\].

Note: We can also solve this question by considering the LHS of the given question.
\[E=\dfrac{1}{\sec A-\tan A}\]
By multiplying the above expression by \[\dfrac{\sec A+\tan A}{\sec A+\tan A}\], we get,
\[E=\dfrac{\sec A+\tan A}{\left( \sec A-\tan A \right)\left( \sec A+\tan A \right)}\]
\[E=\dfrac{\sec A+\tan A}{{{\sec }^{2}}A-{{\tan }^{2}}A}\]
We know that \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\]. So, we get,
\[E=\sec A+\tan A\]
E = RHS
So, LHS = RHS
Hence proved.