Question

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

Answer 1

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.