Question

How do you verify $\left( \sec +\tan \right)\left( \sec -\tan \right)=1$ ?

Answer 1

Hint: For answering this question we need to verify the given expression. For verification we will substitute a value and check whether the right hand side and left hand side are equal or not. Firstly we will conveniently write the given equation as $\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1$

Complete step by step answer:
For verifying the given expression we can substitute some angle and verify if the right hand side and left hand side of the given expression is correct or not.
The given expression is $\left( \sec +\tan \right)\left( \sec -\tan \right)=1$ which can be conveniently written as $\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1$ .
Let us assume $\theta $ as $\dfrac{\pi }{4}$ by substituting this value in the right hand side of the expression we will have $\left( \sec \dfrac{\pi }{4}+\tan \dfrac{\pi }{4} \right)\left( \sec \dfrac{\pi }{4}-\tan \dfrac{\pi }{4} \right)$ .
From the basic concepts of trigonometric ratios which give us the values for some specific trigonometric ratios of some specific angles we are having $\sec \dfrac{\pi }{4}=\sqrt{2}$ and $\tan \dfrac{\pi }{4}=1$ .
By substituting these values in the right hand side of the expression we will have $\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)$ .
By further simplifying this using the simple arithmetic concept $\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$ we will have $\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)={{\left( \sqrt{2} \right)}^{2}}-{{1}^{2}}$ .
By further simplifying this by performing simple arithmetic calculations we will have $2-1=1$ which is equal to the left hand side of the given expression hence we can say that the given expression is valid.

Note: Sometimes for some substitutions the expressions may be valid but not for all so to avoid this type of mistakes it would be better if we try to prove this expression as a theorem. As we know $\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$ we can say that the given expression can be written as
$\begin{align}
  & \left( \sec +\tan \right)\left( \sec -\tan \right)=1 \\
 & \Rightarrow {{\sec }^{2}}-{{\tan }^{2}}=1 \\
\end{align}$
As we know that $\sec =\dfrac{1}{\cos }$ and $\tan =\dfrac{\sin }{\cos }$ . We also know that ${{\sin }^{2}}+{{\cos }^{2}}=1\Rightarrow 1-{{\sin }^{2}}={{\cos }^{2}}$. From this we can prove the given expression is valid.