Question

Let \[A=\left\{ \theta :\tan \theta +\sec \theta =\sqrt{2}\sec \theta \right\}\] and \[B=\left\{ \theta :\sec \theta -\tan \theta =\sqrt{2}\tan \theta \right\}\] be 2 sets. Then
A. $A=B$
B. $A\subset B$
C. $A\ne B$
D. $B\subset A$

Answer 1

Hint: We first simplify the given sets. We change the trigonometric ratios in one form. We use the rationalization process to get the same surds form. As the conditions for both sets are equal, we can treat them as the equal sets.

Complete step by step answer:
We need to simplify the equations \[\tan \theta +\sec \theta =\sqrt{2}\sec \theta \] and \[\sec \theta -\tan \theta =\sqrt{2}\tan \theta \].
We take \[\tan \theta +\sec \theta =\sqrt{2}\sec \theta \] and sec ratio in one side.
\[\begin{align}
  & \tan \theta +\sec \theta =\sqrt{2}\sec \theta \\
 & \Rightarrow \tan \theta =\left( \sqrt{2}-1 \right)\sec \theta \\
 & \Rightarrow \dfrac{\tan \theta }{\sec \theta }=\sin \theta =\left( \sqrt{2}-1 \right) \\
\end{align}\]
Now we take \[\sec \theta -\tan \theta =\sqrt{2}\tan \theta \] and tan ratio in one side.
\[\begin{align}
  & \sec \theta -\tan \theta =\sqrt{2}\tan \theta \\
 & \Rightarrow \sec \theta =\left( \sqrt{2}+1 \right)\tan \theta \\
 & \Rightarrow \dfrac{\tan \theta }{\sec \theta }=\sin \theta =\dfrac{1}{\left( \sqrt{2}+1 \right)} \\
\end{align}\]
We now need to rationalize the given form of \[\dfrac{1}{\left( \sqrt{2}+1 \right)}\]. The given surds expression is \[\dfrac{1}{\left( \sqrt{2}+1 \right)}\].
We have to apply the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We multiply \[\left( \sqrt{2}-1 \right)\] to both denominator and the numerator. This is the conjugate form of \[\left( \sqrt{2}+1 \right)\].
Now the dfraction becomes \[\dfrac{1}{\left( \sqrt{2}+1 \right)}=\dfrac{\left( \sqrt{2}-1 \right)}{\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)}\].
We assume the values $a=\sqrt{2};b=1$ for ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Therefore, $\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)={{\left( \sqrt{2} \right)}^{2}}-{{1}^{2}}$.
Simplifying we get $\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)={{\left( \sqrt{2} \right)}^{2}}-{{1}^{2}}=2-1=1$.
The dfraction becomes \[\dfrac{1}{\left( \sqrt{2}+1 \right)}=\left( \sqrt{2}-1 \right)\].
The given sets \[A=\left\{ \theta :\tan \theta +\sec \theta =\sqrt{2}\sec \theta \right\}\] and \[B=\left\{ \theta :\sec \theta -\tan \theta =\sqrt{2}\tan \theta \right\}\] can also be expressed as \[A=\left\{ \theta :\sin \theta =\left( \sqrt{2}-1 \right) \right\}\] and \[B=\left\{ \theta :\sin \theta =\dfrac{1}{\left( \sqrt{2}+1 \right)}=\left( \sqrt{2}-1 \right) \right\}\] respectively.
Both of them are equal. Therefore, $A=B$. The correct option is option (A).

Note:
Instead of multiplying \[\left( \sqrt{2}-1 \right)\], we can also form 1 in numerator as $1=2-1$. We try to form the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ where we take \[1=2-1={{\left( \sqrt{2} \right)}^{2}}-{{1}^{2}}\].
Now we break it to get ${{\left( \sqrt{2} \right)}^{2}}-{{1}^{2}}=\left( \sqrt{2}-1 \right)\left( \sqrt{2}+1 \right)$.
Then we can eliminate the part of \[\left( \sqrt{2}-1 \right)\] to simplify.