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

The correct option is A

$A=B$

Explanation for the correct option:

Simplifying the given equations using trigonometric identities:

The given set $A=\left\{\theta :\tan \theta +sec\theta =\sqrt{2}sec\theta \right\}$

$$\begin{gathered} \because \tan \theta +\sec \theta =\sqrt{2}\sec \theta \\ \Rightarrow \frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }=\frac{\sqrt{2}}{\cos \theta }\left[\because sec\theta =\frac{1}{cos\theta }andtan\theta =\frac{sin\theta }{cos\theta }\right] \\ \Rightarrow \sin \theta +1=\sqrt{2} \\ \Rightarrow \sin \theta =\sqrt{2}–1 \end{gathered}$$

So the set can be written as

$A=\left\{\theta :\sin \theta =\sqrt{2}–1\right\}$

Also, the given set $B=\left\{\theta :sec\theta -\tan \theta =\sqrt{2}\tan \theta \right\}$

$$\begin{gathered} \because sec\theta -\tan \theta =\sqrt{2}\tan \theta \\ \Rightarrow \frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }=\sqrt{2}\frac{\sin \theta }{\cos \theta }\mathbf{}\left[\because \mathrm{sec\theta }=\frac{1}{\mathrm{cos\theta }}\mathrm{and}\mathrm{tan\theta }=\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}\right] \\ \Rightarrow 1-\sin \theta =\sqrt{2}\sin \theta \\ \Rightarrow \left(\sqrt{2}+1\right)\sin \theta =1 \\ \Rightarrow \sin \theta =\frac{1}{\left(\sqrt{2}+1\right)} \\ \mathrm{Multiplying}\mathrm{and}\mathrm{Dividng}\mathrm{by}\left(\sqrt{2}-1\right) \\ \Rightarrow \sin \theta =\frac{1}{\left(\sqrt{2}+1\right)}\left[\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}\right] \\ \Rightarrow \sin \theta =\frac{\left(\sqrt{2}-1\right)}{\left({\left(\sqrt{2}\right)}^2-1\right)} \\ \Rightarrow \sin \theta =\frac{\left(\sqrt{2}-1\right)}{\left(2-1\right)} \\ \Rightarrow \sin \theta =\sqrt{2}-1 \end{gathered}$$

So this set can be written as

$B=\left\{\theta :\sin \theta =\sqrt{2}–1\right\}$

Thus, $A=B$

Therefore, the correct answer is option (A).