Question

Let P=\left\\{ \theta :\sin \theta -\cos \theta =\sqrt{2}\cos \theta \right\\} and Q=\left\\{ \sin \theta +\cos \theta =\sqrt{2}\sin \theta \right\\} be two sets. Then:
(a) $P\subset Q$ and $Q-P\ne 0$
(b) $Q\subset P$
(c) $P\subset Q$
(d) $P=Q$

Answer 1

Hint: Solve the equations given inside the brackets for defining the sets in a simple version of equations. Use the relation $\dfrac{\sin \theta }{\cos \theta }=\tan \theta$ to simplify the equations.
Apply rationalization wherever required. Rationalization means converting the irrational number to a rational by multiplying and dividing by a number.

Complete step-by-step answer:
Here, we have two sets P and Q are defined as
P=\left\\{ \theta :\sin \theta -\cos \theta =\sqrt{2}\cos \theta \right\\} ….......................................(i)
Q=\left\\{ \sin \theta +\cos \theta =\sqrt{2}\sin \theta \right\\} …..............................................(ii)
Now, we can observe that sets P and Q are defined in the equations written inside the sets to find the elements of set P and set Q.
So, equation given in set P as
$\sin \theta -\cos \theta =\sqrt{2}\cos \theta$
Transfer $\cos \theta$ to other side, we get
$\sin \theta =\cos \theta +\sqrt{2}\cos \theta$
$\Rightarrow \sin \theta =\cos \theta \left( \sqrt{2}+1 \right)$
On dividing the whole equation by $\cos \theta$ , we get
$\dfrac{\sin \theta }{\cos \theta }=\dfrac{\cos \theta }{\cos \theta }\left( \sqrt{2}+1 \right)$
$\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\sqrt{2}+1$
Now, we can replace $\dfrac{\sin \theta }{\cos \theta }$ by $\tan \theta$ using the identity given as
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ …………………………………(iii)
So, we get
$\tan \theta =\sqrt{2}+1$ ………………………………(iv)
Now, we can solve the expression given in the set Q, i.e. equation (ii); so, we have
$\sin \theta +\cos \theta =\sqrt{2}\sin \theta$
Transfer $\sin \theta$ to other side, we get
$\cos \theta =\sqrt{2}\sin \theta -\sin \theta$
$\Rightarrow \cos \theta =\sin \theta \left( \sqrt{2}-1 \right)$
Now, divide the whole equation by $\cos \theta$ , we get
$\dfrac{\cos \theta }{\cos \theta }=\dfrac{\sin \theta }{\cos \theta }\left( \sqrt{2}-1 \right)$
$\Rightarrow \dfrac{\sin \theta }{\cos \theta }\left( \sqrt{2}-1 \right)=1$
Now, we can replace $\dfrac{\sin \theta }{\cos \theta }$ by $\tan \theta$ by the identity expressed in equation (iii); so, we get
$\tan \theta \left( \sqrt{2}-1 \right)=1$
$\tan \theta =\dfrac{1}{\sqrt{2}-1}$
Now, we can rationalized $\dfrac{1}{\sqrt{2}-1}$ by multiplying in the denominator and numerator by the conjugate of $\sqrt{2}-1$ , i.e. $\sqrt{2}+1$ which removes the irrational form in denominator.
So, we get
$\tan \theta =\dfrac{1}{\left( \sqrt{2}-1 \right)}\times \left( \dfrac{\sqrt{2}+1}{\sqrt{2}+1} \right)$
Now, we can use the algebraic identity of ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ to simplify the denominator. So, we get
$\tan \theta =\dfrac{\sqrt{2}+1}{{{\left( \sqrt{2} \right)}^{2}}-{{\left( 1 \right)}^{2}}}$
$\tan \theta =\dfrac{\sqrt{2}+1}{2-1}$
$\tan \theta =\sqrt{2}+1$ ……………………………………(v)
Now, we can observe that the equations of set P and Q are the same after simplification. So,we can represent the sets P and Q as
P=\left\\{ \theta :\tan \theta =\sqrt{2}+1 \right\\}
Q=\left\\{ \theta :\tan \theta =\sqrt{2}+1 \right\\}
Hence, sets P and Q should be equal. So, option ‘D’ $\left( P=Q \right)$ is the correct answer.

Note: One may go wrong if he/she will not rationalize $\dfrac{1}{\sqrt{2}-1}$ which is value of $\tan \theta$ for set Q, he/she may answer $P\ne Q$ , because $\tan \theta =\left( \sqrt{2}+1 \right)$ and $\tan \theta =\dfrac{1}{\sqrt{2}-1}$ are not looking same at very first time. But if we rationalize the term $\dfrac{1}{\sqrt{2}-1}$ , we get $P=Q$ . So, be careful with this step. Most of the students may give answers as $P\ne Q$ . So, take care with this step of the solution.
Another approach for the question would be that we can square the equations of Set P and Q both. So, we get
$P\to {{\left( \sin \theta -\cos \theta \right)}^{2}}={{\left( \sqrt{2}\cos \theta \right)}^{2}}$
$Q\to {{\left( \sin \theta +\cos \theta \right)}^{2}}={{\left( \sqrt{2}\sin \theta \right)}^{2}}$
Simplify both the equations using the trigonometric identities:-
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,\cos 2\theta =2{{\cos }^{2}}\theta -1=1-2{{\sin }^{2}}\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta$
We don’t need to find the exact values of $\theta$ , we can compare the sets P and Q only by simplifying them. So, don’t go for calculating exact values of $\theta$ from the equations $\tan \theta =\sqrt{2}+1$ and $\tan \theta =\dfrac{1}{\sqrt{2}-1}$. Hence, be careful with it