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