Question

Let \[P = \{ \theta :\sin \theta - \cos \theta = \sqrt 2 \cos \theta \} \] and \[Q = \{ \theta :\sin \theta + \cos \theta = \sqrt 2 \sin \theta \} \] be two sets then
A. \[P \subset Q\]and \[Q - P \ne \phi \]
B.\[Q \not\subset P\]
C. \[P \not\subset Q\]
D. \[P = Q\]

Answer 1

Hint: We solve for the values of \[\theta \] from both sets. Shift all similar functions on one side and calculate the value of tangent of angle in both sets. Rationalize the term formed in the second set. Check if the values of P and Q are equal, unequal, subsets or not.
* \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]

Complete step by step answer:
We are given two sets \[P = \{ \theta :\sin \theta - \cos \theta = \sqrt 2 \cos \theta \} \]and \[Q = \{ \theta :\sin \theta + \cos \theta = \sqrt 2 \sin \theta \} \]
We first solve for set P
\[P = \{ \theta :\sin \theta - \cos \theta = \sqrt 2 \cos \theta \} \]
\[ \Rightarrow \sin \theta - \cos \theta = \sqrt 2 \cos \theta \]
Shift all cosine values to RHS
\[ \Rightarrow \sin \theta = \sqrt 2 \cos \theta + \cos \theta \]
\[ \Rightarrow \sin \theta = (\sqrt 2 + 1)\cos \theta \]
Divide both sides by cosine of angle
\[ \Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{(\sqrt 2 + 1)\cos \theta }}{{\cos \theta }}\]
Cancel same terms from numerator and denominator
\[ \Rightarrow \tan \theta = (\sqrt 2 + 1)\]
Take inverse tangent on both sides of the equation
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}(\sqrt 2 + 1)\]
Cancel inverse function by function
\[ \Rightarrow \theta = {\tan ^{ - 1}}(\sqrt 2 + 1)\] … (1)
Now we solve for set Q
\[Q = \{ \theta :\sin \theta + \cos \theta = \sqrt 2 \sin \theta \} \]
\[ \Rightarrow \sin \theta + \cos \theta = \sqrt 2 \sin \theta \]
Shift all sine values to RHS
\[ \Rightarrow \cos \theta = \sqrt 2 \sin \theta - \sin \theta \]
\[ \Rightarrow \cos \theta = (\sqrt 2 - 1)\sin \theta \]
Divide both sides by sine of angle
\[ \Rightarrow \dfrac{{\cos \theta }}{{\sin \theta }} = \dfrac{{(\sqrt 2 - 1)\sin \theta }}{{\sin \theta }}\]
Take reciprocal on both sides
\[ \Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\sin \theta }}{{(\sqrt 2 - 1)\sin \theta }}\]
Cancel same terms from numerator and denominator
\[ \Rightarrow \tan \theta = \dfrac{1}{{(\sqrt 2 - 1)}}\]
Rationalize RHS
\[ \Rightarrow \tan \theta = \dfrac{1}{{\sqrt 2 - 1}} \times \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}}\]
Use the formula \[(a - b)(a + b) = {a^2} - {b^2}\]
\[ \Rightarrow \tan \theta = \dfrac{{\sqrt 2 + 1}}{{{{\left( {\sqrt 2 } \right)}^2} - {{\left( 1 \right)}^2}}}\]
\[ \Rightarrow \tan \theta = \dfrac{{\sqrt 2 + 1}}{{2 - 1}}\]
\[ \Rightarrow \tan \theta = \dfrac{{\sqrt 2 + 1}}{1}\]
\[ \Rightarrow \tan \theta = \sqrt 2 + 1\]
Take inverse tangent on both sides of the equation
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}(\sqrt 2 + 1)\]
Cancel inverse function by function
\[ \Rightarrow \theta = {\tan ^{ - 1}}(\sqrt 2 + 1)\] … (2)
From both equations (1) and (2) we have \[\theta = {\tan ^{ - 1}}(\sqrt 2 + 1)\]
\[ \Rightarrow P = Q\]

\[\therefore \]Option D is correct.

Note: Many students make the mistake of leaving the value of tangent of angle in fraction form which is wrong, keep in mind we always have to have value in denominator not as an irrational number i.e. of kind under root, exponential form etc. Here we have terms under root in the denominator so we have to rationalize it in order to form an answer and then compare the values. Also, when taking inverse functions on both sides, we take the same functions inverse in order to cancel out a function and obtain the value of angle.