Question

If $P=\theta :sin\theta –cos\theta =\sqrt{2}cos\theta \mathrm{and}Q=\theta :sin\theta +cos\theta =\sqrt{2}sin\theta$are two sets. Then,

• A. $P\subset Q\mathrm{and}\mathrm{Q}-\mathrm{P}\ne \mathrm{\varphi }$
• B. $Q\not\subset P$
• C. $P\not\subset Q$
• D. $P=Q$

Answer 1

The correct option is D

$P=Q$

Step 1: Given sets

$P=\theta :sin\theta –cos\theta =\sqrt{2}cos\theta ......\left(1\right)$

$Q=\theta :sin\theta +cos\theta =\sqrt{2}sin\theta ......\left(2\right)$

Step 2: Formation to find the correct answer

From equation (1)

$\because P=\theta :sin\theta –cos\theta =\sqrt{2}cos\theta$

Take only one side to solve the expression

$\therefore sin\theta –cos\theta =\sqrt{2}cos\theta$

In the above equation take the cos term at LHS

$\begin{array}{rcl}& \Rightarrow & sin\theta =\sqrt{2}cos\theta +cos\theta \\ & \Rightarrow & sin\theta =\left(\sqrt{2}+1\right)cos\theta \\ & \Rightarrow & \frac{\sin \theta }{\cos \theta }=\left(\sqrt{2}+1\right)\end{array}$

$\begin{array}{rcl}& \Rightarrow & \tan \theta =\left(\sqrt{2}+1\right).....\left(3\right)\end{array}$

From equation (2)

$\because Q=\theta :sin\theta +cos\theta =\sqrt{2}sin\theta$

Take only one side to solve the expression

$\therefore sin\theta +cos\theta =\sqrt{2}sin\theta$

In the above equation take the sin term at LHS

$\begin{array}{rcl}& \Rightarrow & sin\theta +cos\theta =\sqrt{2}sin\theta \\ & \Rightarrow & \sqrt{2}sin\theta -\sin \theta =cos\theta \\ & \Rightarrow & \left(\sqrt{2}-1\right)\sin \theta =\cos \theta \\ & \Rightarrow & \frac{\sin \theta }{\cos \theta }=\frac{1}{\left(\sqrt{2}-1\right)}\\ & \Rightarrow & \tan \theta =\frac{1}{\left(\sqrt{2}-1\right)}\end{array}$

Multiply by $\sqrt{2}+1$in numerator and denominator

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & \tan \theta =\frac{1}{\left(\sqrt{2}-1\right)}\times \frac{\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\end{array} \\ \begin{array}{rcl}& \Rightarrow & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \tan \end{array}\begin{array}{rcl}& & \theta \end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \left(\sqrt{2}+1\right)\end{array}.......\left(4\right) \end{gathered}$$

Step 3: From equations (3) and (4)

therefore $P=Q$

Hence option (D) is the correct answer.