Question

If $\cos \theta -\sin \theta =\frac{1}{5}$, where $0<\theta <\frac{\pi }{2}$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& \frac{(\cos \theta +\sin \theta )}{2}& \left(p\right)& \frac{4}{5}\\ \text{(2)}& \sin 2\theta & \left(q\right)& \frac{7}{10}\\ \text{(3)}& \cos 2\theta & \left(r\right)& \frac{24}{25}\\ \text{(4)}& \cos \theta & \left(s\right)& \frac{7}{25}\end{array}$

Which of the following is the correct combination?

• A. $1\to s,2\to p,3\to r,4\to q$
• B. $1\to p,2\to s,3\to r,4\to p$
• C. $1\to q,2\to s,3\to r,4\to p$
• D. $1\to q,2\to r,3\to s,4\to p$

Answer 1

The correct option is D $1\to q,2\to r,3\to s,4\to p$

$\cos \theta -\sin \theta =\frac{1}{5}\cdots \left(1\right)$
Squaring equation $\left(1\right)$, we get
$1-\sin 2\theta =\frac{1}{25}$
$\Rightarrow \sin 2\theta =\frac{24}{25}$
$2\to r$

Now,
$(\cos \theta +\sin \theta {)}^2=(\cos \theta -\sin \theta {)}^2+4\sin \theta \cos \theta$
$=\frac{1}{25}+2\sin 2\theta$
$=\frac{1}{25}+\frac{48}{25}$
$=\frac{49}{25}$
$\Rightarrow (\cos \theta +\sin \theta {)}^2=\frac{49}{25}$
$\Rightarrow (\cos \theta +\sin \theta )=\pm \frac{7}{5}$
Since $\theta \in (0,\frac{\pi }{2})$
$\Rightarrow \cos \theta +\sin \theta =\frac{7}{5}\cdots \left(2\right)$
$\Rightarrow \frac{(\cos \theta +\sin \theta )}{2}=\frac{7}{10}$
$1\to q$

Adding equation $\left(1\right)$ and $\left(2\right)$, we get
$2\cos \theta =\frac{8}{5}\Rightarrow \cos \theta =\frac{4}{5}$
$4\to p$

Now,
$\cos 2\theta =2{\cos }^2\theta -1\Rightarrow \cos 2\theta =2\times \frac{16}{25}-1\Rightarrow \cos 2\theta =\frac{7}{25}$
$3\to s$

Hence,
$1\to q,2\to r,3\to s,4\to p$