Q62 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics
If $|\sin \alpha-\cos \alpha|=\frac{3}{4}$, then $|\sec 2 \alpha-\tan 2 \alpha|=$
- A$\frac{12}{17}$
- B$\frac{4}{\sqrt{23}}$
- Correct answerC$\frac{3}{\sqrt{23}}$
- D$\frac{7}{\sqrt{23}}$
Solution
Given $|\sin \alpha-\cos \alpha|=\frac{3}{4}$
$$
\sin \alpha-\cos \alpha=\frac{3}{4}
$$
Take square both sides,
$$
\sin ^2 \alpha+\cos ^2 \alpha-\sin 2 \alpha=\frac{9}{16}
$$
$1-\frac{9}{16}=\sin 2 \alpha$
$$
\sin 2 \alpha=\frac{7}{16}=\frac{P}{H}
$$
Here, $\mathrm{P}=7, \mathrm{H}=16$, then, $\mathrm{B}=3 \sqrt{23}$
$$
\cos 2 \alpha=\frac{B}{H}=\frac{3 \sqrt{23}}{16}
$$
$$
\begin{aligned}
& \tan 2 \alpha=\frac{\sin 2 \alpha}{\cos 2 \alpha}=\frac{\frac{7}{16}}{\frac{3 \sqrt{23}}{16}}=\frac{7}{3 \sqrt{23}} \\
& \sec 2 \alpha=\frac{1}{\cos 2 \alpha}=\frac{16}{3 \sqrt{23}}
\end{aligned}
$$
Now, $|\sec 2 \alpha-\tan 2 \alpha|=\left|\frac{16}{3 \sqrt{23}}-\frac{7}{3 \sqrt{23}}\right|$
$$
=\frac{9}{3 \sqrt{23}}=\frac{3}{\sqrt{23}}
$$
Therefore, option (c) is correct.
Hence, the correct answer is option: 'C'.