Q60 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
If $1-\cot 23^{\circ}=\frac{x}{1-\cot 22^{\circ}}$, then $x=$
- A$1$
- Correct answerB$2$
- C$\frac{1}{2}$
- D$3$
Solution
Given, $1-\cot 23^{\circ}=\frac{x}{1-\cot 22^{\circ}}$
$\begin{aligned} & \Rightarrow x=\left(1-\cot 23^{\circ}\right)\left(1-\cot 22^{\circ}\right) \\ & \Rightarrow x=\frac{\left(\sin 23^{\circ}-\cos 23^{\circ}\right)}{\sin 23^{\circ}} \times \frac{\left(\sin 22^{\circ}-\cos 22^{\circ}\right)}{\sin 22^{\circ}}\end{aligned}$
$\left(\sin 23^{\circ} \cdot \sin 22^{\circ}+\cos 23^{\circ} \cdot \cos 22^{\circ}\right)$
$\begin{aligned} & \Rightarrow x=\frac{-\left(\sin 23^{\circ} \cos 22^{\circ}+\cos 23^{\circ} \sin 22^{\circ}\right)}{\sin 23^{\circ} \cdot \sin 22^{\circ}} \\ & \Rightarrow x=\frac{\cos \left(23^{\circ}-22^{\circ}\right)-\sin \left(23^{\circ}+22^{\circ}\right)}{\sin 23^{\circ} \cdot \sin 22^{\circ}}\end{aligned}$
$\Rightarrow x=\frac{2\left(\cos 1^{\circ}-\sin 45^{\circ}\right)}{2 \sin 23^{\circ} \sin 22^{\circ}}=\frac{2\left(\cos 1^{\circ}-1 / \sqrt{2}\right)}{\left(\cos 1^{\circ}-\cos 45^{\circ}\right)}$
$=\frac{2\left(\cos 1^{\circ}-1 / \sqrt{2}\right)}{\left(\cos 1^{\circ}-1 / \sqrt{2}\right)}=2$
Hence, the correct answer is option: 'B'.