MCQ
If $\tan A - \tan B = x$ and $\cot B - \cot A = y,$ then $\cot (A - B) = $
- A$\frac{1}{x} + y$
- B$\frac{1}{{xy}}$
- C$\frac{1}{x} - \frac{1}{y}$
- ✓$\frac{1}{x} + \frac{1}{y}$
$= \frac{{1 + \tan A\,\,\tan B}}{{\tan A - \tan B}}$
$ = \frac{1}{{\tan A - \tan B}} + \frac{{\tan A\,\,\tan B}}{{\tan A - \tan B}} $
$= \frac{1}{x} + \frac{1}{y}$.
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$1.$ The equation of circle $\mathrm{C}$ is
$(A)$ $(x-2 \sqrt{3})^2+(y-1)^2=1$
$(B)$ $(x-2 \sqrt{3})^2+\left(y+\frac{1}{2}\right)^2=1$
$(C)$ $(x-\sqrt{3})^2+(y+1)^2=1$
$(D)$ $(x-\sqrt{3})^2+(y-1)^2=1$
$2.$ Points $E$ and $F$ are given by
$(A)$ $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$
$(B)$ $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right),(\sqrt{3}, 0)$
$(C)$ $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$(D)$ $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$3.$ Equation of the sides $Q R, R P$ are
$(A)$ $y=\frac{2}{\sqrt{3}} x+1, y=-\frac{2}{\sqrt{3}} x-1$
$(B)$ $y=\frac{1}{\sqrt{3}} x, y=0$
$(C)$ $y=\frac{\sqrt{3}}{2} x+1, y=-\frac{\sqrt{3}}{2} x-1$
$(D)$ $y=\sqrt{3} x, y=0$
Give the answer question $1,2$ and $3.$