MCQ
$\frac{{{{\sin }^2}A - {{\sin }^2}B}}{{\sin A\cos A - \sin B\cos B}} = $
- A$\tan (A - B)$
- ✓$\tan (A + B)$
- C$\cot (A - B)$
- D$\cot (A + B)$
$= \frac{{2\,\sin \,(A + B)\,\sin \,(A - B)}}{{\sin \,2A - \sin \,2B}}$
$ = \frac{{2\,\sin \,(A + B)\,\sin \,(A - B)}}{{2\,\cos \,(A + B)\,\sin \,(A - B)}} $
$= \tan \,(A + B)$.
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$(A)$ $\arg (-1- i )=\frac{\pi}{4}$, where $i =\sqrt{-1}$
$(B)$ The function $f: R \rightarrow(-\pi, \pi]$, defined by $f(t)=\arg (-1+i t)$ for all $t \in R$, is continuous at all points of $R$, where $i=\sqrt{-1}$
$(C)$ For any two non-zero complex numbers $z_1$ and $z_2$, $\arg \left(\left(\frac{z_1}{z_2}\right)-\arg \left(z_1\right)+\arg \left(z_2\right)\right.$ is an integer multiple of $2 \pi$.
$(D)$ For any three given distinct complex numbers, $z_1, z_2$ and $z_3$, the locus of the point $z$ satisfying the condition $\arg \left(\frac{\left( z - z _1\right)\left( z _2- z _3\right)}{\left( z - z _3\right)\left( z _2- z _1\right)}\right)=\pi$, lies on a straight line