MCQ
જો $x = \sec \,\phi - \tan \phi,y = {\rm{cosec}}\phi+ \cot \phi,$ તો
- A$x = \frac{{y + 1}}{{y - 1}}$
- ✓$x = \frac{{y - 1}}{{y + 1}}$
- C$y = \frac{{1 - x}}{{1 + x}}$
- Dએકપણ નહી
$ = \frac{{1 - \sin \,\phi}}{{\cos \,\phi}}\,.\,\frac{{1 + \cos \,\phi}}{{\sin \,\phi}}$
$ \Rightarrow \,xy + 1 = \frac{{1 - \sin \,\phi+ \cos \,\phi- \sin \,\phi\,\cos \,\phi+ \sin \phi \cos \phi}}{{\cos \phi\sin \phi}}$
$= \frac{{1 - \sin \,\phi+ \cos \,\phi}}{{\cos \,\phi\sin \,\phi}}$…..$(i)$
$x - y = (\sec \,\phi- \tan \,\phi) - (\cos ec\,\phi+ \cot \,\phi)$ $ = \frac{{1 - \sin \,\phi}}{{\cos \,\phi}} - \frac{{1 + \cos \,\phi}}{{\sin \,\phi}}$
$= \frac{{\sin \,\phi- {{\sin }^2}\phi- \cos \,\phi- {{\cos }^2}\phi}}{{\cos \,\phi\,\sin \,\phi}}$
$ = \frac{{\sin \,\phi - \cos \,\phi- 1}}{{\cos \,\phi \,\sin \,\phi}}$…..$(ii)$
Adding $(i)$ and $(ii)$ we get,
$xy + 1 + (x - y) = 0$
$ \Rightarrow x = \frac{{y - 1}}{{y + 1}}$.
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$\mathrm{S}_{1} =\left\{\mathrm{z} \in \mathrm{C}|| \mathrm{z}-3-\left.2 \mathrm{i}\right|^{2}=8\right\}$
$\mathrm{S}_{2} =\{\mathrm{z} \in \mathrm{C} \mid \operatorname{Re}(\mathrm{z}) \geq 5\}$ અને
$\mathrm{S}_{3} =\{\mathrm{z} \in \mathrm{C} \| \mathrm{z}-\bar{z} \mid \geq 8\}$
તો $\mathrm{S}_{1} \cap \mathrm{S}_{2} \cap \mathrm{S}_{3}$ માં ઘટકોની સંખ્યા મેળવો.