MCQ
$\sqrt{\frac{1 - sin A}{1 + sin A}}= ......... - \frac{\pi}{2} < A < \frac{\pi}{2}$
- A$sec A + tan A$
- ✓$sec A - tan A$
- C$-sec A + tan A$
- D$-(sec A + tan A)$
$\sqrt{\frac{1 - sin A}{1 + sin A}} \times \sqrt{\frac{1 - sin A}{1 - sin A}}$
$= \frac{1 - sin A}{\sqrt{1 - sin^2 A}} = \frac{1 - sin A}{|cos A|}$
$= \frac{1 - sin A}{cos A} \left[\left(-\frac{\pi}{2} < A < \frac{\pi}{2}\right) \ \ cos A > 0\right]$ હોવાથી
$= sec A - tan A$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\mathrm{S}_{1}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}-2| \leq 1\}$ અને
$\mathrm{S}_{2}=\{\mathrm{z} \in \mathrm{C}: \mathrm{z}(1+\mathrm{i})+\overline{\mathrm{z}}(1-\mathrm{i}) \geq 4\}$
આપેલ હોય તો $z \in \mathrm{S}_{1} \cap \mathrm{S}_{2}$ માટે $\left|z-\frac{5}{2}\right|^{2}$ ની મહતમ કિમંત મેળવો.