MCQ
If $\text{y}=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\tan\frac{\text{x}}{2}\Big),\text{a}>\text{b}>0,$then:
- A$\text{y}_1=\frac{-1}{\text{a}+\text{b}\cos\text{x}}$
- ✓$\text{y}_2=\frac{\text{b}\sin\text{x}}{(\text{a}+\text{b}\cos\text{x})^2}$
- C$\text{y}_1=\frac{1}{\text{a}-\text{b}\cos\text{x}}$
- D$\text{y}_2=\frac{-\text{b}\sin\text{x}}{(\text{a}-\text{b}\cos\text{x})^2}$