Q66 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics
If $\tan \alpha=\frac{-12}{5}, \cot \beta=\frac{7}{24}, \alpha$ does not belong to second quadrant and $\beta$ does not belong to first quadrant, then $\sqrt{13} \sin \frac{\alpha}{2}+\cos \frac{\beta}{2}+\tan \frac{\alpha}{2} \cot \frac{\beta}{2}=$
- A$\frac{31}{10}$
- Correct answerB$\frac{19}{10}$
- C$\frac{21}{10}$
- D$\frac{-9}{10}$
Solution
$\tan \alpha=\frac{-12}{5} \Rightarrow \alpha$ belongs to 4 th quadrant $\cot \beta=\frac{7}{24} \Rightarrow \beta$ belongs to 3rd quadrant
$$
\begin{aligned}
\tan \alpha= & \frac{2 \tan \frac{\alpha}{2}}{1-\tan ^2 \frac{\alpha}{2}} \Rightarrow \frac{\alpha}{2} \in\left(135^{\circ}, 180^{\circ}\right) \\
\Rightarrow & \frac{-12}{5}\left(1-\tan ^2 \frac{\alpha}{2}\right)=2 \tan \frac{\alpha}{2} \\
\Rightarrow & \frac{-6}{5}+\frac{6}{5} \tan ^2 \frac{\alpha}{2}=\tan \frac{\alpha}{2} \\
\Rightarrow & 6 \tan ^2 \frac{\alpha}{2}-5 \tan \frac{\alpha}{2}-6=0 \\
\Rightarrow & 6 \tan ^2 \frac{\alpha}{2}-9 \tan \frac{\alpha}{2}+4 \tan \frac{\alpha}{2}-6=0 \\
\Rightarrow & \left(3 \tan \frac{\alpha}{2}+2\right)\left(2 \tan \frac{\alpha}{2}-3\right)=0 \\
\Rightarrow & \tan \frac{\alpha}{2}=\frac{-2}{3} \\
& \sin \frac{\alpha}{2}=\frac{2}{\sqrt{13}} \\
& \cot \beta=-\frac{7}{24} \Rightarrow \cos \beta=\frac{-7}{25} \\
& \frac{\beta}{2} \in 2 \text { nd quadrant } \\
& \cos \beta=2 \cos 2 \frac{\beta}{2}-1 \Rightarrow \frac{-7}{25}+1=2 \cos 2 \frac{\beta}{2} \\
\Rightarrow & \cos \frac{\beta}{2}=\frac{-3}{5} \Rightarrow \cot \frac{\beta}{2}=\frac{-3}{4} \\
\therefore \quad & \sqrt{13} \sin \frac{\alpha}{2}+\cos \frac{\beta}{2}+\tan \frac{\alpha}{2} \cot \frac{\beta}{2} \\
& =\sqrt{13} \cdot \frac{2}{\sqrt{13}}+\left(\frac{-3}{5}\right)+\left(\frac{-2}{3}\right)\left(\frac{-3}{4}\right) \\
= & 2-\frac{3}{5}+\frac{1}{2}=\frac{20-6+5}{10}=\frac{19}{10} .
\end{aligned}
$$
Hence, the correct answer is option: 'B'.