If tan alpha = frac{-12}{5},cot beta = frac{7 | vedclass

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(B) Given $	an \alpha = \frac{-12}{5}$. Since $\alpha$ is not in the second quadrant,$\alpha$ must be in the fourth quadrant. Thus,$\frac{\alpha}{2}

Vedclass · Accepted Answer

$\frac{19}{10}$

Vedclass · Answer

(B) Given $	an \alpha = \frac{-12}{5}$. Since $\alpha$ is not in the second quadrant,$\alpha$ must be in the fourth quadrant. Thus,$\frac{\alpha}{2} \in (135^{\circ}, 180^{\circ})$.Using $	an \alpha = \frac{2 	an \frac{\alpha}{2}}{1 - 	an^2 \frac{\alpha}{2}}$,we get $\frac{-12}{5} = \frac{2 	an \frac{\alpha}{2}}{1 - 	an^2 \frac{\alpha}{2}}$.Solving $6 	an^2 \frac{\alpha}{2} - 5 	an \frac{\alpha}{2} - 6 = 0$,we find $	an \frac{\alpha}{2} = \frac{-2}{3}$. Since $\frac{\alpha}{2}$ is in the second quadrant,$\sin \frac{\alpha}{2} = \frac{2}{\sqrt{13}}$.Given $\cot \beta = \frac{7}{24}$. Since $\beta$ is not in the first quadrant,$\beta$ must be in the third quadrant. Thus,$\cos \beta = \frac{-7}{25}$.Using $\cos \beta = 2 \cos^2 \frac{\beta}{2} - 1$,we get $\frac{-7}{25} = 2 \cos^2 \frac{\beta}{2} - 1$,so $\cos^2 \frac{\beta}{2} = \frac{9}{25}$. Since $\frac{\beta}{2}$ is in the second quadrant,$\cos \frac{\beta}{2} = \frac{-3}{5}$ and $\sin \frac{\beta}{2} = \frac{4}{5}$,so $\cot \frac{\beta}{2} = \frac{-3}{4}$.Substituting these values: $\sqrt{13} (\frac{2}{\sqrt{13}}) + (\frac{-3}{5}) + (\frac{-2}{3})(\frac{-3}{4}) = 2 - \frac{3}{5} + \frac{1}{2} = \frac{20 - 6 + 5}{10} = \frac{19}{10}$.

If $\tan \alpha = \frac{-12}{5}$,$\cot \beta = \frac{7}{24}$,$\alpha$ does not belong to the second quadrant,and $\beta$ does not belong to the first quadrant,then $\sqrt{13} \sin \frac{\alpha}{2} + \cos \frac{\beta}{2} + \tan \frac{\alpha}{2} \cot \frac{\beta}{2} = $

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