If 540^{circ}

Question

(D) Given $540^{\circ} < A < 630^{\circ}$,which is $3\pi < A < \frac{7\pi}{2}$.In this interval,$A$ lies in the third quadrant,so $\cos A < 0$ and $

Vedclass · Accepted Answer

$-\frac{18}{5}$

Vedclass · Answer

(D) Given $540^{\circ} In this interval,$A$ lies in the third quadrant,so $\cos A  0$.Since $|\cos A| = \frac{5}{13}$ and $\cos A Using $	an A = \sqrt{\sec^2 A - 1}$,we get $	an A = \sqrt{(\frac{13}{5})^2 - 1} = \sqrt{\frac{169}{25} - 1} = \sqrt{\frac{144}{25}} = \frac{12}{5}$.For $	an \frac{A}{2}$,we use the formula $\cos A = \frac{1 - 	an^2(A/2)}{1 + 	an^2(A/2)}$.$-\frac{5}{13} = \frac{1 - 	an^2(A/2)}{1 + 	an^2(A/2)} \Rightarrow -5 - 5	an^2(A/2) = 13 - 13	an^2(A/2)$.$8	an^2(A/2) = 18 \Rightarrow 	an^2(A/2) = \frac{18}{8} = \frac{9}{4}$.Since $540^{\circ} In the fourth quadrant,$	an \frac{A}{2} Therefore,$	an \frac{A}{2} 	an A = (-\frac{3}{2}) 	imes (\frac{12}{5}) = -\frac{18}{5}$.

If $540^{\circ} < A < 630^{\circ}$ and $|\cos A| = \frac{5}{13}$,then $\tan \frac{A}{2} \tan A = $

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