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 = $

Similar Questions

$(4 \cos^2 9^{\circ} - 3)(4 \cos^2 27^{\circ} - 3) = $

$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $

If $\cos A = \frac{3}{4}$,then $32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) = $

Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ given that $\cos x = -\frac{1}{3}$ and $x$ is in quadrant $III$.

If $\sin \theta - \cos \theta = \frac{1}{\sqrt{3}}$,then $\sin(2\theta) + \cos(4\theta) + \sin(6\theta) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module