If 90^{circ}

Question

(A) Given $\sin \alpha = \frac{\sqrt{3}}{2}$ where $90^{\circ} < \alpha < 180^{\circ}$ (Second Quadrant).Since $\sin \alpha$ is positive and $\cos \al

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) Given $\sin \alpha = \frac{\sqrt{3}}{2}$ where $90^{\circ} Since $\sin \alpha$ is positive and $\cos \alpha$ is negative in the second quadrant,$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \frac{3}{4}} = -\frac{1}{2}$.Thus,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\sqrt{3}/2}{-1/2} = -\sqrt{3}$.Given $\sin \beta = -\frac{\sqrt{3}}{2}$ where $180^{\circ} Since $\sin \beta$ is negative and $\cos \beta$ is negative in the third quadrant,$\cos \beta = -\sqrt{1 - \sin^2 \beta} = -\sqrt{1 - \frac{3}{4}} = -\frac{1}{2}$.Thus,$	an \beta = \frac{\sin \beta}{\cos \beta} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$.Substituting these values into the expression:$\frac{4 \sin \alpha - 3 	an \beta}{	an \alpha + \sin \beta} = \frac{4(\frac{\sqrt{3}}{2}) - 3(\sqrt{3})}{-\sqrt{3} + (-\frac{\sqrt{3}}{2})} = \frac{2\sqrt{3} - 3\sqrt{3}}{-\frac{3\sqrt{3}}{2}} = \frac{-\sqrt{3}}{-\frac{3\sqrt{3}}{2}} = \frac{2}{3}$.

If $90^{\circ} < \alpha < 180^{\circ}$,$\sin \alpha = \frac{\sqrt{3}}{2}$ and $180^{\circ} < \beta < 270^{\circ}$,$\sin \beta = -\frac{\sqrt{3}}{2}$,then $\frac{4 \sin \alpha - 3 \tan \beta}{\tan \alpha + \sin \beta} = $

Similar Questions

If $\sec \theta(\cos \theta+\sin \theta)=\sqrt{2},$ then what is the value of $\frac{2 \sin \theta}{\cos \theta-\sin \theta} ?$

If $\tan x = (\sin 45^{\circ})(\cos 45^{\circ}) + \sin 30^{\circ}$,then the value of $x$ is (in $^{\circ}$)

If $A + B + C = 180^\circ,$ then $\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C - 1} = $

If $\sin \alpha = 1/\sqrt{5}$ and $\sin \beta = 3/5$,then $\beta - \alpha$ lies in the interval

If $x=a(\sin \theta+\cos \theta)$ and $y=b(\sin \theta-\cos \theta)$,then the value of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module