Question - Trigonometric Ratios in the Second Quadrant
Solution:
The question gives us the cosine of an angle (θ) and the range for the angle, which is the second quadrant ($$\frac{\pi}{2} < \theta < \pi$$), and asks us to find the values of the remaining trigonometric ratios.Given:$$\cos \theta = -\frac{\sqrt{3}}{2}$$Since the cosine of θ is negative and the angle is in the second quadrant, sine will be positive (as sine is positive in the second quadrant).Let's use the Pythagorean identity to find sine:$$\sin^2 \theta + \cos^2 \theta = 1$$Substitute the given value of cos θ:$$\sin^2 \theta + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1$$$$\sin^2 \theta + \frac{3}{4} = 1$$Solving for $$\sin^2 \theta$$:$$\sin^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$$Taking the square root to find sin θ yields two possible values, +1/2 and -1/2. Since we are in the second quadrant and sine is positive:$$\sin \theta = \frac{1}{2}$$The tangent of θ is the ratio of sine to cosine:$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$$$\tan \theta = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$$$$\tan \theta = -\frac{1}{\sqrt{3}}$$$$\tan \theta = -\frac{\sqrt{3}}{3}$$For the reciprocals:$$\csc \theta = \frac{1}{\sin \theta} = 2$$$$\sec \theta = \frac{1}{\cos \theta} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$$$$\cot \theta = \frac{1}{\tan \theta} = -\sqrt{3}$$So, the six trigonometric ratios for the given angle θ in the second quadrant are:$$\sin \theta = \frac{1}{2}, \cos \theta = -\frac{\sqrt{3}}{2}, \tan \theta = -\frac{\sqrt{3}}{3}$$$$\csc \theta = 2, \sec \theta = -\frac{2\sqrt{3}}{3}, \cot \theta = -\sqrt{3}$$
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