2. Without using a calculator, state whether the following ratios are positive or negative: \( \begin{array}{lll}\text { a) } \cos 325^{\circ} & \text { b) } \tan 126^{\circ} & \text { c) } \sin 154^{\circ}\end{array} \) 3. In which quadrant will \( x \) lie if: \( \begin{array}{ll}\text { a) } \sin x>0 \text { and } \tan x<0 & \text { b) } \tan x>0 \text { and } \cos x<0 \\ \text { c) } \tan x<0 \text { and } 180^{\circ}0 \text { and } \sin x<0 \text { ? }\end{array} \) 4. If \( 12 \tan C+5=0 \) and \( C \in\left(90^{\circ} ; 270^{\circ}\right) \), find the value of \( 12 \tan C-13 \cos C \) with the aid of a diagram and without using a calculator.

Alright, I need to tackle these trigonometry problems without using a calculator. Let's break them down one by one. **Problem 2:** a) \( \cos 325^{\circ} \) b) \( \tan 126^{\circ} \) c) \( \sin 154^{\circ} \) First, I recall that cosine and sine functions have specific signs in each quadrant: - **First Quadrant (0°-90°):** All trigonometric functions are positive. - **Second Quadrant (90°-180°):** Sine is positive, cosine and tangent are negative. - **Third Quadrant (180°-270°):** Tangent is positive, sine and cosine are negative. - **Fourth Quadrant (270°-360°):** Cosine is positive, sine and tangent are negative. For **a) \( \cos 325^{\circ} \):** 325° is in the fourth quadrant. In the fourth quadrant, cosine is positive. So, \( \cos 325^{\circ} \) is **positive**. For **b) \( \tan 126^{\circ} \):** 126° is in the second quadrant. In the second quadrant, tangent is negative. So, \( \tan 126^{\circ} \) is **negative**. For **c) \( \sin 154^{\circ} \):** 154° is in the second quadrant. In the second quadrant, sine is positive. So, \( \sin 154^{\circ} \) is **positive**. **Problem 3:** a) \( \sin x > 0 \) and \( \tan x < 0 \) b) \( \tan x > 0 \) and \( \cos x < 0 \) c) \( \tan x < 0 \) and \( 180^{\circ} < x < 360^{\circ} \) d) \( \cos x > 0 \) and \( \sin x < 0 \) Let's analyze each case: a) \( \sin x > 0 \) and \( \tan x < 0 \): - \( \sin x > 0 \) in the first and second quadrants. - \( \tan x < 0 \) in the second and fourth quadrants. - The only quadrant where both conditions are satisfied is the **second quadrant**. b) \( \tan x > 0 \) and \( \cos x < 0 \): - \( \tan x > 0 \) in the first and third quadrants. - \( \cos x < 0 \) in the second and third quadrants. - The only quadrant where both conditions are satisfied is the **third quadrant**. c) \( \tan x < 0 \) and \( 180^{\circ} < x < 360^{\circ} \): - \( \tan x < 0 \) in the second and fourth quadrants. - The range \( 180^{\circ} < x < 360^{\circ} \) covers the third and fourth quadrants. - \( \tan x < 0 \) in the **fourth quadrant**. d) \( \cos x > 0 \) and \( \sin x < 0 \): - \( \cos x > 0 \) in the first and fourth quadrants. - \( \sin x < 0 \) in the third and fourth quadrants. - The only quadrant where both conditions are satisfied is the **fourth quadrant**. **Problem 4:** Given \( 12 \tan C + 5 = 0 \) and \( C \in (90^{\circ}, 270^{\circ}) \), find the value of \( 12 \tan C - 13 \cos C \) with the aid of a diagram and without using a calculator. First, solve for \( \tan C \): \( 12 \tan C + 5 = 0 \) \( 12 \tan C = -5 \) \( \tan C = -\frac{5}{12} \) Since \( C \) is in the second quadrant (90° < C < 270°), and tangent is negative there, this makes sense. Now, find \( \cos C \). Using the identity: \( \tan C = \frac{\sin C}{\cos C} \) \( -\frac{5}{12} = \frac{\sin C}{\cos C} \) \( \sin C = -\frac{5}{12} \cos C \) Using the Pythagorean identity: \( \sin^2 C + \cos^2 C = 1 \) \( \left(-\frac{5}{12} \cos C\right)^2 + \cos^2 C = 1 \) \( \frac{25}{144} \cos^2 C + \cos^2 C = 1 \) \( \frac{169}{144} \cos^2 C = 1 \) \( \cos^2 C = \frac{144}{169} \) \( \cos C = \pm \frac{12}{13} \) Since \( C \) is in the second quadrant, cosine is negative: \( \cos C = -\frac{12}{13} \) Now, find \( \sin C \): \( \sin C = -\frac{5}{12} \cos C = -\frac{5}{12} \times -\frac{12}{13} = \frac{5}{13} \) Now, compute \( 12 \tan C - 13 \cos C \): \( 12 \tan C = 12 \times -\frac{5}{12} = -5 \) \( -13 \cos C = -13 \times -\frac{12}{13} = 12 \) So, \( -5 + 12 = 7 \) Therefore, the value is **7**.