3. If \( 3 \tan \theta-4=0 \) and \( \cos \theta<0 \) calculate without the use of a calculator and with the aid of a diagram the value of \( \begin{array}{ll}\text { (a) } \frac{\sin \theta}{\cos \theta} & \text { (b) } 10 \sin \theta-25 \cos ^{2} \theta\end{array} \) 4. If \( \sin A=\frac{2 \sqrt{6}}{5} \) and \( A \in\left[90^{\circ} ; 360^{\circ}\right] \) calculate without the use of a calculator and with the aid of a diagram the value of \( 15 \tan A \cdot \cos A \) *5. If \( \sin \alpha=\frac{3}{5} \) with \( \alpha \in\left[90^{\circ} ; 270^{\circ}\right] \) and \( \cos \beta=\frac{-12}{13} \) with \( \beta \in\left[0^{\circ} ; 180^{\circ}\right] \) calculate without the use of a calculator and with the aid of a diagram the value of \( \cos \alpha+\tan \beta \)

We will solve each part step‐by‐step, drawing on triangle diagrams and the relevant sign conventions. ──────────────────────── Problem 3: ──────────────────────── We are given   3 tan θ – 4 = 0      (1) and the condition cos θ < 0. Step 1. Solve (1):   3 tan θ – 4 = 0 ⟹ tan θ = 4/3. Step 2. Determine the quadrant:   Since tan θ = sin θ/cos θ is positive, sine and cosine must have the same sign. However, the condition cos θ < 0 tells us that cosine is negative. Thus, sin θ must also be negative. This is true only in Quadrant III. Step 3. Construct a right‐angled triangle:   Assume the sides are scaled as follows:    Opposite = 4, Adjacent = 3.   Then, by the Pythagorean theorem,    Hypotenuse = √(4² + 3²) = 5.   Since we are in QIII, both sine and cosine are negative:    sin θ = –4⁄5 and cos θ = –3⁄5. Now we calculate: (a) sin θ/cos θ:   (sin θ)/(cos θ) = (–4/5)/(–3/5) = 4/3 (which confirms our tan θ value). (b) 10 sin θ – 25 cos² θ:   First, cos² θ = (–3/5)² = 9/25.   Then,    10 sin θ – 25 cos² θ = 10(–4/5) – 25(9/25)              = –8 – 9 = –17. ──────────────────────── Problem 4: ──────────────────────── We are given   sin A = (2√6)/5    and A ∈ [90°, 360°]. Step 1. Determine the quadrant:   Since sin A > 0, A could lie in Quadrant I or II. But as A is at least 90°, it must be in Quadrant II where sine is positive and cosine is negative. Step 2. Find cos A:   Using the Pythagorean identity,    cos² A = 1 – sin² A = 1 – ( (2√6/5)² )          = 1 – (24/25) = 1/25.   In Quadrant II, cosine is negative, so:    cos A = –1/5. Step 3. Find tan A:   tan A = sin A/cos A = (2√6/5)/(–1/5) = –2√6. Step 4. Now calculate 15 tan A · cos A:   15 tan A · cos A = 15 (–2√6)(–1/5) = 15 · (2√6/5) = (15/5)(2√6) = 3(2√6) = 6√6. ──────────────────────── Problem 5: ──────────────────────── We are given   sin α = 3/5  with α ∈ [90°, 270°],   cos β = –12/13 with β ∈ [0°, 180°]. Step 1. Determine the quadrant for α:   For α ∈ [90°, 270°] the angle is either in Quadrant II or III. Because sin α is positive, α must lie in Quadrant II. In Quadrant II, cosine is negative.   Thus, from the triangle:    sin α = 3/5 ⟹ cos α = –4/5 (using the 3-4-5 triangle). Step 2. Determine the quadrant for β:   β ∈ [0°, 180°] means β is either in Quadrant I or II. Given cos β is negative, β must lie in Quadrant II.   Now, compute sin β:    sin² β = 1 – cos² β = 1 – (144/169) = 25/169 ⟹ sin β = 5/13 (the sine in QII is positive).   So, tan β = sin β/cos β = (5/13)/(–12/13) = –5/12. Step 3. Now compute cos α + tan β:   cos α + tan β = (–4/5) + (–5/12)   Find a common denominator (60):    –4/5 = –48/60 and –5/12 = –25/60.   Thus, cos α + tan β = –48/60 – 25/60 = –73/60. ──────────────────────── Final Answers: ──────────────────────── 3. (a) sin θ/cos θ = 4/3     (b) 10 sin θ – 25 cos²θ = –17 4. 15 tan A · cos A = 6√6 5. cos α + tan β = –73/60 These results are obtained without using a calculator, with the aid of appropriate diagrams and sign analyses in the relevant quadrants.