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 th value of \( \cos \alpha+\tan \beta \). UCTION FORMITAE

Step 1: Find cos α We are given sin α = 3/5 and that α lies in [90°, 270°]. Since sine is positive in this interval, α must be in Quadrant II, where cosine is negative. Using the identity:   sin²α + cos²α = 1 Substitute sin α:   (3/5)² + cos²α = 1   9/25 + cos²α = 1   cos²α = 1 - 9/25 = 16/25 Since α is in Quadrant II, cos α = –√(16/25) = –4/5. Step 2: Find tan β We are given cos β = –12/13 and that β lies in [0°, 180°]. Since cosine is negative in this interval, β is in Quadrant II where sine is positive. Compute sin β using:   sin²β + cos²β = 1   sin²β + (–12/13)² = 1   sin²β + 144/169 = 1   sin²β = 1 - 144/169 = 25/169   sin β = 5/13  (since sine is positive in Quadrant II) Then, tan β = sin β / cos β = (5/13) / (–12/13) = –5/12. Step 3: Compute cos α + tan β Now sum:   cos α + tan β = (–4/5) + (–5/12) Find a common denominator (60):   (–4/5) = –48/60   (–5/12) = –25/60 So,   cos α + tan β = –48/60 – 25/60 = –73/60 Final Answer: cos α + tan β = –73/60. Diagram (for visualization): • For angle α (in Quadrant II):   Imagine a right triangle where the opposite side is 3 and the hypotenuse is 5. Since α is in QII, the adjacent side is negative, –4. • For angle β (in Quadrant II):   Imagine a right triangle where the adjacent side is –12 and the hypotenuse is 13. The opposite side is 5. These triangles help illustrate the sign choices and the computed values.