A) Prove that (1+cot θ)/(1+tan θ) = cot θ. b) Show that cos2 x – sin2 x = 1 – 2sin2 x. c) Determine the acute angle sec⁻¹ (2.3164).

*   **Step 1:** Anza na upande wa kushoto (LHS):
    $$
    \text{LHS} = \cos^2 x - \sin^2 x
    $$
*   **Step 2:** Tumia utambulisho wa Pythagorean $\sin^2 x + \cos^2 x = 1$, ambayo inamaanisha $\cos^2 x = 1 - \sin^2 x$.
*   **Step 3:** Badilisha $\cos^2 x$ katika LHS.
    $$
    \text{LHS} = (1 - \sin^2 x) - \sin^2 x
    $$
*   **Step 4:** Rahisisha usemi.
    $$
    \text{LHS} = 1 - 2\sin^2 x
    $$
*   **Jibu la Mwisho:** Kwa kuwa LHS = RHS, uthibitisho umekamilika.
    $$
    \boxed{\mathbf{\cos^2 x - \sin^2 x = 1 - 2\sin^2 x}}
    $$

*   **Step 1:** Acha $\theta = \sec^{-1} (2.3164)$. Hii inamaanisha $\sec \theta = 2.3164$.
*   **Step 2:** Tumia uhusiano $\sec \theta = \frac{1}{\cos \theta}$ ili kupata $\cos \theta$.
    $$
    \cos \theta = \frac{1}{2.3164}
    $$
*   **Step 3:** Kokotoa thamani ya $\frac{1}{2.3164}$.
    $$
    \cos \theta \approx 0.43170436
    $$
*   **Step 4:** Tumia kazi ya inverse cosine ($\cos^{-1}$) kupata $\theta$.
    $$
    \theta = \cos^{-1}(0.43170436)
    $$
    $$
    \theta \approx 64.41^\circ
    $$
*   **Jibu la Mwisho:**
    $$
    \boxed{\mathbf{64.41^\circ}}
    $$

*   **Step 1:** Rahisisha nambari ya juu kwanza kwa kuzidisha coefficients na kuongeza viashiria vya vigezo vinavyofanana.
    $$
    3a^{-7}b^3 \times 2ab^{-1} = (3 \times 2) \times (a^{-7} \times a^1) \times (b^3 \times b^{-1})
    $$
    $$
    = 6a^{(-7+1)}b^{(3-1)}
    $$
    $$
    = 6a^{-6}b^2
    $$
*   **Step 2:** Gawanya matokeo ya hatua ya 1 na nambari ya chini $6a^{-1}$.
    $$
    \frac{6a^{-6}b^2}{6a^{-1}} = \left(\frac{6}{6}\right) \times \left(\frac{a^{-6}}{a^{-1}}\right) \times b^2
    $$
    $$
    = 1 \times a^{(-6 - (-1))} \times b^2
    $$
    $$
    = a^{(-6+1)}b^2
    $$
    $$
    = a^{-5}b^2
    $$
*   **Jibu la Mwisho:**
    $$
    \boxed{\mathbf{a^{-5}b^2}}
    $$

*   **Step 1:** Kokotoa $(3.52)^2$.
    $$
    (3.52)^2 = 12.3904
    $$
*   **Step 2:** Kokotoa $\sqrt{45}$.
    $$
    \sqrt{45} \approx 6.708203932
    $$
*   **Step 3:** Jumlisha matokeo ya hatua ya 1 na 2.
    $$
    12.3904 + 6.708203932 = 19.098603932
    $$
*   **Step 4:** Zungusha jibu kwa sehemu 4 za desimali.
    $$
    19.0986
    $$
*   **Jibu la Mwisho:**
    $$
    \boxed{\mathbf{19.0986}}
    $$

*   **Step 1:** Fomula ya eneo la mstatili ni Urefu $\times$ Upana.
    $$
    (x+4)(x-2) = 48
    $$
*   **Step 2:** Panua usemi na upange upya kuwa equation ya quadratic.
    $$
    x^2 - 2x + 4x - 8 = 48
    $$
    $$
    x^2 + 2x - 8 = 48
    $$
    $$
    x^2 + 2x - 8 - 48 = 0
    $$
    $$
    x^2 + 2x - 56 = 0
    $$
*   **Step 3:** Tatua equation ya quadratic. Tutatumia fomula ya quadratic $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Hapa $a=1, b=2, c=-56$.
    $$
    x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-56)}}{2(1)}
    $$
    $$
    x = \frac{-2 \pm \sqrt{4 + 224}}{2}
    $$
    $$
    x = \frac{-2 \pm \sqrt{228}}{2}
    $$
    $$
    x = \frac{-2 \pm 15.09966}{2}
    $$
*   **Step 4:** Pata maadili mawili ya $x$.
    $$
    x_1 = \frac{-2 + 15.09966}{2} = \frac{13.09966}{2} \approx 6.5498
    $$
    $$
    x_2 = \frac{-2 - 15.09966}{2} = \frac{-17.09966}{2} \approx -8.5498
    $$
*   **Step 5:** Kwa kuwa urefu hauwezi kuwa hasi, tunachukua $x \approx 6.55$.
*   **Step 6:** Kokotoa vipimo vya shamba.
    $$
    \text{Urefu} = x+4 = 6.55 + 4 = 10.55 \text{ m}
    $$
    $$
    \text{Upana} = x-2 = 6.55 - 2 = 4.55 \text{ m}
    $$
*   **Jibu la Mwisho:**
    $$
    \boxed{\mathbf{\text{Urefu} \approx 10.55 \text{ m, Upana} \approx 4.55 \text{ m}}}
    $$

*   **Step 1:** Kokotoa urefu wa slant ($l$) kwa kutumia Theorem ya Pythagoras.
    $$
    l = \sqrt{r^2 + h^2}
    $$
    $$
    l = \sqrt{(3 \text{ m})^2 + (4 \text{ m})^2}
    $$
    $$
    l = \sqrt{9 \text{ m}^2 + 16 \text{ m}^2}
    $$
    $$
    l = \sqrt{25 \text{ m}^2}
    $$
    $$
    l = 5 \text{ m}
    $$
*   **Step 2:** Kokotoa eneo la uso ($A$) la koni.
    $$
    A = \pi r^2 + \pi r l
    $$
    $$
    A = \pi (3 \text{ m})^2 + \pi (3 \text{ m})(5 \text{ m})
    $$
    $$
    A = 9\pi \text{ m}^2 + 15\pi \text{ m}^2
    $$
    $$
    A = 24\pi \text{ m}^2
    $$
*   **Step 3:** Kokotoa ujazo ($V$) wa koni.
    $$
    V = \frac{1}{3}\pi r^2 h
    $$
    $$
    V = \frac{1}{3}\pi (3 \text{ m})^2 (4 \text{ m})
    $$
    $$
    V = \frac{1}{3}\pi (9 \text{ m}^2) (4 \text{ m})
    $$
    $$
    V = 12\pi \text{ m}^3
    $$
*   **Jibu la Mwisho:**
    $$
    \boxed{\mathbf{\text{Eneo la uso} = 24\pi \text{ m}^2, \text{Ujazo} = 12\pi \text{ m}^3}}
    $$

*   **Step 1:** Kokotoa eneo la uso ($A$) la tufe.
    $$
    A = 4\pi r^2
    $$
    $$
    A = 4\pi (3 \text{ m})^2
    $$
    $$
    A = 4\pi (9 \text{ m}^2)
    $$
    $$
    A = 36\pi \text{ m}^2
    $$
*   **Step 2:** Kokotoa ujazo ($V$) wa tufe.
    $$
    V = \frac{4}{3}\pi r^3
    $$
    $$
    V = \frac{4}{3}\pi (3 \text{ m})^3
    $$
    $$
    V = \frac{4}{3}\pi (27 \text{ m}^3)
    $$
    $$
    V = 4\pi (9 \text{ m}^3)
    $$
    $$
    V = 36\pi \text{ m}^3
    $$
*   **Jibu la Mwisho:**
    $$
    \boxed{\mathbf{\text{Eneo la uso} = 36\pi \text{ m}^2, \text{Ujazo} = 36\pi \text{ m}^3}}
    $$

*   **Step 1:** Badilisha kasi kutoka km/h kwenda m/s. (Kumbuka: $1 \text{ km/h} = \frac{5}{18} \text{ m/s}$)
    $$
    \text{Kasi ya Lori A} (V_A) = 54 \text{ km/h} = 54 \times \frac{5}{18} \text{ m/s} = 15 \text{ m/s}
    $$
    $$
    \text{Kasi ya Lori B} (V_B) = 72 \text{ km/h} = 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}
    $$
*   **Step 2:** Kokotoa kasi ya jamaa ya Lori B kwa Lori A.
    $$
    \text{Kasi ya Jamaa} (V_{rel}) = V_B - V_A = 20 \text{ m/s} - 15 \text{ m/s} = 5 \text{ m/s}
    $$
*   **Step 3:** Kokotoa umbali jumla ambao Lori B linahitaji kusafiri ili kulipita kabisa Lori A. Umbali huu ni jumla ya urefu wa malori yote mawili.
    $$
    \text{Umbali} (D) = \text{Urefu wa Lori A} + \text{Urefu wa Lori B} = 12 \text{ m} + 18 \text{ m} = 30 \text{ m}
    $$
*   **Step 4:** Kokotoa muda unaohitajika.
    $$
    \text{Muda} (T) = \frac{\text{Umbali}}{\text{Kasi ya Jamaa}} = \frac{30 \text{ m}}{5 \text{ m/s}} = 6 \text{ s}
    $$
*   **Jibu la Mwisho:**
    $$
    \boxed{\mathbf{6 \text{ sekunde}}}
    $$

*   **Step 1:** Badilisha kasi kutoka km/h kwenda m/s.
    $$
    \text{Kasi ya Basi} (V_{bus}) = 72 \text{ km/h} = 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}
    $$
    $$
    \text{Kasi ya Lori} (V_{truck}) = 54 \text{ km/h} = 54 \times \frac{5}{18} \text{ m/s} = 15 \text{ m/s}
    $$
*   **Step 2:** Kokotoa kasi ya jamaa ya basi kwa lori.
    $$
    \text{Kasi ya Jamaa} (V_{rel}) = V_{bus} - V_{truck} = 20 \text{ m/s} - 15 \text{ m/s} = 5 \text{ m/s}
    $$
*   **Step 3:** Umbali wa awali kati ya basi na lori ni 200 m.
    $$
    \text{Umbali} (D) = 200 \text{ m}
    $$
*   **Step 4:** Kokotoa muda unaohitajika kwa basi kulifikia lori.
    $$
    \text{Muda} (T) = \frac{\text{Umbali}}{\text{Kasi ya Jamaa}} = \frac{200 \text{ m}}{5 \text{ m/s}} = 40 \text{ s}
    $$
*   **Jibu la Mwisho:**
    $$
    \boxed{\mathbf{40 \text{ sekunde}}}
    $$