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.
    $$\boxed{\cos^2 x - \sin^2 x = 1 - 2\sin^2 x}$$

*   **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:**
    $$\boxed{64.41^\circ}$$

*   **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:**
    $$\boxed{a^{-5b^2}}$$

*   **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:**
    $$\boxed{19.0986}$$

*   **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:**
    $$\boxed{\text{Urefu \approx 10.55  m, Upana \approx 4.55  m}}$$

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

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

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

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