5. Diketahui $$ an \alpha=-\frac{2}{3} $$ dan $$ \alpha $$ adalah sudut di kuadran kedua. Hitunglah: a. $$ \frac{\sin \left(90^{\circ}-\alpha ight)-\cos \left(180^{\circ}-\alpha ight)}{ an \left(270^{\circ}+\alpha ight)+\cot \left(360^{\circ}-\alpha ight)} $$ b. $$ \frac{ an \left(90^{\circ}+\alpha ight)+\cos \left(180^{\circ}+\alpha ight)}{\sin \left(270^{\circ}-\alpha ight)-\cot \left(180^{\circ}-\alpha ight)} $$

Question

5. Diketahui $$ 	an \alpha=-\frac{2}{3} $$ dan $$ \alpha $$ adalah sudut di kuadran kedua. Hitunglah: a. $$ \frac{\sin \left(90^{\circ}-\alphaight)-\cos \left(180^{\circ}-\alphaight)}{	an \left(270^{\circ}+\alphaight)+\cot \left(360^{\circ}-\alphaight)} $$ b. $$ \frac{	an \left(90^{\circ}+\alphaight)+\cos \left(180^{\circ}+\alphaight)}{\sin \left(270^{\circ}-\alphaight)-\cot \left(180^{\circ}-\alphaight)} $$

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Diketahui $$ \tan \alpha = -\frac{2}{3} $$ dan $$ \alpha $$ berada di kuadran kedua. Kita akan menghitung bagian a dan b sebagai berikut:

### a. 
$$
\frac{\sin \left(90^{\circ}-\alpha\right)-\cos \left(180^{\circ}-\alpha\right)}{\tan \left(270^{\circ}+\alpha\right)+\cot \left(360^{\circ}-\alpha\right)}
$$

**Langkah-langkah:**

1. **Identitas Trigonometri:**
   - $$\sin(90^{\circ} - \alpha) = \cos \alpha$$
   - $$\cos(180^{\circ} - \alpha) = -\cos \alpha$$
   - $$\tan(270^{\circ} + \alpha) = \frac{\sin(270^{\circ} + \alpha)}{\cos(270^{\circ} + \alpha)} = \frac{3/\sqrt{13}}{2/\sqrt{13}} = \frac{3}{2}$$
   - $$\cot(360^{\circ} - \alpha) = \cot(-\alpha) = -\cot \alpha = \frac{3}{2}$$

2. **Menghitung Nilai Sine dan Cosine:**
   - Karena $$\tan \alpha = -\frac{2}{3}$$ dan $$\alpha$$ di kuadran II:
     $$
     \sin \alpha = \frac{2}{\sqrt{13}}, \quad \cos \alpha = -\frac{3}{\sqrt{13}}
     $$

3. **Substitusi ke Dalam Rumus:**
   $$
   \text{Pembilang} = \cos \alpha - (-\cos \alpha) = -\frac{3}{\sqrt{13}} - \frac{3}{\sqrt{13}} = -\frac{6}{\sqrt{13}}
   $$
   $$
   \text{Penyebut} = \frac{3}{2} + \frac{3}{2} = 3
   $$
   $$
   \text{Hasil} = \frac{-\frac{6}{\sqrt{13}}}{3} = -\frac{2}{\sqrt{13}} = -\frac{2\sqrt{13}}{13}
   $$

**Jawaban a:**
$$
-\frac{2\sqrt{13}}{13}
$$

### b. 
$$
\frac{\tan \left(90^{\circ}+\alpha\right)+\cos \left(180^{\circ}+\alpha\right)}{\sin \left(270^{\circ}-\alpha\right)-\cot \left(180^{\circ}-\alpha\right)}
$$

**Langkah-langkah:**

1. **Identitas Trigonometri:**
   - $$\tan(90^{\circ} + \alpha) = -\cot \alpha = \frac{3}{2}$$
   - $$\cos(180^{\circ} + \alpha) = -\cos \alpha = \frac{3}{\sqrt{13}}$$
   - $$\sin(270^{\circ} - \alpha) = -\cos \alpha = \frac{3}{\sqrt{13}}$$
   - $$\cot(180^{\circ} - \alpha) = -\cot \alpha = \frac{3}{2}$$

2. **Substitusi ke Dalam Rumus:**
   $$
   \text{Pembilang} = \frac{3}{2} + \frac{3}{\sqrt{13}}
   $$
   $$
   \text{Penyebut} = \frac{3}{\sqrt{13}} - \frac{3}{2}
   $$
   
3. **Menyederhanakan Pecahan:**
   $$
   \frac{\frac{3}{2} + \frac{3}{\sqrt{13}}}{\frac{3}{\sqrt{13}} - \frac{3}{2}} = \frac{3\left(\frac{1}{2} + \frac{1}{\sqrt{13}}\right)}{3\left(\frac{1}{\sqrt{13}} - \frac{1}{2}\right)} = \frac{\frac{1}{2} + \frac{1}{\sqrt{13}}}{\frac{1}{\sqrt{13}} - \frac{1}{2}}
   $$
   $$
   = \frac{\frac{\sqrt{13} + 2}{2\sqrt{13}}}{\frac{2 - \sqrt{13}}{2\sqrt{13}}} = \frac{\sqrt{13} + 2}{2 - \sqrt{13}} = \frac{(\sqrt{13} + 2)(2 + \sqrt{13})}{(2 - \sqrt{13})(2 + \sqrt{13})} = \frac{4\sqrt{13} + 17}{-9} = -\frac{4\sqrt{13} + 17}{9}
   $$

**Jawaban b:**
$$
-\frac{4\sqrt{13} + 17}{9}
$$
a. $$ -\frac{2\sqrt{13}}{13} $$ b. $$ -\frac{4\sqrt{13} + 17}{9} $$

Diketahui dan adalah sudut
di kuadran kedua. Hitunglah:
a.
b.

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Solución

a.

b.

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Diketahui dan adalah sudut di kuadran kedua. Hitunglah: a. b.