(d) If $$ \sin \mathrm{A}=\frac{12}{13} $$ and $$ \mathrm{A} \in\left(90^{\circ} ; 270^{\circ} ight) $$, determine without the use of a calculator and with the aid of a diagram the value of: $$ \begin{array}{lll} ext { (1) } an \mathrm{A} & ext { (2) } \cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A} & ext { (3) } 13 \cos \mathrm{~A}+5 an \mathrm{~A} \ ext { (e) If } \cos \mathrm{B}=\frac{-8}{17} ext { and } \mathrm{B} \in\left(180^{\circ} ; 360^{\circ} ight) ext {, determine without the use of a calculator and with the } \ ext { aid of a diagram the value of: } \ \begin{array}{lll} ext { (1) } \sin \mathrm{B} & ext { (2) } \frac{1}{ an ^{2} \mathrm{~B}}-\frac{1}{\sin ^{2} \mathrm{~B}} & ext { (3) } 17 \sin \mathrm{~B}-8 an \mathrm{~B}\end{array}\end{array} $$

Question

(d) If $$ \sin \mathrm{A}=\frac{12}{13} $$ and $$ \mathrm{A} \in\left(90^{\circ} ; 270^{\circ}ight) $$, determine without the use of a calculator and with the aid of a diagram the value of: $$ \begin{array}{lll}	ext { (1) } 	an \mathrm{A} & 	ext { (2) } \cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A} & 	ext { (3) } 13 \cos \mathrm{~A}+5 	an \mathrm{~A} \ 	ext { (e) If } \cos \mathrm{B}=\frac{-8}{17} 	ext { and } \mathrm{B} \in\left(180^{\circ} ; 360^{\circ}ight) 	ext {, determine without the use of a calculator and with the } \ 	ext { aid of a diagram the value of: } \ \begin{array}{lll}	ext { (1) } \sin \mathrm{B} & 	ext { (2) } \frac{1}{	an ^{2} \mathrm{~B}}-\frac{1}{\sin ^{2} \mathrm{~B}} & 	ext { (3) } 17 \sin \mathrm{~B}-8 	an \mathrm{~B}\end{array}\end{array}> $$

UpStudy AI · Accepted Answer

**(d) Given**:  
$$
\sin A=\frac{12}{13},\quad A\in\left(90^{\circ},270^{\circ}\right)
$$

Since $$\sin A > 0$$ and $$A$$ lies between $$90^{\circ}$$ and $$270^{\circ}$$, it must be in the second quadrant. In the second quadrant, $$\cos A$$ is negative.

1. **Find $$\cos A$$**:  
   Using the identity  
   $$
   \sin^2 A+\cos^2 A=1,
   $$  
   we have  
   $$
   \cos^2 A=1-\sin^2 A=1-\left(\frac{12}{13}\right)^2=1-\frac{144}{169}=\frac{25}{169}.
   $$  
   Since $$A$$ is in the second quadrant,  
   $$
   \cos A=-\frac{5}{13}.
   $$

2. **(1) Compute $$\tan A$$**:  
   $$
   \tan A=\frac{\sin A}{\cos A}=\frac{\frac{12}{13}}{-\frac{5}{13}}=-\frac{12}{5}.
   $$

3. **(2) Compute $$\cos^2A-\sin^2A$$**:  
   $$
   \cos^2A-\sin^2A=\frac{25}{169}-\frac{144}{169}=-\frac{119}{169}.
   $$

4. **(3) Compute $$13\cos A+5\tan A$$**:  
   $$
   13\cos A+5\tan A=13\left(-\frac{5}{13}\right)+5\left(-\frac{12}{5}\right)=-5-12=-17.
   $$

---

**(e) Given**:  
$$
\cos B=\frac{-8}{17},\quad B\in\left(180^{\circ},360^{\circ}\right)
$$

For angles in this interval, cosine is negative in the third quadrant and positive only in the fourth. Since $$\cos B$$ is negative, $$B$$ is in the third quadrant where both sine and cosine are negative.

1. **Find $$\sin B$$**:  
   Using the Pythagorean identity,  
   $$
   \sin^2B=1-\cos^2B=1-\left(\frac{-8}{17}\right)^2=1-\frac{64}{289}=\frac{225}{289}.
   $$  
   In the third quadrant,  
   $$
   \sin B=-\frac{15}{17}.
   $$

2. **(1) $$\sin B$$** is already found:  
   $$
   \sin B=-\frac{15}{17}.
   $$

3. **Find $$\tan B$$**:  
   $$
   \tan B=\frac{\sin B}{\cos B}=\frac{-\frac{15}{17}}{-\frac{8}{17}}=\frac{15}{8}.
   $$

4. **(2) Compute $$\frac{1}{\tan^2B}-\frac{1}{\sin^2B}$$**:  
   First, calculate $$\tan^2B$$ and $$\sin^2B$$:  
   $$
   \tan^2B=\left(\frac{15}{8}\right)^2=\frac{225}{64},\quad \sin^2B=\frac{225}{289}.
   $$
   Then,  
   $$
   \frac{1}{\tan^2B}=\frac{64}{225},\quad \frac{1}{\sin^2B}=\frac{289}{225}.
   $$
   So,  
   $$
   \frac{1}{\tan^2B}-\frac{1}{\sin^2B}=\frac{64}{225}-\frac{289}{225}=-\frac{225}{225}=-1.
   $$

5. **(3) Compute $$17\sin B-8\tan B$$**:  
   $$
   17\sin B-8\tan B=17\left(-\frac{15}{17}\right)-8\left(\frac{15}{8}\right)=-15-15=-30.
   $$

---

**Final Answers**:

**(d)**  
1. $$\tan A=-\frac{12}{5}$$  
2. $$\cos^2 A-\sin^2 A=-\frac{119}{169}$$  
3. $$13\cos A+5\tan A=-17$$

**(e)**  
1. $$\sin B=-\frac{15}{17}$$  
2. $$\frac{1}{\tan^2B}-\frac{1}{\sin^2B}=-1$$  
3. $$17\sin B-8\tan B=-30$$
**(d)** 1. $$\tan A = -\frac{12}{5}$$ 2. $$\cos^2 A - \sin^2 A = -\frac{119}{169}$$ 3. $$13\cos A + 5\tan A = -17$$ **(e)** 1. $$\sin B = -\frac{15}{17}$$ 2. $$\frac{1}{\tan^2 B} - \frac{1}{\sin^2 B} = -1$$ 3. $$17\sin B - 8\tan B = -30$$

(d) If and , determine without the use of a calculator and with the
aid of a diagram the value of:

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(d) If and , determine without the use of a calculator and with the aid of a diagram the value of: