1. For the questions below, $$ A $$ is acute. You may not use a calculator. a) If $$ \sin A=\frac{4}{5} $$, find the value of $$ \cot A+ an A $$. b) If $$ \sin A=\frac{3}{8} $$, find the value of $$ \operatorname{cosec} A-\sec A $$. c) If $$ 13 \cos A=12 $$, find the value of $$ \sec A+\sin A $$. d) If $$ 9 \cos A-7=0 $$, find the value of $$ \operatorname{cosec} A $$. e) If $$ \cos A=\frac{a}{b} $$, prove that $$ \frac{\sin A \cdot an A}{\sec A-1}=\frac{a+b}{b} $$.

Question

1. For the questions below, $$ A $$ is acute. You may not use a calculator. a) If $$ \sin A=\frac{4}{5} $$, find the value of $$ \cot A+	an A $$. b) If $$ \sin A=\frac{3}{8} $$, find the value of $$ \operatorname{cosec} A-\sec A $$. c) If $$ 13 \cos A=12 $$, find the value of $$ \sec A+\sin A $$. d) If $$ 9 \cos A-7=0 $$, find the value of $$ \operatorname{cosec} A $$. e) If $$ \cos A=\frac{a}{b} $$, prove that $$ \frac{\sin A \cdot 	an A}{\sec A-1}=\frac{a+b}{b} $$.

UpStudy AI · Accepted Answer

**(a)**  
Given that  
$$
\sin A=\frac{4}{5},
$$  
we calculate $$\cos A$$ using  
$$
\cos A=\sqrt{1-\sin^2 A}=\sqrt{1-\left(\frac{4}{5}\right)^2}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}.
$$  
Now,  
$$
\tan A=\frac{\sin A}{\cos A}=\frac{4/5}{3/5}=\frac{4}{3},\quad \cot A=\frac{1}{\tan A}=\frac{3}{4}.
$$  
Thus,  
$$
\cot A+\tan A=\frac{3}{4}+\frac{4}{3}=\frac{3\cdot3+4\cdot4}{12}=\frac{9+16}{12}=\frac{25}{12}.
$$

---

**(b)**  
Given  
$$
\sin A=\frac{3}{8},
$$  
find $$\cos A$$ using  
$$
\cos A=\sqrt{1-\sin^2 A}=\sqrt{1-\left(\frac{3}{8}\right)^2}=\sqrt{1-\frac{9}{64}}=\sqrt{\frac{55}{64}}=\frac{\sqrt{55}}{8}.
$$  
Then,  
$$
\operatorname{cosec} A=\frac{1}{\sin A}=\frac{8}{3},\quad \sec A=\frac{1}{\cos A}=\frac{8}{\sqrt{55}}.
$$  
Therefore,  
$$
\operatorname{cosec} A-\sec A=\frac{8}{3}-\frac{8}{\sqrt{55}}.
$$

---

**(c)**  
We are given  
$$
13\cos A=12\quad\Longrightarrow\quad \cos A=\frac{12}{13}.
$$  
Since $$A$$ is acute,  
$$
\sin A=\sqrt{1-\cos^2 A}=\sqrt{1-\left(\frac{12}{13}\right)^2}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}.
$$  
Now,  
$$
\sec A=\frac{1}{\cos A}=\frac{13}{12}.
$$  
Thus,  
$$
\sec A+\sin A=\frac{13}{12}+\frac{5}{13}=\frac{13\cdot13+5\cdot12}{12\cdot13}=\frac{169+60}{156}=\frac{229}{156}.
$$

---

**(d)**  
We have  
$$
9\cos A-7=0\quad\Longrightarrow\quad \cos A=\frac{7}{9}.
$$  
Then,  
$$
\sin A=\sqrt{1-\cos^2 A}=\sqrt{1-\left(\frac{7}{9}\right)^2}=\sqrt{1-\frac{49}{81}}=\sqrt{\frac{32}{81}}=\frac{4\sqrt{2}}{9}.
$$  
So,  
$$
\operatorname{cosec} A=\frac{1}{\sin A}=\frac{9}{4\sqrt{2}}=\frac{9\sqrt{2}}{8}.
$$

---

**(e)**  
Assume  
$$
\cos A=\frac{a}{b}.
$$  
We need to prove that  
$$
\frac{\sin A\cdot \tan A}{\sec A-1}=\frac{a+b}{b}.
$$

**Step 1.** Express $$\tan A$$ and $$\sec A$$ in terms of $$\sin A$$ and $$\cos A$$:  
$$
\tan A=\frac{\sin A}{\cos A},\quad \sec A=\frac{1}{\cos A}.
$$

**Step 2.** Substitute into the left-hand side:  
$$
\frac{\sin A\cdot \tan A}{\sec A-1}=\frac{\sin A\cdot\frac{\sin A}{\cos A}}{\frac{1}{\cos A}-1}=\frac{\frac{\sin^2 A}{\cos A}}{\frac{1-\cos A}{\cos A}}=\frac{\sin^2 A}{1-\cos A}.
$$

**Step 3.** Use the Pythagorean identity:  
$$
\sin^2 A=1-\cos^2 A=(1-\cos A)(1+\cos A).
$$

Substitute this into the expression:  
$$
\frac{\sin^2 A}{1-\cos A}=\frac{(1-\cos A)(1+\cos A)}{1-\cos A}=1+\cos A.
$$

**Step 4.** Replace $$\cos A$$ with $$\frac{a}{b}$$:  
$$
1+\cos A=1+\frac{a}{b}=\frac{b+a}{b}.
$$

Thus,  
$$
\frac{\sin A\cdot \tan A}{\sec A-1}=\frac{a+b}{b},
$$  
which is what we needed to prove.
**(a)** $$ \cot A + \tan A = \frac{25}{12} $$ **(b)** $$ \operatorname{cosec} A - \sec A = \frac{8}{3} - \frac{8}{\sqrt{55}} $$ **(c)** $$ \sec A + \sin A = \frac{229}{156} $$ **(d)** $$ \operatorname{cosec} A = \frac{9\sqrt{2}}{8} $$ **(e)** $$ \frac{\sin A \cdot \tan A}{\sec A - 1} = \frac{a + b}{b} $$

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