Answer :
Let's tackle each problem step by step.
### Problem 5
Given that [tex]\(\tan \theta = \frac{1}{\sqrt{3}}\)[/tex], we need to find the value of:
[tex]\[ \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
1. Substitute [tex]\(\tan \theta\)[/tex] with [tex]\(\frac{1}{\sqrt{3}}\)[/tex]:
[tex]\[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} \][/tex]
2. Simplify the expression:
[tex]\[ = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} \][/tex]
Therefore, the value is [tex]\(\sqrt{3}\)[/tex].
### Problem 6
Given that [tex]\(\sin \theta = \frac{4}{5}\)[/tex], we need to find the value of:
[tex]\[ 3 \sec \theta - 5 \cos \theta \][/tex]
1. First, determine [tex]\(\cos \theta\)[/tex] using the Pythagorean identity:
[tex]\[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
2. Find [tex]\(\sec \theta\)[/tex] which is the reciprocal of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \][/tex]
3. Substitute into the given expression:
[tex]\[ 3 \sec \theta - 5 \cos \theta = 3 \times \frac{5}{3} - 5 \times \frac{3}{5} = 5 - 3 = 2 \][/tex]
Therefore, the value is 2.
### Problem 7
Given that [tex]\(4 \tan \theta = 3\)[/tex], show that:
[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{6}{11} \][/tex]
1. First, solve for [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{3}{4} \][/tex]
2. Use the identity [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex], we can let:
[tex]\[ \sin \theta = 3k \quad \text{and} \quad \cos \theta = 4k \][/tex]
3. Normalize using [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[ (3k)^2 + (4k)^2 = 1 \quad \Rightarrow \quad 9k^2 + 16k^2 = 1 \quad \Rightarrow \quad 25k^2 = 1 \quad \Rightarrow \quad k = \frac{1}{5} \][/tex]
So, [tex]\(\sin \theta = \frac{3}{5}\)[/tex] and [tex]\(\cos \theta = \frac{4}{5}\)[/tex].
4. Substitute these values into the given expression:
[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{4 \cdot \frac{3}{5} + 3 \cdot \frac{4}{5}}{8 \cdot \frac{3}{5} + 5 \cdot \frac{4}{5}} = \frac{\frac{12}{5} + \frac{12}{5}}{\frac{24}{5} + \frac{20}{5}} = \frac{\frac{24}{5}}{\frac{44}{5}} = \frac{24}{44} = \frac{6}{11} \][/tex]
Hence, the expression simplifies to [tex]\(\frac{6}{11}\)[/tex], proving the equality.
### Problem 8
Given that [tex]\(\cos \theta = \frac{\sqrt{3}}{2}\)[/tex], show that:
[tex]\[ 4 \cos^3 \theta - 3 \cos \theta = 0 \][/tex]
1. Substitute [tex]\(\cos \theta\)[/tex] into the expression:
[tex]\[ 4 \left(\frac{\sqrt{3}}{2}\right)^3 - 3 \left(\frac{\sqrt{3}}{2}\right) \][/tex]
2. Calculate each term individually:
[tex]\[ \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{(\sqrt{3})^3}{2^3} = \frac{3\sqrt{3}}{8} \][/tex]
3. Now plug it back into the original equation:
[tex]\[ 4 \cdot \frac{3\sqrt{3}}{8} - 3 \cdot \frac{\sqrt{3}}{2} = \frac{12\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} = \frac{12\sqrt{3}}{8} - \frac{12\sqrt{3}}{8} = 0 \][/tex]
Therefore, we have shown that [tex]\(4 \cos^3 \theta - 3 \cos \theta = 0\)[/tex].
### Problem 5
Given that [tex]\(\tan \theta = \frac{1}{\sqrt{3}}\)[/tex], we need to find the value of:
[tex]\[ \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
1. Substitute [tex]\(\tan \theta\)[/tex] with [tex]\(\frac{1}{\sqrt{3}}\)[/tex]:
[tex]\[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} \][/tex]
2. Simplify the expression:
[tex]\[ = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} \][/tex]
Therefore, the value is [tex]\(\sqrt{3}\)[/tex].
### Problem 6
Given that [tex]\(\sin \theta = \frac{4}{5}\)[/tex], we need to find the value of:
[tex]\[ 3 \sec \theta - 5 \cos \theta \][/tex]
1. First, determine [tex]\(\cos \theta\)[/tex] using the Pythagorean identity:
[tex]\[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
2. Find [tex]\(\sec \theta\)[/tex] which is the reciprocal of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \][/tex]
3. Substitute into the given expression:
[tex]\[ 3 \sec \theta - 5 \cos \theta = 3 \times \frac{5}{3} - 5 \times \frac{3}{5} = 5 - 3 = 2 \][/tex]
Therefore, the value is 2.
### Problem 7
Given that [tex]\(4 \tan \theta = 3\)[/tex], show that:
[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{6}{11} \][/tex]
1. First, solve for [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{3}{4} \][/tex]
2. Use the identity [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex], we can let:
[tex]\[ \sin \theta = 3k \quad \text{and} \quad \cos \theta = 4k \][/tex]
3. Normalize using [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[ (3k)^2 + (4k)^2 = 1 \quad \Rightarrow \quad 9k^2 + 16k^2 = 1 \quad \Rightarrow \quad 25k^2 = 1 \quad \Rightarrow \quad k = \frac{1}{5} \][/tex]
So, [tex]\(\sin \theta = \frac{3}{5}\)[/tex] and [tex]\(\cos \theta = \frac{4}{5}\)[/tex].
4. Substitute these values into the given expression:
[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{4 \cdot \frac{3}{5} + 3 \cdot \frac{4}{5}}{8 \cdot \frac{3}{5} + 5 \cdot \frac{4}{5}} = \frac{\frac{12}{5} + \frac{12}{5}}{\frac{24}{5} + \frac{20}{5}} = \frac{\frac{24}{5}}{\frac{44}{5}} = \frac{24}{44} = \frac{6}{11} \][/tex]
Hence, the expression simplifies to [tex]\(\frac{6}{11}\)[/tex], proving the equality.
### Problem 8
Given that [tex]\(\cos \theta = \frac{\sqrt{3}}{2}\)[/tex], show that:
[tex]\[ 4 \cos^3 \theta - 3 \cos \theta = 0 \][/tex]
1. Substitute [tex]\(\cos \theta\)[/tex] into the expression:
[tex]\[ 4 \left(\frac{\sqrt{3}}{2}\right)^3 - 3 \left(\frac{\sqrt{3}}{2}\right) \][/tex]
2. Calculate each term individually:
[tex]\[ \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{(\sqrt{3})^3}{2^3} = \frac{3\sqrt{3}}{8} \][/tex]
3. Now plug it back into the original equation:
[tex]\[ 4 \cdot \frac{3\sqrt{3}}{8} - 3 \cdot \frac{\sqrt{3}}{2} = \frac{12\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} = \frac{12\sqrt{3}}{8} - \frac{12\sqrt{3}}{8} = 0 \][/tex]
Therefore, we have shown that [tex]\(4 \cos^3 \theta - 3 \cos \theta = 0\)[/tex].
