Answer :
To solve the given problem step-by-step, let's start with the given equation and use fundamental trigonometric identities. Here is how it is done:
1. Given Equation:
[tex]\[ 5 \tan \theta = 2 \][/tex]
From this, we can express [tex]\(\tan \theta\)[/tex] as:
[tex]\[ \tan \theta = \frac{2}{5} \][/tex]
2. Using the Identity for [tex]\(\tan \theta\)[/tex]:
The tangent of an angle [tex]\(\theta\)[/tex] is the ratio of the sine and cosine of that angle:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Substituting [tex]\(\tan \theta = \frac{2}{5}\)[/tex], we get:
[tex]\[ \frac{\sin \theta}{\cos \theta} = \frac{2}{5} \][/tex]
This implies:
[tex]\[ \sin \theta = \frac{2}{5} \cos \theta \][/tex]
3. Using the Pythagorean Identity:
We know from the Pythagorean identity that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Substituting [tex]\(\sin \theta = \frac{2}{5} \cos \theta\)[/tex]:
[tex]\[ \left(\frac{2}{5} \cos \theta\right)^2 + \cos^2 \theta = 1 \][/tex]
Simplifying this equation:
[tex]\[ \frac{4}{25} \cos^2 \theta + \cos^2 \theta = 1 \][/tex]
Combine like terms:
[tex]\[ \frac{4}{25} \cos^2 \theta + \frac{25}{25} \cos^2 \theta = 1 \][/tex]
[tex]\[ \left(\frac{4 + 25}{25}\right) \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{29}{25} \cos^2 \theta = 1 \][/tex]
Solving for [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos^2 \theta = \frac{25}{29} \][/tex]
[tex]\[ \cos \theta = \frac{5}{\sqrt{29}} \][/tex]
4. Finding [tex]\(\sin \theta\)[/tex]:
Using [tex]\(\sin \theta = \frac{2}{5} \cos \theta\)[/tex]:
[tex]\[ \sin \theta = \frac{2}{5} \cdot \frac{5}{\sqrt{29}} = \frac{2}{\sqrt{29}} \][/tex]
5. Substitute [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] into the Expression:
We need to find the value of:
[tex]\[ \frac{10 \sin \theta - 2 \cos \theta}{5 \sin \theta + 3 \cos \theta} \][/tex]
Substituting the values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sin \theta = \frac{2}{\sqrt{29}}, \quad \cos \theta = \frac{5}{\sqrt{29}} \][/tex]
First, calculate the numerator:
[tex]\[ 10 \sin \theta - 2 \cos \theta = 10 \left(\frac{2}{\sqrt{29}}\right) - 2 \left(\frac{5}{\sqrt{29}}\right) \][/tex]
[tex]\[ = \frac{20}{\sqrt{29}} - \frac{10}{\sqrt{29}} = \frac{20 - 10}{\sqrt{29}} = \frac{10}{\sqrt{29}} \][/tex]
Then, calculate the denominator:
[tex]\[ 5 \sin \theta + 3 \cos \theta = 5 \left(\frac{2}{\sqrt{29}}\right) + 3 \left(\frac{5}{\sqrt{29}}\right) \][/tex]
[tex]\[ = \frac{10}{\sqrt{29}} + \frac{15}{\sqrt{29}} = \frac{10 + 15}{\sqrt{29}} = \frac{25}{\sqrt{29}} \][/tex]
6. Final Result:
Combine the numerator and the denominator:
[tex]\[ \frac{10 \sin \theta - 2 \cos \theta}{5 \sin \theta + 3 \cos \theta} = \frac{\frac{10}{\sqrt{29}}}{\frac{25}{\sqrt{29}}} = \frac{10}{25} = \frac{2}{5} \][/tex]
Hence, the final value is:
[tex]\[ \boxed{0.4} \][/tex]
1. Given Equation:
[tex]\[ 5 \tan \theta = 2 \][/tex]
From this, we can express [tex]\(\tan \theta\)[/tex] as:
[tex]\[ \tan \theta = \frac{2}{5} \][/tex]
2. Using the Identity for [tex]\(\tan \theta\)[/tex]:
The tangent of an angle [tex]\(\theta\)[/tex] is the ratio of the sine and cosine of that angle:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Substituting [tex]\(\tan \theta = \frac{2}{5}\)[/tex], we get:
[tex]\[ \frac{\sin \theta}{\cos \theta} = \frac{2}{5} \][/tex]
This implies:
[tex]\[ \sin \theta = \frac{2}{5} \cos \theta \][/tex]
3. Using the Pythagorean Identity:
We know from the Pythagorean identity that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Substituting [tex]\(\sin \theta = \frac{2}{5} \cos \theta\)[/tex]:
[tex]\[ \left(\frac{2}{5} \cos \theta\right)^2 + \cos^2 \theta = 1 \][/tex]
Simplifying this equation:
[tex]\[ \frac{4}{25} \cos^2 \theta + \cos^2 \theta = 1 \][/tex]
Combine like terms:
[tex]\[ \frac{4}{25} \cos^2 \theta + \frac{25}{25} \cos^2 \theta = 1 \][/tex]
[tex]\[ \left(\frac{4 + 25}{25}\right) \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{29}{25} \cos^2 \theta = 1 \][/tex]
Solving for [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos^2 \theta = \frac{25}{29} \][/tex]
[tex]\[ \cos \theta = \frac{5}{\sqrt{29}} \][/tex]
4. Finding [tex]\(\sin \theta\)[/tex]:
Using [tex]\(\sin \theta = \frac{2}{5} \cos \theta\)[/tex]:
[tex]\[ \sin \theta = \frac{2}{5} \cdot \frac{5}{\sqrt{29}} = \frac{2}{\sqrt{29}} \][/tex]
5. Substitute [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] into the Expression:
We need to find the value of:
[tex]\[ \frac{10 \sin \theta - 2 \cos \theta}{5 \sin \theta + 3 \cos \theta} \][/tex]
Substituting the values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sin \theta = \frac{2}{\sqrt{29}}, \quad \cos \theta = \frac{5}{\sqrt{29}} \][/tex]
First, calculate the numerator:
[tex]\[ 10 \sin \theta - 2 \cos \theta = 10 \left(\frac{2}{\sqrt{29}}\right) - 2 \left(\frac{5}{\sqrt{29}}\right) \][/tex]
[tex]\[ = \frac{20}{\sqrt{29}} - \frac{10}{\sqrt{29}} = \frac{20 - 10}{\sqrt{29}} = \frac{10}{\sqrt{29}} \][/tex]
Then, calculate the denominator:
[tex]\[ 5 \sin \theta + 3 \cos \theta = 5 \left(\frac{2}{\sqrt{29}}\right) + 3 \left(\frac{5}{\sqrt{29}}\right) \][/tex]
[tex]\[ = \frac{10}{\sqrt{29}} + \frac{15}{\sqrt{29}} = \frac{10 + 15}{\sqrt{29}} = \frac{25}{\sqrt{29}} \][/tex]
6. Final Result:
Combine the numerator and the denominator:
[tex]\[ \frac{10 \sin \theta - 2 \cos \theta}{5 \sin \theta + 3 \cos \theta} = \frac{\frac{10}{\sqrt{29}}}{\frac{25}{\sqrt{29}}} = \frac{10}{25} = \frac{2}{5} \][/tex]
Hence, the final value is:
[tex]\[ \boxed{0.4} \][/tex]
