If \(\rm tan A = {3 \over 8},\) then the value of \({3 \sin A + 2 \cos A} \over 3 \sin A - 2 \cos A\) is:
Finding the Value of a Trigonometric Expression
The question asks us to find the value of the expression \( \frac{3 \sin A + 2 \cos A}{3 \sin A - 2 \cos A} \) given that \( \tan A = \frac{3}{8} \).
We are given the value of \( \tan A \). Recall that the trigonometric identity \( \tan A = \frac{\sin A}{\cos A} \) relates sine, cosine, and tangent. We can use this relationship to simplify the given expression.
Transforming the Expression using tan A
The expression we need to evaluate is:
\[ \frac{3 \sin A + 2 \cos A}{3 \sin A - 2 \cos A} \]To introduce \( \tan A \) into this expression, we can divide both the numerator and the denominator by \( \cos A \). This is valid as long as \( \cos A \ne 0 \). If \( \cos A = 0 \), then \( \tan A \) would be undefined, which contradicts the given information \( \tan A = \frac{3}{8} \).
Dividing the numerator by \( \cos A \):
\[ \frac{3 \sin A + 2 \cos A}{\cos A} = \frac{3 \sin A}{\cos A} + \frac{2 \cos A}{\cos A} = 3 \tan A + 2 \]Dividing the denominator by \( \cos A \):
\[ \frac{3 \sin A - 2 \cos A}{\cos A} = \frac{3 \sin A}{\cos A} - \frac{2 \cos A}{\cos A} = 3 \tan A - 2 \]So the original expression transforms into:
\[ \frac{3 \tan A + 2}{3 \tan A - 2} \]Substituting the Value of tan A
Now we substitute the given value \( \tan A = \frac{3}{8} \) into the transformed expression:
\[ \frac{3 \left( \frac{3}{8} \right) + 2}{3 \left( \frac{3}{8} \right) - 2} \]First, calculate the products in the numerator and the denominator:
\[ 3 \left( \frac{3}{8} \right) = \frac{3 \times 3}{8} = \frac{9}{8} \]Substitute this back into the expression:
\[ \frac{\frac{9}{8} + 2}{\frac{9}{8} - 2} \]Next, we need to add and subtract the fractions. To do this, we write the integer 2 as a fraction with a denominator of 8:
\[ 2 = \frac{2 \times 8}{8} = \frac{16}{8} \]Now substitute \( 2 = \frac{16}{8} \) into the expression:
\[ \frac{\frac{9}{8} + \frac{16}{8}}{\frac{9}{8} - \frac{16}{8}} \]Perform the addition in the numerator and the subtraction in the denominator:
Numerator: \( \frac{9}{8} + \frac{16}{8} = \frac{9 + 16}{8} = \frac{25}{8} \)
Denominator: \( \frac{9}{8} - \frac{16}{8} = \frac{9 - 16}{8} = \frac{-7}{8} \)
So the expression becomes:
\[ \frac{\frac{25}{8}}{\frac{-7}{8}} \]To divide these fractions, we multiply the numerator by the reciprocal of the denominator:
\[ \frac{25}{8} \times \frac{8}{-7} \]We can cancel out the 8 in the numerator and denominator:
\[ \frac{25}{\cancel{8}} \times \frac{\cancel{8}}{-7} = \frac{25}{-7} \]The value of the expression is \( - \frac{25}{7} \).
Conclusion
Given \( \tan A = \frac{3}{8} \), the value of \( \frac{3 \sin A + 2 \cos A}{3 \sin A - 2 \cos A} \) is \( - \frac{25}{7} \).
| Step | Process | Result |
|---|---|---|
| 1 | Transform expression by dividing numerator and denominator by \( \cos A \) | \( \frac{3 \tan A + 2}{3 \tan A - 2} \) |
| 2 | Substitute \( \tan A = \frac{3}{8} \) | \( \frac{3(\frac{3}{8}) + 2}{3(\frac{3}{8}) - 2} \) |
| 3 | Simplify using fraction arithmetic | \( \frac{\frac{9}{8} + \frac{16}{8}}{\frac{9}{8} - \frac{16}{8}} = \frac{\frac{25}{8}}{\frac{-7}{8}} \) |
| 4 | Perform fraction division | \( \frac{25}{8} \times \frac{8}{-7} = - \frac{25}{7} \) |
Revision Table: Key Trigonometric Identities and Fraction Operations
- Tangent Identity: \( \tan A = \frac{\sin A}{\cos A} \)
- Adding/Subtracting Fractions: \( \frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd} \). If denominators are same: \( \frac{a}{b} \pm \frac{c}{b} = \frac{a \pm c}{b} \)
- Dividing Fractions: \( \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \times \frac{d}{c} \)
Additional Information: Finding sin A and cos A from tan A
If you are given \( \tan A \), you can also find the values of \( \sin A \) and \( \cos A \) directly, although it is not necessary for this specific problem. One way is to use the identity \( 1 + \tan^2 A = \sec^2 A \). Since \( \sec A = \frac{1}{\cos A} \), you can find \( \cos A \). Once you have \( \cos A \), you can use \( \sin A = \tan A \times \cos A \) to find \( \sin A \).
Alternatively, you can construct a right-angled triangle. If \( \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{8} \), you can assume the opposite side is 3 units and the adjacent side is 8 units. Using the Pythagorean theorem (\(\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2\)), the hypotenuse would be \( \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73} \). Then \( \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{\sqrt{73}} \) and \( \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{8}{\sqrt{73}} \). You could substitute these values into the original expression, which would lead to the same result but involve more complex algebra with square roots.
Similar Questions
Simplify the following.
\(\frac{\sin^3 α + \cos^3 α}{\sin α + \cos α}\)
Find the value of the following expression.
12(sin4 θ + cos4 θ) + 18(sin6 θ + cos6 θ) + 78 sin2 θ cos2 θ
Find the exact value of cos 120°.
Find the value of \(\frac{2}{3}\)tan2 60° + 3 cos2 30° − 2 sec2 30° − \(\frac{3}{4}\)cot2 60°.
Simplify \(\frac{1+sint}{4-4sint}-\frac{1-sint}{4+4sint}\)
Find the value of tan (−1125°).
ΔABC is a right triangle. If ∠B = 90° and tan A = \(\frac{1}{\sqrt2}\), then the value of sin A cos C + cos A sin C is :
If \(\rm \frac{21\ cosA+3\ sinA}{3\ cosA+4\ sinA}\) = 2, then find the value of cot A
In a right triangle for an acute angle x, if sin x = \(\frac{3}{7}\), then find the value of cosx.
If \(\rm cos x + sec x = {7 \over2 \sqrt3}\), then the value of cos2x + sec2x will be ________.
Important Questions from Trigonometry
The value of 5 sin 14° sec 76° + 3 cot 15° cot 75° + 2 tan 45° is:
If two complimentary angles are in the ratio of 4 : 5, find the greater angle.
If \(\frac{\sin\spaceθ \space+\space \cos\spaceθ} {\sin \spaceθ \space-\space \cos \spaceθ} = \frac{\sqrt3 \space-\space 1}{\sqrt3 \space+\space 1} \) , then the angle θ is
If tan α = 1/2, tan β = 1/3, then find α + β.
Simplify: sin (A + B) sin (A – B)
