If \(\rm \frac{21\ cosA+3\ sinA}{3\ cosA+4\ sinA}\) = 2, then find the value of cot A
Solving Trigonometric Equations to Find cot A
The problem asks us to find the value of \(\cot A\) given a specific trigonometric equation relating \(\cos A\) and \(\sin A\). The given equation is:
\(\frac{21\ cosA+3\ sinA}{3\ cosA+4\ sinA}\) = 2
Step-by-Step Solution to Find cot A
To solve for \(\cot A\), we first need to eliminate the fraction and rearrange the terms. We can start by multiplying both sides of the equation by the denominator \((3\ \cos A + 4\ \sin A)\):
\(21\ \cos A + 3\ \sin A = 2 \times (3\ \cos A + 4\ \sin A)\)
Now, distribute the 2 on the right side:
\(21\ \cos A + 3\ \sin A = 6\ \cos A + 8\ \sin A\)
Next, we want to group the terms involving \(\cos A\) and the terms involving \(\sin A\) on different sides of the equation. Let's move the \(\cos A\) term from the right side to the left side by subtracting \(6\ \cos A\) from both sides:
\(21\ \cos A - 6\ \cos A + 3\ \sin A = 8\ \sin A\)
\(15\ \cos A + 3\ \sin A = 8\ \sin A\)
Now, move the \(\sin A\) term from the left side to the right side by subtracting \(3\ \sin A\) from both sides:
\(15\ \cos A = 8\ \sin A - 3\ \sin A\)
\(15\ \cos A = 5\ \sin A\)
Recall that \(\cot A\) is defined as \(\frac{\cos A}{\sin A}\). To get \(\frac{\cos A}{\sin A}\), we can divide both sides of the equation \(15\ \cos A = 5\ \sin A\) by \(5\ \sin A\) (assuming \(\sin A \neq 0\), which must be true for the original expression to be defined and for \(\cot A\) to exist):
\(\frac{15\ \cos A}{5\ \sin A} = \frac{5\ \sin A}{5\ \sin A}\)
Simplify both sides:
\(3 \frac{\cos A}{\sin A} = 1\)
Substitute \(\cot A = \frac{\cos A}{\sin A}\) into the equation:
\(3\ \cot A = 1\)
Finally, solve for \(\cot A\) by dividing both sides by 3:
\(\cot A = \frac{1}{3}\)
Thus, the value of \(\cot A\) is \(\frac{1}{3}\).
| Step | Action | Equation |
|---|---|---|
| 1 | Given Equation | \(\frac{21\ cosA+3\ sinA}{3\ cosA+4\ sinA}\) = 2 |
| 2 | Cross-multiply | \(21\ \cos A + 3\ \sin A = 2(3\ \cos A + 4\ \sin A)\) |
| 3 | Expand the right side | \(21\ \cos A + 3\ \sin A = 6\ \cos A + 8\ \sin A\) |
| 4 | Group \(\cos A\) and \(\sin A\) terms | \(21\ \cos A - 6\ \cos A = 8\ \sin A - 3\ \sin A\) |
| 5 | Simplify terms | \(15\ \cos A = 5\ \sin A\) |
| 6 | Divide by \(5\ \sin A\) | \(\frac{15\ \cos A}{5\ \sin A} = \frac{5\ \sin A}{5\ \sin A}\) |
| 7 | Simplify and use \(\cot A = \frac{\cos A}{\sin A}\) | \(3\ \cot A = 1\) |
| 8 | Solve for \(\cot A\) | \(\cot A = \frac{1}{3}\) |
Revision Table: Key Trigonometric Ratios
Understanding the basic trigonometric ratios is essential for solving problems like this. The three main ratios are sine, cosine, and tangent. Their reciprocals are cosecant, secant, and cotangent.
| Ratio | Definition (Right Triangle) | Relation to Other Ratios |
|---|---|---|
| \(\sin A\) | Opposite / Hypotenuse | \(1/\csc A\) |
| \(\cos A\) | Adjacent / Hypotenuse | \(1/\sec A\) |
| \(\tan A\) | Opposite / Adjacent | \(\sin A / \cos A\), \(1/\cot A\) |
| \(\cot A\) | Adjacent / Opposite | \(\cos A / \sin A\), \(1/\tan A\) |
| \(\sec A\) | Hypotenuse / Adjacent | \(1/\cos A\) |
| \(\csc A\) | Hypotenuse / Opposite | \(1/\sin A\) |
Additional Information on Solving Trigonometric Equations
Trigonometric equations are equations that involve trigonometric functions of a variable. Solving them means finding the values of the variable that satisfy the equation. In many cases, like this problem, we use algebraic manipulation along with trigonometric identities to simplify the equation and solve for a specific trigonometric ratio.
- Look for ways to express all trigonometric functions in terms of a single function or ratio (like \(\tan A\) or \(\cot A\)).
- Use fundamental identities such as \(\sin^2 A + \cos^2 A = 1\), \(\tan A = \frac{\sin A}{\cos A}\), and \(\cot A = \frac{\cos A}{\sin A}\).
- Algebraic techniques like cross-multiplication, factorization, and rearranging terms are crucial.
- Always check for restrictions on the variable (e.g., domain) or conditions under which certain operations (like division) are valid (e.g., divisor is not zero). In this case, dividing by \(\sin A\) requires \(\sin A \neq 0\).
This problem is a straightforward application of algebraic manipulation to a trigonometric equation to find a specific ratio, \(\cot A\).
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