What is the value of following? \(\rm cot \left[sin^{-1} \frac{3}{5}+cot^{-1}\frac{3}{2} \right]\)
Evaluating Trigonometric Expressions with Inverse Functions
The question asks for the value of the trigonometric expression \( \cot \left[\sin^{-1} \frac{3}{5}+\cot^{-1}\frac{3}{2} \right] \). This involves the sum of two inverse trigonometric functions inside a cotangent function. To solve this, we can use trigonometric identities and the properties of inverse trigonometric functions.
Let the expression inside the cotangent function be the sum of two angles. Let:
- \( A = \sin^{-1} \frac{3}{5} \)
- \( B = \cot^{-1} \frac{3}{2} \)
The problem then becomes finding the value of \( \cot(A+B) \).
Step 1: Convert Inverse Sine to Tangent
For the first angle, \( A = \sin^{-1} \frac{3}{5} \), we have \( \sin A = \frac{3}{5} \). Since the value \(\frac{3}{5}\) is positive, and the principal value branch of \( \sin^{-1}x \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), the angle \( A \) must lie in the first quadrant \( \left(0, \frac{\pi}{2}\right] \). In the first quadrant, all trigonometric ratios are positive.
We can find \( \cos A \) using the identity \( \sin^2 A + \cos^2 A = 1 \):
\( \cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25-9}{25} = \frac{16}{25} \)
Since \( A \) is in the first quadrant, \( \cos A \) is positive:
\( \cos A = \sqrt{\frac{16}{25}} = \frac{4}{5} \)
Now, we can find \( \tan A \):
\( \tan A = \frac{\sin A}{\cos A} = \frac{3/5}{4/5} = \frac{3}{4} \)
Step 2: Convert Inverse Cotangent to Tangent
For the second angle, \( B = \cot^{-1} \frac{3}{2} \), we have \( \cot B = \frac{3}{2} \). Since the value \(\frac{3}{2}\) is positive, and the principal value branch of \( \cot^{-1}x \) is \( (0, \pi) \), the angle \( B \) must lie in the first quadrant \( \left(0, \frac{\pi}{2}\right) \). In the first quadrant, all trigonometric ratios are positive.
We can find \( \tan B \) directly using the identity \( \tan B = \frac{1}{\cot B} \):
\( \tan B = \frac{1}{3/2} = \frac{2}{3} \)
Step 3: Use the Tangent Addition Formula
We need to evaluate \( \cot(A+B) \). It's easier to calculate \( \tan(A+B) \) first and then take its reciprocal. The tangent addition formula is:
\( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Substitute the values of \( \tan A \) and \( \tan B \) we found:
\( \tan(A+B) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left(\frac{3}{4}\right)\left(\frac{2}{3}\right)} \)
Calculate the numerator:
\( \frac{3}{4} + \frac{2}{3} = \frac{3 \times 3 + 2 \times 4}{4 \times 3} = \frac{9 + 8}{12} = \frac{17}{12} \)
Calculate the denominator:
\( 1 - \left(\frac{3}{4}\right)\left(\frac{2}{3}\right) = 1 - \frac{6}{12} = 1 - \frac{1}{2} = \frac{2-1}{2} = \frac{1}{2} \)
Now, substitute these back into the tangent formula:
\( \tan(A+B) = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \times \frac{2}{1} = \frac{17 \times 2}{12} = \frac{34}{12} = \frac{17}{6} \)
Step 4: Find the Cotangent of the Sum
Finally, we need to find \( \cot(A+B) \), which is the reciprocal of \( \tan(A+B) \):
\( \cot(A+B) = \frac{1}{\tan(A+B)} = \frac{1}{17/6} = \frac{6}{17} \)
Thus, the value of \( \cot \left[\sin^{-1} \frac{3}{5}+\cot^{-1}\frac{3}{2} \right] \) is \( \frac{6}{17} \).
| Inverse Function | Given Value | Equivalent Tangent |
|---|---|---|
| \( \sin^{-1} \frac{3}{5} \) | \( \sin A = \frac{3}{5} \) | \( \tan A = \frac{3}{4} \) |
| \( \cot^{-1} \frac{3}{2} \) | \( \cot B = \frac{3}{2} \) | \( \tan B = \frac{2}{3} \) |
Revision Table: Key Trigonometric Concepts
| Concept | Formula/Property |
|---|---|
| Pythagorean Identity | \( \sin^2 \theta + \cos^2 \theta = 1 \) |
| Tangent from Sine and Cosine | \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) |
| Tangent and Cotangent Relation | \( \tan \theta = \frac{1}{\cot \theta} \) |
| Principal Value Branch of \( \sin^{-1}x \) | \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) |
| Principal Value Branch of \( \cot^{-1}x \) | \( (0, \pi) \) |
| Tangent Addition Formula | \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \) |
| Cotangent and Tangent Relation | \( \cot(A+B) = \frac{1}{\tan(A+B)} \) |
Additional Information: Understanding Inverse Trigonometric Functions
Inverse trigonometric functions, also known as arc functions, are the inverses of the basic trigonometric functions (sine, cosine, tangent, etc.). They are used to find the angle when the value of the trigonometric ratio is known. For example, \( \sin^{-1}x \) gives the angle whose sine is \( x \). Since trigonometric functions are periodic, their inverses are multi-valued. To make them single-valued, we define principal value branches.
- \( \sin^{-1}x \): The angle \( \theta \) in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) such that \( \sin \theta = x \).
- \( \cos^{-1}x \): The angle \( \theta \) in \( [0, \pi] \) such that \( \cos \theta = x \).
- \( \tan^{-1}x \): The angle \( \theta \) in \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) such that \( \tan \theta = x \).
- \( \cot^{-1}x \): The angle \( \theta \) in \( (0, \pi) \) such that \( \cot \theta = x \).
When working with sums or differences of inverse trigonometric functions, it is often helpful to convert them to the same type of inverse function (like \( \tan^{-1} \)) or find the tangent/cotangent of the individual angles and then use the sum/difference formulas for tangent or cotangent.
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