If A is an acute angle and cot A + cosec A = 3, then the value of sin A is:

Given: cot A + cosec A = 3, and A is an acute angle.

Step 1: Express cot A and cosec A in terms of sin A:

cot A = \frac{\cos A}{\sin A}, \quad \csc A = \frac{1}{\sin A}

Step 2: Use identity \cos A = \sqrt{1 - \sin^2 A} (since A is acute, cos A > 0).

Substitute into the equation:

\frac{\sqrt{1 - \sin^2 A}}{\sin A} + \frac{1}{\sin A} = 3

Step 3: Let x = sin A. Then:

\frac{\sqrt{1 - x^2}}{x} + \frac{1}{x} = 3

Multiply both sides by x:

\sqrt{1 - x^2} + 1 = 3x

Step 4: Isolate the square root:

\sqrt{1 - x^2} = 3x - 1

Step 5: Square both sides:

1 - x^2 = (3x - 1)^2 = 9x^2 - 6x + 1

Step 6: Simplify:

1 - x^2 = 9x^2 - 6x + 1

Bring all terms to one side:

-x^2 - 9x^2 + 6x + 1 - 1 = 0

-10x^2 + 6x = 0

Factor:

2x(-5x + 3) = 0

Solutions: x = 0 or x = 3/5

Step 7: Check validity:

x = 0: sin A = 0 → A = 0°, but cot A and cosec A are undefined. Invalid.

x = 3/5: sin A = 3/5 → check in original equation:

cos A = \sqrt{1 - (3/5)^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5

cot A = (4/5)/(3/5) = 4/3, cosec A = 5/3

cot A + cosec A = 4/3 + 5/3 = 9/3 = 3. Valid.

Thus, sin A = 3/5.

हिन्दी उत्तर:

दिया गया है: cot A + cosec A = 3, और A एक न्यून कोण है।

चरण 1: cot A और cosec A को sin A के रूप में व्यक्त करें:

cot A = \frac{\cos A}{\sin A}, \quad \csc A = \frac{1}{\sin A}

चरण 2: पहचान का उपयोग करें \cos A = \sqrt{1 - \sin^2 A} (क्योंकि A न्यून कोण है, cos A > 0)।

समीकरण में प्रतिस्थापित करें:

\frac{\sqrt{1 - \sin^2 A}}{\sin A} + \frac{1}{\sin A} = 3

चरण 3: मान लें x = sin A। तब:

\frac{\sqrt{1 - x^2}}{x} + \frac{1}{x} = 3

दोनों तरफ x से गुणा करें:

\sqrt{1 - x^2} + 1 = 3x

चरण 4: वर्गमूल को अलग करें:

\sqrt{1 - x^2} = 3x - 1

चरण 5: दोनों तरफ वर्ग करें:

1 - x^2 = (3x - 1)^2 = 9x^2 - 6x + 1

चरण 6: सरल करें:

1 - x^2 = 9x^2 - 6x + 1

सभी पदों को एक तरफ लाएँ:

-x^2 - 9x^2 + 6x + 1 - 1 = 0

-10x^2 + 6x = 0

गुणनखंड करें:

2x(-5x + 3) = 0

समाधान: x = 0 या x = 3/5

चरण 7: वैधता की जाँच करें:

x = 0: sin A = 0 → A = 0°, लेकिन cot A और cosec A परिभाषित नहीं हैं। अवैध।

x = 3/5: sin A = 3/5 → मूल समीकरण में जाँच करें:

cos A = \sqrt{1 - (3/5)^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5

cot A = (4/5)/(3/5) = 4/3, cosec A = 5/3

cot A + cosec A = 4/3 + 5/3 = 9/3 = 3. वैध।

इसलिए, sin A = 3/5।