Tangents are drawn from origin to the curve `y = sin+cos x`·Then their points of contact lie on the curve

Options

A. `(1)/(x^(2))+(2)/(y^(2))=1`
B. `(2)/(x^(2))-(1)/(y^(2))=1`
C. `(2)/(x^(2))+(1)/(y^(2))=1`
D. `(2)/(y^(2))-(1)/(x^(2))=1`

Explanation

To find the locus of the points of contact of tangents drawn from the origin to the curve \( y = \sin x + \cos x \), we will follow these steps:

### Step 1: Identify the curve and point of tangency
The given curve is:
\[ y = \sin x + \cos x \]
Let the point of tangency be \( (h, k) \). Since the tangents are drawn from the origin \( (0, 0) \), we will find the slope of the tangent at the point \( (h, k) \).

### Step 2: Find the slope of the curve at the point of tangency
To find the slope of the curve at \( (h, k) \), we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \cos x - \sin x \]
At the point \( (h, k) \), the slope is:
\[ \text{slope} = \cos h - \sin h \]

### Step 3: Find the slope of the line from the origin to the point of tangency
The slope of the line connecting the origin \( (0, 0) \) to the point \( (h, k) \) is given by:
\[ \text{slope} = \frac{k - 0}{h - 0} = \frac{k}{h} \]

### Step 4: Set the slopes equal
Since both expressions represent the slope of the tangent line, we equate them:
\[ \frac{k}{h} = \cos h - \sin h \]
This can be rearranged to:
\[ k = h(\cos h - \sin h) \]
This is our first equation.

### Step 5: Substitute \( k \) in terms of \( h \) into the curve equation
Since the point \( (h, k) \) lies on the curve, we substitute \( k \) into the curve equation:
\[ k = \sin h + \cos h \]
Substituting for \( k \):
\[ h(\cos h - \sin h) = \sin h + \cos h \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ h \cos h - h \sin h - \sin h - \cos h = 0 \]
This can be simplified to:
\[ h \cos h - \cos h = h \sin h + \sin h \]
Factoring out common terms:
\[ \cos h (h - 1) = \sin h (h + 1) \]

### Step 7: Find the locus
To find the locus, we will express \( k \) in terms of \( h \):
From the earlier equation:
\[ k = h(\cos h - \sin h) \]
We can square both sides and use the Pythagorean identity:
\[ k^2 = h^2(\cos^2 h - 2\cos h \sin h + \sin^2 h) \]
Using \( \cos^2 h + \sin^2 h = 1 \):
\[ k^2 = h^2(1 - 2\cos h \sin h) \]
Now, we will replace \( h \) and \( k \) with \( x \) and \( y \):
\[ y^2 = x^2(1 - 2\cos h \sin h) \]

### Step 8: Final equation for the locus
After simplifying and rearranging, we can find the final equation for the locus of the points of contact.