Locus of the point `(sec h theta, tan h theta)` is

Options

A. `x^(2)+y^(2)=1`
B. `x^(2)-y^(2)=1`
C. `x^(2)+y^(2)+1=0`
D. `x^(2)-y^(2)=x+y`

Answer

To find the locus of the point \((\sec h \theta, \tan h \theta)\), we will follow these steps:

### Step 1: Define the variables
Let:
- \(x = \sec h \theta\)
- \(y = \tan h \theta\)

### Step 2: Use the hyperbolic identity
We know the hyperbolic identity:
\[ \sec h^2 \theta - \tan h^2 \theta = 1 \]
This can be rewritten using our definitions of \(x\) and \(y\):
\[ x^2 - y^2 = 1 \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ x^2 - y^2 = 1 \]

### Conclusion
The locus of the point \((\sec h \theta, \tan h \theta)\) is given by the equation:
\[ x^2 - y^2 = 1 \]

### Final Answer
Thus, the correct option is:
**Option 2: \(x^2 - y^2 = 1\)**.

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