What is the value of `sqrt(sec^(2)A+"cosec"^(2)A)` ?

Options

A. `secA+"cosecA`
B. cos A+ sin A
C. tan A + cot A
D. sec A + cot A

Answer

To find the value of \( \sqrt{\sec^2 A + \csc^2 A} \), we can follow these steps:

### Step 1: Recall the definitions of secant and cosecant
We know that:
\[ \sec^2 A = 1 + \tan^2 A \]
\[ \csc^2 A = 1 + \cot^2 A \]

### Step 2: Substitute the identities into the expression
Now we can substitute these identities into the expression:
\[ \sec^2 A + \csc^2 A = (1 + \tan^2 A) + (1 + \cot^2 A) \]

### Step 3: Simplify the expression
This simplifies to:
\[ \sec^2 A + \csc^2 A = 2 + \tan^2 A + \cot^2 A \]

### Step 4: Use the identity for \( \tan^2 A + \cot^2 A \)
We know that:
\[ \tan^2 A + \cot^2 A = \frac{\sin^2 A}{\cos^2 A} + \frac{\cos^2 A}{\sin^2 A} \]
This can be rewritten as:
\[ \tan^2 A + \cot^2 A = \frac{\sin^4 A + \cos^4 A}{\sin^2 A \cos^2 A} \]
Using the identity \( \sin^4 A + \cos^4 A = (\sin^2 A + \cos^2 A)^2 - 2\sin^2 A \cos^2 A = 1 - 2\sin^2 A \cos^2 A \):
\[ \tan^2 A + \cot^2 A = \frac{1 - 2\sin^2 A \cos^2 A}{\sin^2 A \cos^2 A} \]

### Step 5: Substitute back into the expression
Now substituting back, we have:
\[ \sec^2 A + \csc^2 A = 2 + \tan^2 A + \cot^2 A = 2 + \frac{1 - 2\sin^2 A \cos^2 A}{\sin^2 A \cos^2 A} \]

### Step 6: Find the square root
Now we need to find:
\[ \sqrt{\sec^2 A + \csc^2 A} \]
We can simplify this further:
\[ \sqrt{2 + \tan^2 A + \cot^2 A} = \sqrt{2 + \frac{1 - 2\sin^2 A \cos^2 A}{\sin^2 A \cos^2 A}} \]

However, we can also use the identity directly:
\[ \tan^2 A + \cot^2 A = \tan^2 A + \frac{1}{\tan^2 A} = \left(\tan A + \frac{1}{\tan A}\right)^2 - 2 \]
Thus,
\[ \sec^2 A + \csc^2 A = 2 + \tan^2 A + \cot^2 A = 2 + \left(\tan A + \cot A\right)^2 - 2 = \left(\tan A + \cot A\right)^2 \]

### Final Result
Finally, we have:
\[ \sqrt{\sec^2 A + \csc^2 A} = \tan A + \cot A \]