What is the simplified value of `((cosecA)/(cotA+tanA))^(2)` ?

Options

A. `2cos^(2)A`
B. `1-sin^(2)A`
C. `sec^(2)A`
D. `secAtanA`

Answer

To simplify the expression \(\left(\frac{\csc A}{\cot A + \tan A}\right)^{2}\), we will follow these steps:

### Step 1: Rewrite the trigonometric functions in terms of sine and cosine.
We know that:
- \(\csc A = \frac{1}{\sin A}\)
- \(\cot A = \frac{\cos A}{\sin A}\)
- \(\tan A = \frac{\sin A}{\cos A}\)

Substituting these into the expression, we have:
\[ \frac{\csc A}{\cot A + \tan A} = \frac{\frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}} \]

### Step 2: Simplify the denominator.
The denominator \(\cot A + \tan A\) can be simplified as follows:
\[ \cot A + \tan A = \frac{\cos A}{\sin A} + \frac{\sin A}{\cos A} \]
To combine these fractions, we need a common denominator, which is \(\sin A \cos A\):
\[ \cot A + \tan A = \frac{\cos^2 A + \sin^2 A}{\sin A \cos A} \]
Using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\), we can simplify this to:
\[ \cot A + \tan A = \frac{1}{\sin A \cos A} \]

### Step 3: Substitute back into the expression.
Now substituting back into our original expression:
\[ \frac{\csc A}{\cot A + \tan A} = \frac{\frac{1}{\sin A}}{\frac{1}{\sin A \cos A}} = \frac{1}{\sin A} \cdot \frac{\sin A \cos A}{1} = \cos A \]

### Step 4: Square the result.
Now we need to square the result:
\[ \left(\cos A\right)^{2} = \cos^2 A \]

### Final Answer:
Thus, the simplified value of \(\left(\frac{\csc A}{\cot A + \tan A}\right)^{2}\) is:
\[ \cos^2 A \]

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