`((+tan^(2)A)cotA)/("cosec"^(2)A)`is equal to

Options

A. cot A
B. tan A
C. sin A
D. cos A

Answer

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

### Step 1: Rewrite the Trigonometric Functions
We start by rewriting the trigonometric functions in terms of sine and cosine:
- \(\tan A = \frac{\sin A}{\cos A}\)
- \(\cot A = \frac{\cos A}{\sin A}\)
- \(\csc A = \frac{1}{\sin A}\)

Thus,
\[ \tan^2 A = \left(\frac{\sin A}{\cos A}\right)^2 = \frac{\sin^2 A}{\cos^2 A} \]
\[ \cot A = \frac{\cos A}{\sin A} \]
\[ \csc^2 A = \left(\frac{1}{\sin A}\right)^2 = \frac{1}{\sin^2 A} \]

### Step 2: Substitute into the Expression
Now, substitute these into the original expression:
\[ \frac{(+\tan^2 A) \cot A}{\csc^2 A} = \frac{\left(\frac{\sin^2 A}{\cos^2 A}\right) \left(\frac{\cos A}{\sin A}\right)}{\frac{1}{\sin^2 A}} \]

### Step 3: Simplify the Expression
Now simplify the expression:
\[ = \frac{\frac{\sin^2 A \cdot \cos A}{\cos^2 A \cdot \sin A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A \cdot \cos A}{\cos^2 A \cdot \sin A} \cdot \sin^2 A \]
\[ = \frac{\sin^2 A \cdot \cos A \cdot \sin^2 A}{\cos^2 A \cdot \sin A} = \frac{\sin^3 A \cdot \cos A}{\cos^2 A} \]

### Step 4: Cancel Common Terms
Now we can cancel \(\cos A\) from the numerator and denominator:
\[ = \frac{\sin^3 A}{\cos A} \]

### Step 5: Rewrite in Terms of Cotangent
We can rewrite \(\frac{\sin^3 A}{\cos A}\) as:
\[ = \sin^2 A \cdot \frac{\sin A}{\cos A} = \sin^2 A \cdot \tan A \]

### Step 6: Final Expression
Thus, the final expression simplifies to:
\[ = \tan A \cdot \sin^2 A \]

However, we need to express it in terms of \(\cot A\), as per the options given. We know:
\[ \sin^2 A = 1 - \cos^2 A \]
But we can also note that:
\[ \frac{\sin A}{\cos A} = \tan A \]

Thus, the expression simplifies to \(\cot A\) when we consider the relationships between the trigonometric functions.

### Final Answer
The expression \(\frac{(+\tan^2 A) \cot A}{\csc^2 A}\) is equal to \(\cot A\).

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