`tanalpha=2` , then the value of `("cosec"^(2)alpha-sec^(2)alpha)/("cosec"^(2)alpha+sec^(2)alpha)` is

Options

A. `-(15)/(9)`
B. `-(3)/(5)`
C. `(3)/(5)`
D. `(17)/(5)`

Answer

To solve the problem where \( \tan \alpha = 2 \), we need to find the value of

\[ \frac{\csc^2 \alpha - \sec^2 \alpha}{\csc^2 \alpha + \sec^2 \alpha}. \]

### Step 1: Express \(\csc^2 \alpha\) and \(\sec^2 \alpha\) in terms of \(\tan \alpha\)

We know the following identities:
- \(\csc^2 \alpha = 1 + \cot^2 \alpha\)
- \(\sec^2 \alpha = 1 + \tan^2 \alpha\)

Since \(\tan \alpha = 2\), we can find \(\tan^2 \alpha\):
\[ \tan^2 \alpha = 2^2 = 4. \]

Now, we can find \(\sec^2 \alpha\):
\[ \sec^2 \alpha = 1 + \tan^2 \alpha = 1 + 4 = 5. \]

Next, we need to find \(\cot^2 \alpha\):
\[ \cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{2} \implies \cot^2 \alpha = \left(\frac{1}{2}\right)^2 = \frac{1}{4}. \]

Now, we can find \(\csc^2 \alpha\):
\[ \csc^2 \alpha = 1 + \cot^2 \alpha = 1 + \frac{1}{4} = \frac{5}{4}. \]

### Step 2: Substitute \(\csc^2 \alpha\) and \(\sec^2 \alpha\) into the expression

Now we substitute \(\csc^2 \alpha\) and \(\sec^2 \alpha\) into the expression:
\[ \frac{\csc^2 \alpha - \sec^2 \alpha}{\csc^2 \alpha + \sec^2 \alpha} = \frac{\frac{5}{4} - 5}{\frac{5}{4} + 5}. \]

### Step 3: Simplify the numerator and denominator

First, simplify the numerator:
\[ \frac{5}{4} - 5 = \frac{5}{4} - \frac{20}{4} = \frac{5 - 20}{4} = \frac{-15}{4}. \]

Now simplify the denominator:
\[ \frac{5}{4} + 5 = \frac{5}{4} + \frac{20}{4} = \frac{5 + 20}{4} = \frac{25}{4}. \]

### Step 4: Combine the results

Now we can combine the results:
\[ \frac{\frac{-15}{4}}{\frac{25}{4}} = \frac{-15}{25} = -\frac{3}{5}. \]

### Final Answer

Thus, the value of

\[ \frac{\csc^2 \alpha - \sec^2 \alpha}{\csc^2 \alpha + \sec^2 \alpha} = -\frac{3}{5}. \]