If `alpha+beta=90^(@)` , then the expression `(tanalpha)/(tanbeta)+sin^(2)alpha+sin^(2)beta` is equal to :

Options

A. `sec^(2)beta`
B. `tan^(2)alpha`
C. `tan^(2)beta`
D. `sec^(2)alpha`

Answer

To solve the expression \(\frac{\tan \alpha}{\tan \beta} + \sin^2 \alpha + \sin^2 \beta\) given that \(\alpha + \beta = 90^\circ\), we can follow these steps:

### Step 1: Use the identity for \(\tan\)
Since \(\alpha + \beta = 90^\circ\), we can express \(\beta\) as:
\[ \beta = 90^\circ - \alpha \]
Using the tangent identity:
\[ \tan(90^\circ - \theta) = \cot(\theta) \]
we can rewrite \(\tan \beta\):
\[ \tan \beta = \tan(90^\circ - \alpha) = \cot \alpha = \frac{1}{\tan \alpha} \]

### Step 2: Substitute \(\tan \beta\) in the expression
Now substitute \(\tan \beta\) into the expression:
\[ \frac{\tan \alpha}{\tan \beta} = \frac{\tan \alpha}{\cot \alpha} = \tan^2 \alpha \]
So, the expression becomes:
\[ \tan^2 \alpha + \sin^2 \alpha + \sin^2 \beta \]

### Step 3: Use the identity for \(\sin^2\)
Using the identity \(\sin(90^\circ - \theta) = \cos(\theta)\), we can express \(\sin^2 \beta\):
\[ \sin^2 \beta = \sin^2(90^\circ - \alpha) = \cos^2 \alpha \]
Now, we can rewrite the expression:
\[ \tan^2 \alpha + \sin^2 \alpha + \cos^2 \alpha \]

### Step 4: Simplify the expression
We know that \(\sin^2 \alpha + \cos^2 \alpha = 1\). Therefore, we can simplify the expression:
\[ \tan^2 \alpha + 1 \]

### Step 5: Use the identity for \(\tan^2\)
Using the identity:
\[ \tan^2 \alpha = \sec^2 \alpha - 1 \]
we can rewrite the expression as:
\[ \tan^2 \alpha + 1 = \sec^2 \alpha \]

### Final Result
Thus, the final result is:
\[ \sec^2 \alpha \]