If `tan^(2)alpha=1+2tan^(2)beta`, then `sqrt(2)cosalpha+cosbeta` is equal to (if `0ltalphalt90^(@)and90^(@)ltbetalt180^(@))`

Options

A. `2sqrt(2)cosalpha`
B. 0
C. `2cosbeta`
D. 1

Answer

To solve the problem, we start with the given equation:

1. **Given Equation**:
\[ \tan^2 \alpha = 1 + 2 \tan^2 \beta \]

2. **Rearranging the Equation**:
We can rewrite the equation as:
\[ \tan^2 \alpha - 2 \tan^2 \beta = 1 \]

3. **Using Trigonometric Identities**:
We know that:
\[ 1 + \tan^2 \theta = \sec^2 \theta \]
Therefore, we can express \(\tan^2 \alpha\) and \(\tan^2 \beta\) in terms of secant:
\[ \sec^2 \alpha = 1 + \tan^2 \alpha \quad \text{and} \quad \sec^2 \beta = 1 + \tan^2 \beta \]

4. **Substituting into the Equation**:
From the equation \(\tan^2 \alpha = 1 + 2 \tan^2 \beta\), we can write:
\[ \sec^2 \alpha = 2 \sec^2 \beta \]

5. **Taking Square Roots**:
Taking the square root of both sides gives us:
\[ \sec \alpha = \sqrt{2} \sec \beta \]

6. **Expressing in terms of Cosine**:
Since \(\sec \theta = \frac{1}{\cos \theta}\), we can rewrite the equation as:
\[ \frac{1}{\cos \alpha} = \sqrt{2} \cdot \frac{1}{\cos \beta} \]
This simplifies to:
\[ \cos \beta = \sqrt{2} \cos \alpha \]

7. **Finding the Expression**:
We need to find the value of:
\[ \sqrt{2} \cos \alpha + \cos \beta \]
Substituting \(\cos \beta\) from the previous step:
\[ \sqrt{2} \cos \alpha + \sqrt{2} \cos \alpha = 2 \sqrt{2} \cos \alpha \]

8. **Final Result**:
Thus, we conclude that:
\[ \sqrt{2} \cos \alpha + \cos \beta = 2 \cos \beta \]