If `sqrt(2)tan2theta=sqrt(6)and0^(@)ltthetalt45^(@)`, then the value of `sintheta+sqrt(3)costheta-1tan^(2)theta` is

Options

A. `(2)/(3)`
B. `(4)/(3)`
C. 2
D. `(8)/(3)`

Answer

To solve the problem step by step, we start with the given equation and conditions.

### Step 1: Given Equation
We have:
\[ \sqrt{2} \tan(2\theta) = \sqrt{6} \]
From this, we can isolate \(\tan(2\theta)\):
\[ \tan(2\theta) = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3} \]

### Step 2: Find \(2\theta\)
We know that:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus, we can set:
\[ 2\theta = 60^\circ \]
This gives us:
\[ \theta = \frac{60^\circ}{2} = 30^\circ \]

### Step 3: Calculate \(\sin\theta\) and \(\cos\theta\)
Now we need to find \(\sin(30^\circ)\) and \(\cos(30^\circ)\):
\[ \sin(30^\circ) = \frac{1}{2} \]
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

### Step 4: Substitute into the Expression
We need to evaluate:
\[ \sin\theta + \sqrt{3} \cos\theta - \tan^2\theta \]
First, we calculate \(\tan(30^\circ)\):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \quad \Rightarrow \quad \tan^2(30^\circ) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]

Now substituting into the expression:
\[ \sin(30^\circ) + \sqrt{3} \cos(30^\circ) - \tan^2(30^\circ) \]
Substituting the values:
\[ = \frac{1}{2} + \sqrt{3} \cdot \frac{\sqrt{3}}{2} - \frac{1}{3} \]
Calculating \(\sqrt{3} \cdot \frac{\sqrt{3}}{2}\):
\[ = \frac{3}{2} \]
So we have:
\[ = \frac{1}{2} + \frac{3}{2} - \frac{1}{3} \]
Combining the first two terms:
\[ = \frac{4}{2} - \frac{1}{3} = 2 - \frac{1}{3} \]
To combine these, we convert \(2\) into a fraction:
\[ = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \]

### Final Answer
Thus, the value of \(\sin\theta + \sqrt{3} \cos\theta - \tan^2\theta\) is:
\[ \frac{5}{3} \]