If 3 `sec^(2)theta + tan theta = 7, 0^(@) lt theta lt 90^(@)`, then the value of `(cosec2theta+costheta)/(sin2theta+cottheta)` is:

Options

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

Answer

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

**Step 1:** Start with the equation:
\[ 3 \sec^2 \theta + \tan \theta = 7 \]

**Step 2:** Recall the identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Substituting this into the equation gives:
\[ 3(1 + \tan^2 \theta) + \tan \theta = 7 \]
This simplifies to:
\[ 3 + 3 \tan^2 \theta + \tan \theta = 7 \]

**Step 3:** Rearranging the equation:
\[ 3 \tan^2 \theta + \tan \theta - 4 = 0 \]

**Step 4:** Let \( x = \tan \theta \). The equation becomes:
\[ 3x^2 + x - 4 = 0 \]

**Step 5:** Now, we can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 3, b = 1, c = -4 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \]
\[ x = \frac{-1 \pm \sqrt{1 + 48}}{6} \]
\[ x = \frac{-1 \pm \sqrt{49}}{6} \]
\[ x = \frac{-1 \pm 7}{6} \]

**Step 6:** Calculating the two possible values:
1. \( x = \frac{6}{6} = 1 \)
2. \( x = \frac{-8}{6} \) (not valid as \( \tan \theta \) cannot be negative in the given range)

Thus, \( \tan \theta = 1 \) implies:
\[ \theta = 45^\circ \]

**Step 7:** Now, we need to find the value of:
\[ \frac{\csc 2\theta + \cos \theta}{\sin 2\theta + \cot \theta} \]

**Step 8:** Calculate each term:
- \( \csc 2\theta = \csc 90^\circ = 1 \)
- \( \cos \theta = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \sin 2\theta = \sin 90^\circ = 1 \)
- \( \cot \theta = \cot 45^\circ = 1 \)

**Step 9:** Substitute these values into the expression:
\[ \frac{1 + \frac{1}{\sqrt{2}}}{1 + 1} = \frac{1 + \frac{1}{\sqrt{2}}}{2} \]

**Step 10:** To simplify, multiply numerator and denominator by \( \sqrt{2} \):
\[ = \frac{\sqrt{2} + 1}{2\sqrt{2}} \]

**Step 11:** Finally, we can express this as:
\[ = \frac{2 + \sqrt{2}}{4} \]

Thus, the final answer is:
\[ \frac{2 + \sqrt{2}}{4} \]

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