If `15tan^(2)theta+4sec^(2)theta=23`, then find the value of `(sec theta+"cosec "theta)^(2)-sin^(2)theta`.

Options

A. `(13)/(sqrt(12))`
B. `(13)/(2)`
C. `(15)/(2)`
D. `(15)/(sqrt(2))`

Answer

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

**Step 1: Write down the given equation.**
\[ 15\tan^2\theta + 4\sec^2\theta = 23 \]

**Step 2: Use the identity for \(\sec^2\theta\) and \(\tan^2\theta\).**
We know that:
\[ \sec^2\theta = 1 + \tan^2\theta \]
Let's substitute \(\sec^2\theta\) in the equation:
\[ 15\tan^2\theta + 4(1 + \tan^2\theta) = 23 \]

**Step 3: Simplify the equation.**
Expanding the equation gives:
\[ 15\tan^2\theta + 4 + 4\tan^2\theta = 23 \]
Combine like terms:
\[ (15 + 4)\tan^2\theta + 4 = 23 \]
\[ 19\tan^2\theta + 4 = 23 \]

**Step 4: Isolate \(\tan^2\theta\).**
Subtract 4 from both sides:
\[ 19\tan^2\theta = 19 \]
Now divide by 19:
\[ \tan^2\theta = 1 \]

**Step 5: Find \(\tan\theta\).**
Taking the square root gives:
\[ \tan\theta = 1 \]
This implies:
\[ \theta = 45^\circ \text{ or } \theta = 225^\circ \]

**Step 6: Find \(\sec\theta\) and \(\csc\theta\).**
Using \(\theta = 45^\circ\):
\[ \sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]
\[ \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]

**Step 7: Calculate \( (\sec\theta + \csc\theta)^2 - \sin^2\theta \).**
Now we can find:
\[ \sec\theta + \csc\theta = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \]
Now square this:
\[ (2\sqrt{2})^2 = 4 \cdot 2 = 8 \]

**Step 8: Find \(\sin^2\theta\).**
Since \(\sin 45^\circ = \frac{1}{\sqrt{2}}\):
\[ \sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

**Step 9: Substitute and simplify.**
Now substitute back into the expression:
\[ 8 - \frac{1}{2} = 8 - 0.5 = 7.5 \]

Thus, the final answer is:
\[ \boxed{7.5} \]