If `sec^(2)theta+tan^(2)theta=7` find `sintheta,costheta&tantheta` (`theta` in 1st qudrant).

Answer

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

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

**Step 2: Use the identity**
We know that:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Substituting this into the equation gives:
\[ (1 + \tan^2 \theta) + \tan^2 \theta = 7 \]

**Step 3: Simplify the equation**
This simplifies to:
\[ 1 + 2\tan^2 \theta = 7 \]
Subtract 1 from both sides:
\[ 2\tan^2 \theta = 6 \]
Now, divide by 2:
\[ \tan^2 \theta = 3 \]

**Step 4: Find \(\tan \theta\)**
Taking the square root of both sides, we get:
\[ \tan \theta = \sqrt{3} \]
Since \(\theta\) is in the first quadrant, we take the positive value:
\[ \tan \theta = \sqrt{3} \]

**Step 5: Find \(\sec \theta\)**
Using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\):
\[ \sec^2 \theta = 1 + 3 = 4 \]
Taking the square root:
\[ \sec \theta = 2 \]

**Step 6: Find \(\cos \theta\)**
Since \(\sec \theta = \frac{1}{\cos \theta}\), we have:
\[ \cos \theta = \frac{1}{\sec \theta} = \frac{1}{2} \]

**Step 7: Find \(\sin \theta\)**
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \sin^2 \theta + \left(\frac{1}{2}\right)^2 = 1 \]
\[ \sin^2 \theta + \frac{1}{4} = 1 \]
Subtract \(\frac{1}{4}\) from both sides:
\[ \sin^2 \theta = 1 - \frac{1}{4} = \frac{3}{4} \]
Taking the square root:
\[ \sin \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

**Final Values:**
- \(\sin \theta = \frac{\sqrt{3}}{2}\)
- \(\cos \theta = \frac{1}{2}\)
- \(\tan \theta = \sqrt{3}\)

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