To prove the given trigonometric identity \(\frac{2 \sec \theta + 1}{\cos \theta + 2} = \frac{\sqrt{a^2 + b^2}}{b}\), we start with the known relation \(\cot \theta = \frac{b}{a}\).
1. Express \(\cot \theta\) in terms of \(\cos \theta\) and \(\sin \theta\):
[tex]\[
\cot \theta = \frac{\cos \theta}{\sin \theta}
\][/tex]
Given \(\cot \theta = \frac{b}{a}\), we have:
[tex]\[
\frac{\cos \theta}{\sin \theta} = \frac{b}{a}
\][/tex]
2. Relate \(\cos \theta\) and \(\sin \theta\) using the Pythagorean identity:
[tex]\[
\cos^2 \theta + \sin^2 \theta = 1
\][/tex]
Since \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), we can write:
[tex]\[
\cos \theta = \frac{b}{a} \sin \theta
\][/tex]
3. Solve for \(\sin \theta\) and \(\cos \theta\):
Substitute \(\cos \theta = \frac{b}{a} \sin \theta\) into the Pythagorean identity:
[tex]\[
\left( \frac{b}{a} \sin \theta \right)^2 + \sin^2 \theta = 1
\][/tex]
[tex]\[
\frac{b^2}{a^2} \sin^2 \theta + \sin^2 \theta = 1
\][/tex]
[tex]\[
\left( \frac{b^2 + a^2}{a^2} \right) \sin^2 \theta = 1
\][/tex]
[tex]\[
\sin^2 \theta = \frac{a^2}{a^2 + b^2}
\][/tex]
[tex]\[
\sin \theta = \frac{a}{\sqrt{a^2 + b^2}}
\][/tex]
Therefore:
[tex]\[
\cos \theta = \frac{b}{a} \cdot \frac{a}{\sqrt{a^2 + b^2}} = \frac{b}{\sqrt{a^2 + b^2}}
\][/tex]
4. Express \(\sec \theta\):
\(\sec \theta\) is the reciprocal of \(\cos \theta\):
[tex]\[
\sec \theta = \frac{1}{\cos \theta} = \frac{\sqrt{a^2 + b^2}}{b}
\][/tex]
5. Substitute \(\sec \theta\) and \(\cos \theta\) into the left-hand side:
We need to prove:
[tex]\[
\frac{2 \sec \theta + 1}{\cos \theta + 2} = \frac{\sqrt{a^2 + b^2}}{b}
\][/tex]
Substitute:
[tex]\[
\frac{2 \cdot \frac{\sqrt{a^2 + b^2}}{b} + 1}{\frac{b}{\sqrt{a^2 + b^2}} + 2}
\][/tex]
6. Simplify the left-hand side:
Simplify the numerator:
[tex]\[
2 \cdot \frac{\sqrt{a^2 + b^2}}{b} + 1 = \frac{2\sqrt{a^2 + b^2}}{b} + 1 = \frac{2\sqrt{a^2 + b^2} + b}{b}
\][/tex]
Simplify the denominator:
[tex]\[
\frac{b}{\sqrt{a^2 + b^2}} + 2 = \frac{b + 2\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}
\][/tex]
Now combine:
[tex]\[
\frac{2\sqrt{a^2 + b^2} + b}{b} \cdot \frac{\sqrt{a^2 + b^2}}{b + 2\sqrt{a^2 + b^2}} = \frac{(2\sqrt{a^2 + b^2} + b)\sqrt{a^2 + b^2}}{b(b + 2\sqrt{a^2 + b^2})}
\][/tex]
7. Rewriting and final simplification:
After combining, the left-hand side simplifies due to the trigonometric identities:
[tex]\[
\frac{(2 \sqrt{a^2 + b^2} + b) \cdot \sqrt{a^2 + b^2}}{b \cdot (b + 2 \sqrt{a^2 + b^2})} = \frac{\sqrt{a^2 + b^2}}{b}
\][/tex]
Thus, we have shown that:
[tex]\[
\frac{2 \sec \theta + 1}{\cos \theta + 2} = \frac{\sqrt{a^2 + b^2}}{b}
\][/tex]
which completes the proof.
