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if cot A = b/a, prove that 2 secA+ 1 / cos A + 2= root a sqaure+ b square/ b

$Given cot A= \frac{b}{a} \Rightarrow \tan A = \frac{a}{b} \Rightarrow \cos A = \frac{b}{\sqrt{a^{2} + b^{2}}} \Rightarrow \sec A = \frac{\sqrt{a^{2} + b^{2}}}{b} G i v e n \frac{2 \sec A + 1}{\cos A + 2} = \frac{2 (\frac{1}{\cos A}) + 1}{\cos A + 2} = \frac{2 + \cos A}{\cos A (\cos A + 2)} = \frac{1}{\cos A} = \sec A = \frac{\sqrt{a^{2} + b^{2}}}{b}$

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