If $$\frac{(1 + secA)}{tanA}=x$$, then x is

Using double angle formula, we know that $$cos(2\theta) = cos^2\theta - sin^2\theta$$

=> $$cos(2\theta) = (1 - sin^2\theta) - sin^2\theta$$

=> $$cos(2\theta) = 1 - 2sin^2\theta$$

Replacing $$\theta$$ by $$\frac{A}{2}$$, we get :

=> $$cos A = 1 - 2sin^2(\frac{A}{2})$$

=> $$2sin^2(\frac{A}{2}) = 1 - cosA$$

=> $$sin^2(\frac{A}{2}) = \frac{(1-cosA)}{2}$$

=> $$sin(\frac{A}{2}) = \sqrt{\frac{(1 - cos A)}{2}}$$

Similarly, => $$cos(\frac{A}{2}) = \sqrt{\frac{(1 + cos A)}{2}}$$

Now, to find : $$cot(\frac{A}{2})$$

= $$cos(\frac{A}{2}) \div sin(\frac{A}{2})$$

= $$\sqrt{\frac{(1 + cos A)}{2}}$$ $$\div$$ $$\sqrt{\frac{(1 - cos A)}{2}}$$

= $$\sqrt{\frac{(1 + cos A)}{2}}$$ $$\times$$ $$\sqrt{\frac{2}{(1 - cos A)}}$$

= $$\sqrt{\frac{1+cosA}{1-cosA}}$$

= $$\sqrt{\frac{1+cosA}{1-cosA} \times \frac{1+cosA}{1+cosA}}$$

= $$\sqrt{\frac{(1+cosA)^2}{1-cos^2A}} = \sqrt{\frac{(1+cosA)^2}{sin^2A}}$$

= $$\frac{1+cosA}{sinA}$$

Divide both numerator and denominator by $$(cos A)$$

= $$\frac{1 + secA}{tan A}$$

=> Ans - (A)