# Prove the following: cosA / 1 - sinA = tan (45 + A/2)

$=\frac{{\mathrm{cos}}^{2}\left(\frac{A}{2}\right)-{\mathrm{sin}}^{2}\left(\frac{A}{2}\right)}{{\mathrm{sin}}^{2}\left(\frac{A}{2}\right)+{\mathrm{cos}}^{2}\left(\frac{A}{2}\right)+2\mathrm{sin}\frac{A}{2}\mathrm{cos}\frac{A}{2}}$

$=\frac{\left(\mathrm{cos}\frac{A}{2}+\mathrm{sin}\frac{A}{2}\right)\left(\mathrm{cos}\frac{A}{2}-\mathrm{sin}\frac{A}{2}\right)}{{\left(\mathrm{cos}\frac{A}{2}+\mathrm{sin}\frac{A}{2}\right)}^{2}}$

$=\frac{\mathrm{cos}\frac{A}{2}-\mathrm{sin}\frac{A}{2}}{\mathrm{cos}\frac{A}{2}+\mathrm{sin}\frac{A}{2}}$

$=\frac{1-\mathrm{tan}\frac{A}{2}}{1+\mathrm{tan}\frac{A}{2}}$  (Dividing both numerator and denominator by $\mathrm{cos}\frac{A}{2}$)

$=\mathrm{tan}\left(45°+A\right)$

$=R.H.S$

Hence Proved.

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