Question

Prove that $\frac{\cos A}{1+\sin A}=\frac{\left[cot\left({\displaystyle \frac{A}{2}}\right)-1\right]}{\left[cot{\displaystyle \left(\frac{A}{2}\right)}+1\right]}$

Answer 1

$L.H.S=\frac{\cos A}{1+\sin A}$

$=\frac{\left[2{\cos }^2\left({\displaystyle \frac{A}{2}}\right)–1\right]}{\left(1+2\sin \left({\displaystyle \frac{A}{2}}\right)\cos \left({\displaystyle \frac{A}{2}}\right)\right)}$…….(.$\cos \theta =2{\cos }^2\frac{\theta }{2}-1,\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$)

$=\frac{\left[{\cos }^2\left({\displaystyle \frac{A}{2}}\right)+{\cos }^2\left({\displaystyle \frac{A}{2}}\right)–1\right]}{\left(Si{n}^2{\displaystyle \frac{A}{2}}+{\cos }^2{\displaystyle \frac{A}{2}}+2\sin \left({\displaystyle \frac{A}{2}}\right)\cos \left({\displaystyle \frac{A}{2}}\right)\right)}$…………..(${\sin }^2\theta +{\cos }^2\theta =1$)

$=\frac{\left[{\cos }^2\left({\displaystyle \frac{A}{2}}\right)-{\sin }^2\left({\displaystyle \frac{A}{2}}\right)\right]}{{\left(Sin{\displaystyle \frac{A}{2}}+\cos {\displaystyle \frac{A}{2}}\right)}^2}$………………………(${\left(a+b\right)}^2=a^2+b^2+2ab$)

$=\frac{\left[\cos \left({\displaystyle \frac{A}{2}}\right)+\sin \left({\displaystyle \frac{A}{2}}\right)\right]\left[\cos \left({\displaystyle \frac{A}{2}}\right)-\sin \left({\displaystyle \frac{A}{2}}\right)\right]}{\left(Sin{\displaystyle \frac{A}{2}}+\cos {\displaystyle \frac{A}{2}}\right)\left(\sin {\displaystyle \frac{A}{2}}+\cos {\displaystyle \frac{A}{2}}\right)}$………………..($a^2-b^2=\left(a+b\right)\left(a-b\right)$)

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

Dividing numerator and denominator by $\sin \left(\frac{A}{2}\right)$

$=\frac{\left({\displaystyle \frac{\cos \left({\displaystyle \frac{A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}}-{\displaystyle \frac{\sin \left({\displaystyle \frac{A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}}\right)}{\left({\displaystyle \frac{Sin{\displaystyle \left(\frac{A}{2}\right)}}{\sin \left({\displaystyle \frac{A}{2}}\right)}}+{\displaystyle \frac{\cos {\displaystyle \left(\frac{A}{2}\right)}}{\sin \left({\displaystyle \frac{A}{2}}\right)}}\right)}$…………………………….($\frac{\cos \theta }{\sin \theta }=cot\theta$)

$=\frac{\left({\displaystyle cot\left(\frac{A}{2}\right)}-{\displaystyle 1}\right)}{\left({\displaystyle 1}+{\displaystyle cot\left(\frac{A}{2}\right)}\right)}$

$=R.H.S.$

Hence, it is proved that $\frac{\cos A}{1+\sin A}=\frac{\left[cot\left({\displaystyle \frac{A}{2}}\right)-1\right]}{\left[cot{\displaystyle \left(\frac{A}{2}\right)}+1\right]}$.