Question

( begin{aligned} &left(frac{1+tan theta}{1+cot theta}right)^{2}=left(frac{1+frac{sin theta}{cos theta}}{1+frac{cos theta}{sin theta}}right)^{2} &=left{frac{sin theta(sec theta+sin theta)}{cos theta(sin theta+cos theta)}right]^{2} therefore & tan ^{2} theta therefore quadleft(frac{1+tan theta}{1+cos theta}right)=& frac{1+frac{sin ^{2} theta}{cos ^{2} theta}}{1+frac{cos ^{2} theta}{sin ^{2} theta}}=frac{sin ^{2} thetaleft(cos ^{2} theta+sin ^{2} thetaright)left(sin ^{2} theta+cos ^{2} thetaright)}{cos ^{2} theta} end{aligned} )

# Prove that: a) (1 + tano) (1 + coto) 1 + tan20 1 + cot20

Solution