Question

Prove that

(i) $\frac{\text{sin}2x}{1-\text{cos}2x}=\text{cot}x$

(ii) $\frac{1-\text{cos2x}}{1+\text{cos}2x}={\text{tan}}^{2}x$

Answer 1

LHS$=\frac{\text{sin}2x}{1-\text{cos}2x}=\text{cot}x$

(i)$\text{LHS}=\frac{2\text{sin}x\text{cos}x}{2{\text{sin}}^{2}x}$

$[\because \text{sin}2x=2\text{sin}x\text{cos}x,(1-\text{cos}2x)=2{\text{sin}}^{2}x]$

$=\text{cot}x=\text{RHS}.$

$\text{Hence,}\frac{\text{sin}2x}{1-\text{cos}2x}=\text{cot}x.$

(ii)$\text{LHS}=\frac{1-\text{cos2x}}{1+\text{cos}2x}$

= $\frac{2si{n}^{2}x}{2{\text{cos}}^{2}x}$

$\left[\begin{array}{cc}\because (1-\text{cos2x})& =2{\text{sin}}^{2}x,\\ (1+\text{cos}2x)& =2{\text{cos}}^{2}x\end{array}\right]$

$={\text{tan}}^{2}x=\text{RHS}$

$\text{Hence,}\frac{1-\text{cos}2x}{1+\text{cos}2x}={\text{tan}}^{2}x$