Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 498: 61

See below.

$LHS=cot^2(\frac{\theta}{2})=\frac{1}{tan^2(\frac{\theta}{2})}=(\frac{sin\theta}{1-cos\theta})^2=\frac{sin^2\theta}{(1-cos\theta)^2}=\frac{1-cos^2\theta}{(1-cos\theta)^2}=\frac{1+cos\theta}{1-cos\theta}=\frac{sec\theta+1}{sec\theta-1}=RHS$ Here $\theta=v$