If tan theta+1/tan theta=2;show that:tan^theta+¹/tan^theta=2

We have,
{tex} \tan \theta + \frac { 1 } { \tan \theta } = 2{/tex}

Squaring both sides, we get
{tex}\Rightarrow \left( \tan \theta + \frac { 1 } { \tan \theta } \right) ^ { 2 } = 2 ^ { 2 }{/tex}
{tex}\Rightarrow\quad\tan^2\theta+\frac1{\tan^2\theta}+2\times\tan\theta\times\frac1{\tan\theta}=4{/tex}
{tex}\Rightarrow \quad \tan ^ { 2 } \theta + \frac { 1 } { \tan ^ { 2 } \theta } + 2 = 4{/tex}
{tex}\Rightarrow \quad \tan ^ { 2 } \theta + \frac { 1 } { \tan ^ { 2 } \theta } = 2{/tex}
Alternate method, We have
{tex}\tan \theta + \frac { 1 } { \tan \theta } = 2{/tex}
{tex}\Rightarrow \quad \tan ^ { 2 } \theta + 1 = 2 \tan \theta{/tex}
{tex}\Rightarrow \quad \tan ^ { 2 } \theta - 2 \tan \theta + 1 = 0{/tex}
{tex}\Rightarrow \quad ( \tan \theta - 1 ) ^ { 2 } = 0{/tex}
{tex}\Rightarrow \quad \tan \theta = 1{/tex}
{tex}\therefore \quad \tan ^ { 2 } \theta + \frac { 1 } { \tan ^ { 2 } \theta } = 1 + 1 = 2{/tex}