Question

if cot theta + 1/cot theta = 2 , find the value of cot^2 theta + 1/cot 2 theta

Answer 1

$\cot \theta + \frac{1}{ \cot \theta} = 2 \\ squarin \\ {( \cot \theta + \frac{1}{ \cot \theta}) }^{2} = {2}^{2} = 4 \\ { \cot }^{2} \theta + 2 \times \cot\theta \times \frac{1}{ \cot \theta} + \frac{1}{ { \cot }^{2}\theta } = 4 \\ { \cot }^{2} \theta + 2 + \frac{1}{ { \cot }^{2}\theta } = 4 \\ { \cot }^{2} \theta + \frac{1}{ { \cot }^{2}\theta } = 2$

Answer 2

Solution-

Here given,

$\cot \theta+\frac{1}{\cot \theta} =2$

We have to calculate,

$\cot^2 \theta+\frac{1}{\cot^2 \theta}$

We know that,

$(a+b)^2=a^2+b^2+2ab$

Putting,

$a=\cot \theta,\ and\ b=\frac{1}{\cot \theta}$

$\Rightarrow (\cot \theta+\frac{1}{\cot \theta})^2=(\cot \theta)^2+(\frac{1}{\cot \theta})^2+2\times \cot \theta\times \frac{1}{\cot \theta}$

$\Rightarrow (\cot \theta+\frac{1}{\cot \theta})^2=\cot^2 \theta+\frac{1}{\cot^2 \theta}+2$

$\Rightarrow (2)^2=\cot^2 \theta+\frac{1}{\cot^2 \theta}+2$

$\Rightarrow \cot^2 \theta+\frac{1}{\cot^2 \theta}=4-2$

$\Rightarrow \cot^2 \theta+\frac{1}{\cot^2 \theta}=2$