Question

cot theta -1 upon cot theta + 1 = 1-root 3 upon 1 +root 3

Answer 1

$\large\underline{\sf{Given- }}$

$\red{\rm :\longmapsto\:\dfrac{cot\theta - 1}{cot\theta + 1} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }}$

$\large\underline{\sf{To\:Find - }}$

$\red{\rm :\longmapsto\:\theta }$

$\large\underline{\sf{Solution-}}$

Given that,

$\red{\rm :\longmapsto\:\dfrac{cot\theta - 1}{cot\theta + 1} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }}$

$\rm :\longmapsto\:(cot\theta - 1)(1 + \sqrt{3}) = (cot\theta + 1)(1 - \sqrt{3})$

$\rm :\longmapsto\:cot\theta + \sqrt{3} cot\theta - 1 - \sqrt{3} = cot\theta + 1 - \sqrt{3}cot\theta - \sqrt{3}$

$\rm :\longmapsto\:2 \sqrt{3}cot\theta = 2$

$\rm :\longmapsto\: \sqrt{3}cot\theta = 1$

$\rm :\longmapsto\:cot\theta = \dfrac{1}{ \sqrt{3} }$

$\rm :\longmapsto\:cot\theta = cot60 \degree$

$\bf\implies \:\theta = 60\degree$

Alternative Method :-

Given that,

$\red{\rm :\longmapsto\:\dfrac{cot\theta - 1}{cot\theta + 1} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }}$

$\red{\rm :\longmapsto\:\dfrac{cot\theta (1 - \dfrac{1}{cot\theta }) }{cot\theta (1 + \dfrac{1}{cot\theta } )} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }}$

$\red{\rm :\longmapsto\:\dfrac{1 - tan\theta}{1 + tan\theta} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }}$

On comparing we get

$\red{\rm :\longmapsto\:tan\theta = \sqrt{3}}$

$\red{\rm :\longmapsto\:tan\theta = tan60\degree}$

$\red{\rm :\longmapsto\:\theta = 60\degree}$

Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$