Question

If cos theta - sin theta / cos theta - sin theta = 1- √3 / 1+√3 , find theta ?​

Answer 1

Appropriate Question

$\sf \: If \: \dfrac{cos\theta - sin\theta}{cos\theta + sin\theta} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }, \: find \: \theta$

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

Given Trigonometric equation is

$\rm :\longmapsto\: \dfrac{cos\theta - sin\theta}{cos\theta + sin\theta} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }$

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

$\rm :\longmapsto\:\cancel{cos\theta} - sin\theta + \sqrt{3}cos\theta -\cancel{\sqrt{3}sin\theta} = \cancel{cos\theta} + sin\theta - \sqrt{3}cos\theta - \cancel{\sqrt{3}sin\theta}$

$\rm :\longmapsto\: - sin\theta + \sqrt{3}cos\theta = sin\theta - \sqrt{3}cos\theta$

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

$\rm :\longmapsto\: \sqrt{3}cos\theta = sin\theta$

$\rm :\longmapsto\:\dfrac{sin\theta}{cos\theta} = \sqrt{3}$

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

$\rm :\longmapsto\:tan\theta = tan60 \degree$

$\rm \implies\:\boxed{\tt{ \theta = 60 \degree }}$

Alternative Method

$\rm :\longmapsto\: \dfrac{cos\theta - sin\theta}{cos\theta + sin\theta} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }$

$\rm :\longmapsto\: \dfrac{cos\theta\bigg[1 - \dfrac{sin\theta}{cos\theta} \bigg]}{cos\theta\bigg[1 + \dfrac{sin\theta}{cos\theta} \bigg]} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }$

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

So, on comparing, we get

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

$\rm :\longmapsto\:tan\theta = tan60 \degree$

$\rm \implies\:\boxed{\tt{ \theta = 60 \degree }}$

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More to Learn

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

Answer 2

Appropriate Question

$\sf \: If \: \dfrac{cos\theta - sin\theta}{cos\theta + sin\theta} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }, \: find \: \theta$

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

Given Trigonometric equation is

$\rm :\longmapsto\: \dfrac{cos\theta - sin\theta}{cos\theta + sin\theta} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }$

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

$\rm :\longmapsto\:\cancel{cos\theta} - sin\theta + \sqrt{3}cos\theta -\cancel{\sqrt{3}sin\theta} = \cancel{cos\theta} + sin\theta - \sqrt{3}cos\theta - \cancel{\sqrt{3}sin\theta}$

$\rm :\longmapsto\: - sin\theta + \sqrt{3}cos\theta = sin\theta - \sqrt{3}cos\theta$

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

$\rm :\longmapsto\: \sqrt{3}cos\theta = sin\theta$

$\rm :\longmapsto\:\dfrac{sin\theta}{cos\theta} = \sqrt{3}$

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

$\rm :\longmapsto\:tan\theta = tan60 \degree$

$\rm \implies\:\boxed{\tt{ \theta = 60 \degree }}$

Alternative Method

$\rm :\longmapsto\: \dfrac{cos\theta - sin\theta}{cos\theta + sin\theta} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }$

$\rm :\longmapsto\: \dfrac{cos\theta\bigg[1 - \dfrac{sin\theta}{cos\theta} \bigg]}{cos\theta\bigg[1 + \dfrac{sin\theta}{cos\theta} \bigg]} = \dfrac{1 - \sqrt{3} }{1 + \sqrt{3} }$

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

So, on comparing, we get

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

$\rm :\longmapsto\:tan\theta = tan60 \degree$

$\rm \implies\:\boxed{\tt{ \theta = 60 \degree }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Learn

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