Question

please solve the question in the attachment eith reasons...I'm new to trigonometry​

Answer 1

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

Given that

$\rm :\longmapsto\:\dfrac{1 + cosx}{1 - cosx} = 7 + 4 \sqrt{3}$

$\rm :\longmapsto\:1 + cosx = (7 + 4 \sqrt{3})(1 - cosx)$

$\rm :\longmapsto\:1 + cosx =7 - 7cosx + 4 \sqrt{3} - 4 \sqrt{3}cosx$

$\rm :\longmapsto\:cosx + 7cosx + 4 \sqrt{3}cosx = 7 + 4 \sqrt{3} - 1$

$\rm :\longmapsto\: 8cosx + 4 \sqrt{3}cosx = 6 + 4 \sqrt{3}$

$\rm :\longmapsto\: (8 + 4 \sqrt{3})cosx = 6 + 4 \sqrt{3}$

$\rm :\longmapsto\:cosx = \dfrac{6 + 4 \sqrt{3} }{8 + 4 \sqrt{3} }$

$\rm :\longmapsto\:cosx = \dfrac{3 + 2 \sqrt{3} }{4 + 2 \sqrt{3} }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:cosx = \dfrac{3 + 2 \sqrt{3} }{4 + 2 \sqrt{3} } \times \dfrac{4 - 2 \sqrt{3} }{4 - 2 \sqrt{3} }$

We know,

$\red{\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2}\bigg \}}$

$\rm :\longmapsto\:cosx = \dfrac{12 - 6 \sqrt{3} + 8 \sqrt{3} - 12}{16 - 12}$

$\rm :\longmapsto\:cosx = \dfrac{2 \sqrt{3}}{4}$

$\rm :\longmapsto\:cosx = \dfrac{ \sqrt{3}}{2}$

$\rm :\longmapsto\:cosx = cos30 \degree$

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

Now, Consider

$\rm :\longmapsto\:\dfrac{1 + sinx}{1 - sinx}$

$\rm \: = \:\:\dfrac{1 + sin30 \degree}{1 - sin30 \degree}$

$\rm \: = \:\dfrac{1 + \dfrac{1}{2} }{1 - \dfrac{1}{2} }$

$\rm \: = \:\dfrac{ \dfrac{2 + 1}{2} }{ \dfrac{2 - 1}{2} }$

$\rm \: = \:\dfrac{3}{1}$

$\rm \: = \:3$

Hence,

$\rm :\longmapsto\: \boxed{ \: \: \: \bf{ \: \dfrac{1 + sinx}{1 - sinx} = 3 \: \: \: }}$

Additional Information :-

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