Question

If tan theta = 1 by root 3 theta is acute then find the value of 1+ sin theta

Answer 1

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

Given that,

$\rm \: 0 < \theta < 90\degree$

and

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

From Trigonometric table of standard angles, we have

$\rm \: tan\theta = tan30\degree$

$\rm\implies \:\theta = 30\degree$

Now, Consider

$\rm \: 1 + sin\theta$

$\rm \: = \: 1 + sin30\degree$

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

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

$\rm \: = \: \dfrac{3}{2}$

Hence,

$\rm\implies \: \: \: \boxed{\rm{  \:\rm \: 1 + sin\theta = \: \dfrac{3}{2} \: \: }}$

$\rule{190pt}{2pt}$

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