Question

if tan(a+b)=tana+tanb/1-tana tanb find the value of tan(22.5)​

Answer 1

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

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

$\sf \: \rm :\longmapsto\:tan(a + b) = \dfrac{tana + tanb}{1 - tana \: tanb}$

$\rm :\longmapsto\:Put \: a \: = \: 22.5 \degree \: and \: b \: = \: 22.5\degree \:$

$\sf \: \rm :\longmapsto\:tan(22.5\degree \: + 22.5\degree \: ) = \dfrac{tan22.5\degree \: + tan22.5\degree \: }{1 - tan22.5\degree \: \: tan22.5\degree \: }$

$\rm :\longmapsto\:tan45\degree \: = \dfrac{2tan22.5\degree \: }{1 - {tan}^{2} 22.5\degree \: }$

$\rm :\longmapsto\:1 \: = \dfrac{2tan22.5\degree \: }{1 - {tan}^{2} 22.5\degree \: }$

$\rm :\longmapsto\: \: {2tan22.5\degree \: } = {1 - {tan}^{2} 22.5\degree \: }$

$\rm :\longmapsto\:\: {2tan22.5\degree \: } + {tan}^{2} 22.5\degree \: - 1 = 0$

$\rm :\longmapsto\: {tan}^{2} 22.5\degree \: + 2tan22.5\degree \: - 1 = 0$

Its a quadratic equation in tan22.5°, whose solution is given by

$\rm :\longmapsto\:tan22.5\degree \: = \dfrac{ - 2 \: \pm \: \sqrt{ {2}^{2} - 4 \times 1 \times ( - 1)} }{2 \times 1}$

$\rm :\longmapsto\:tan22.5\degree \: = \dfrac{ - 2 \: \pm \: \sqrt{4 + 4} }{2}$

$\rm :\longmapsto\:tan22.5\degree \: = \dfrac{ - 2 \: \pm \: \sqrt{8} }{2}$

$\rm :\longmapsto\:tan22.5\degree \: = \dfrac{ - 2 \: \pm \: 2 \sqrt{2} }{2}$

$\rm :\longmapsto\:tan22.5\degree \: = - 1 \: \pm \: \sqrt{2}$

$\rm :\longmapsto\:tan22.5\degree \: = \sqrt{2} - 1 \: \: \: or \: \: \: - \sqrt{2} - 1$

$\sf \:As \: tan22.5\degree \: lies \: in \: {1}^{st} \: quadrant \: \therefore \: tan22.5\degree \: > 0$

Thus,

$\rm :\longmapsto\:tan22.5\degree \: = \sqrt{2} - 1$

Additional Information :-

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