Question

if tan alpha is 1/2 and tan beta is 1/3 then alpha and beta are

Answer 1

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

Given that

$\rm :\longmapsto\:tan\alpha = \dfrac{1}{2}$

and

$\rm :\longmapsto\:tan \beta = \dfrac{1}{2}$

We know,

$\green{\rm :\longmapsto\:tan(\alpha + \beta ) = \dfrac{tan\alpha + tan \beta }{1 - tan\alpha tan \beta }}$

On substituting the values, we get

$\rm :\longmapsto\:tan(\alpha + \beta ) = \dfrac{\dfrac{1}{2} + \dfrac{1}{3} }{1 - \dfrac{1}{2} \times \dfrac{1}{3} }$

$\rm :\longmapsto\:tan(\alpha + \beta ) = \dfrac{\dfrac{3 + 2}{6} }{1 - \dfrac{1}{6} }$

$\rm :\longmapsto\:tan(\alpha + \beta ) = \dfrac{\dfrac{5}{6} }{ \dfrac{6 - 1}{6} }$

$\rm :\longmapsto\:tan(\alpha + \beta ) = \dfrac{\dfrac{5}{6} }{ \dfrac{5}{6} }$

$\rm :\longmapsto\:tan(\alpha + \beta ) = 1$

$\rm :\longmapsto\:tan(\alpha + \beta ) = tan\dfrac{\pi}{4}$

We know,

$\boxed{ \rm{ tanx = tany \: \implies \: x = n\pi + y \: \forall \: n \in \: Z}}$

So, using this identity,

$\bf :\longmapsto\:\alpha + \beta = n\pi + \dfrac{\pi}{4} \: \forall \: n \in \: Z$

If solution lies in first quadrant then

$\bf :\longmapsto\:\alpha + \beta = \dfrac{\pi}{4} \:$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2} \: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y \: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$

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

Answer 2

☆SOLUTION☆

tan(a+b)

=(tan a + tan b)/(1-tan a.tanb)

b) =(1/2+1/3)/(1-1/2.1/3)

= 1(5/6)/(1-1/6)

(5/6)/(5/6)a+ b

= tan inverse 1=n/4assume alpha in place of

a and beta in place of b.