Question

Ü) Find the value of
Tan20 + Tan 40 + √3 Tan20° Tan40°​

Answer 1

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

We know,

$\rm :\longmapsto\:60\degree = 40\degree + 20\degree$

Apply tan on both sides, we get

$\rm :\longmapsto\:tan60\degree =tan( 40\degree + 20\degree )$

We know,

$\boxed{ \bf{ \: tan(x + y) = \frac{tanx + tany}{1 - tanxtany}}}$

and

$\boxed{ \bf{ \: tan60\degree = \sqrt{3} }}$

So, on substituting all these values, we get

$\rm :\longmapsto\: \sqrt{3} = \dfrac{tan40\degree + tan20\degree }{1 - tan40\degree tan20\degree }$

$\rm :\longmapsto\: \sqrt{3} - \sqrt{3}tan40\degree tan20\degree = tan40\degree + tan20\degree$

$\rm :\longmapsto\: \sqrt{3} = \sqrt{3}tan40\degree tan20\degree + tan40\degree + tan20\degree$

So,

$\rm :\longmapsto\:tan40\degree + tan20\degree + \sqrt{3}tan40\degree tan20\degree = \sqrt{3}$

Additional Information :-

$\boxed{ \bf{ \: sin(x + y) = sinxcosy + sinycosx}}$

$\boxed{ \bf{ \: sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \bf{ \: cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \bf{ \: cos(x - y) = cosxcosy + sinxsiny}}$

$\boxed{ \bf{ \: tan(x - y) = \frac{tanx + tany}{1 - tanxtany}}}$

$\boxed{ \bf{ \: {sin}^{2}x - {sin}^{2}y = sin(x + y)sin(x - y)}}$

$\boxed{ \bf{ \: {cos}^{2}y - {cos}^{2}x = sin(x + y)sin(x - y)}}$

$\boxed{ \bf{ \: {cos}^{2}x - {sin}^{2}y = cos(x + y)cos(x - y)}}$