Question

please solve this question with easy steps​

Answer 1

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

We know that,

$\rm :\longmapsto\:80\degree = 70\degree + 10\degree$

So,

$\rm :\longmapsto\:tan80\degree =tan( 70\degree + 10\degree)$

We know that,

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

So, using this identity, we get

$\rm :\longmapsto\:tan80\degree = \dfrac{tan70\degree + tan10\degree }{1 - tan70\degree \: tan10\degree }$

$\rm :\longmapsto\:tan70\degree + tan10\degree = tan80\degree - tan80\degree tan70\degree tan10\degree$

Now, we know that

$\underbrace{\boxed{ \sf{ \:tan(180\degree - x) = tanx}}}$

So,

$\boxed{ \sf{ \:tan80\degree = tan(180\degree - 100\degree ) = - tan100\degree }}$

So, substituting this value, we get

$\rm :\longmapsto\:tan70\degree + tan10\degree = - tan100\degree + tan100\degree tan70\degree tan10\degree$

$\rm :\longmapsto\:tan70\degree + tan10\degree + tan100\degree = tan100\degree tan70\degree tan10\degree$

$\rm :\longmapsto\:tan10\degree + tan70\degree + tan100\degree = tan100\degree tan70\degree tan10\degree$

Hence, Proved

Additional Information :-

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

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

$\boxed{ \sf{ \:cos(x - y) = cosxcosy + sinysinx}}$

$\boxed{ \sf{ \:cos(x + y) = cosxcosy - sinysinx}}$

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

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

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