Question

Any one who can give me the solution.
prove that​

Answer 1

Hope this helps !
Thank you !!
Goodnight

Answer 2

$\large\underline{\sf{To\:prove - }}$

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

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

We know that,

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

So,

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

We know,

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

So, using this result,

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

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

Now, we know

$\purple{\bf :\longmapsto\:tan80\degree = tan(180\degree - 100\degree ) = - tan100\degree }$

So, above can be rewritten as

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

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

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

Hence, Proved

Additional Information :-

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

$\red{\boxed{ \bf{ \: sin(x + y) = sinx \: cosy \: + \: siny \: cosx}}}$

$\red{\boxed{ \bf{ \: sin(x - y) = sinx \: cosy \: - \: siny \: cosx}}}$

$\red{\boxed{ \bf{ \: cos(x - y) = cosx \: cosy \: + \: sinx \: siny}}}$

$\red{\boxed{ \bf{ \: cos(x + y) = cosx \: cosy \: - \: sinx \: siny}}}$

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

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