Question

V3+ tan40° + tan80° = V3 tan40°. tan 80°​

Answer 1

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

We know,

$\rm :\longmapsto\:120\degree = 80\degree + 40\degree$

So,

$\rm :\longmapsto\:tan120\degree =tan( 80\degree + 40\degree )$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:tan(180\degree - 60\degree ) = \dfrac{tan80\degree + tan40\degree }{1 - tan80\degree \: tan40\degree }$

$\rm :\longmapsto\: - tan(60\degree ) = \dfrac{tan80\degree + tan40\degree }{1 - tan80\degree \: tan40\degree }$

$\rm :\longmapsto\: - \sqrt{3} = \dfrac{tan80\degree + tan40\degree }{1 - tan80\degree \: tan40\degree }$

$\rm :\longmapsto\:tan80\degree + tan40\degree = - \sqrt{3} + \sqrt{3}tan80\degree tan40\degree$

$\rm\implies \:\:tan80\degree + tan40\degree + \sqrt{3} = \sqrt{3}tan80\degree tan40\degree$

Hence, Proved

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MORE TO KNOW

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

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

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

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

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

$\boxed{\tt{ cot(x + y) = \frac{cotx \: coty \: - \: 1}{coty \: + \: cotx} \: }}$

$\boxed{\tt{ cot(x - y) = \frac{cotx \: coty \: + \: 1}{coty \: - \: cotx} \: }}$

Answer 2

Answer:

$\large\underline{\bf{Question-}}$

V3+ tan40° + tan80° = V3 tan40°. tan 80°

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

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

We know,

$\rm :\longmapsto\:120\degree = 80\degree + 40\degree$

So,

$\rm :\longmapsto\:tan120\degree =tan( 80\degree + 40\degree )$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:tan(180\degree - 60\degree ) = \dfrac{tan80\degree + tan40\degree }{1 - tan80\degree \: tan40\degree }$

$\rm :\longmapsto\: - tan(60\degree ) = \dfrac{tan80\degree + tan40\degree }{1 - tan80\degree \: tan40\degree }$

$\rm :\longmapsto\: - \sqrt{3} = \dfrac{tan80\degree + tan40\degree }{1 - tan80\degree \: tan40\degree }$

$\rm :\longmapsto\:tan80\degree + tan40\degree = - \sqrt{3} + \sqrt{3}tan80\degree tan40\degree$

$\rm\implies \:\:tan80\degree + tan40\degree + \sqrt{3} = \sqrt{3}tan80\degree tan40\degree$

Hence, Proved ✓

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MORE TO KNOW

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

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

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

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

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

$\boxed{\tt{ cot(x + y) = \frac{cotx \: coty \: - \: 1}{coty \: + \: cotx} \: }}$

$\boxed{\tt{ cot(x - y) = \frac{cotx \: coty \: + \: 1}{coty \: - \: cotx} \: }}$

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