Question

Date tan 70°- tan 50° -tan 20 = ton 70 x tans 50 X tan 20​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Prove that

$\rm \: tan70\degree - tan50\degree - tan20\degree = tan70\degree \: tan50\degree \: tan20\degree$

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

We know,

$\rm \: 70\degree = 50\degree + 20\degree$

So,

$\rm \:tan 70\degree = tan(50\degree + 20\degree )$

We know that

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

So, on using this identity, we get

$\rm \: tan70\degree = \frac{tan50\degree + tan20\degree }{1 - tan50\degree \: tan20\degree }$

$\rm \: tan70\degree - tan70\degree tan50\degree tan20\degree = tan50\degree + tan20\degree$

On rearranging the terms, we get

$\rm \: tan70\degree - tan50\degree - tan20\degree = tan70\degree tan50\degree tan20\degree$

Hence, Proved

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Additional Information

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

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

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

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

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

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

Answer 2

Refer the given attachment