Question

maths question solve it fast​

Answer 1

Given Question

Prove that

$\sf \: tan13x - tan9x - tan4x = tan13x \: tan9x \: tan4x$

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

We know that

$\rm :\longmapsto\:13x = 9x + 4x$

$\rm :\longmapsto\:tan13x = tan(9x + 4x)$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:tan13x = \dfrac{tan9x + tan4x}{1 - tan9x \: tan4x}$

$\rm :\longmapsto\:tan13x - tan13xtan9xtan4x = tan9x + tan4x$

On rearrange the terms, we get

$\rm :\longmapsto\:tan13x -tan9x - tan4x = tan13xtan9xtan4x$

Or

$\purple{\boxed{\sf{ tan13x \: tan9x \: tan4x = tan13x - tan9x - tan4x}}}$

Hence, Proved

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MORE TO KNOW

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

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

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

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

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

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

Answer 2

Step-by-step explanation:

Given Question

Prove that

$\sf \: tan13x - tan9x - tan4x = tan13x \: tan9x \: tan4x$

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

We know that

$\rm :\longmapsto\:13x = 9x + 4x$

$\rm :\longmapsto\:tan13x = tan(9x + 4x)$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:tan13x = \dfrac{tan9x + tan4x}{1 - tan9x \: tan4x}$

$\rm :\longmapsto\:tan13x - tan13xtan9xtan4x = tan9x + tan4x$

On rearrange the terms, we get

$\rm :\longmapsto\:tan13x -tan9x - tan4x = tan13xtan9xtan4x$

Or

$\blue{\boxed{\sf{ tan13x \: tan9x \: tan4x = tan13x - tan9x - tan4x}}}$

Hence, Proved

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

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

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

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

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

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

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