Question

Formula of tan (A-B-C). ​

Answer 1

Step-by-step explanation:

Answer: By using the formula of tan (α + β) we can easily expand tan (A + B + C). Let us recall the formula of tan (α + β) = tan α + tan β/1 - tan α tan β. Therefore, the expansion of tan (A + B + C) = tan A + tan B + tan C - tan A tan B tan C/ 1 - tan A tan B- tan C tan A - tan B tan C.

Answer 2

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

$\rm :\longmapsto\:tan(A - B - C)$

Let we assume that A - B - C = D

So, it means we have to find the value of tanD

Consider,

$\rm :\longmapsto\:A - B - C = D$

can be rewritten as

$\rm :\longmapsto\:A - D = B + C$

$\rm :\longmapsto\:tan(A - D)= tan(B + C)$

$\rm :\longmapsto\:\dfrac{tanA - tanD}{1 + tanAtanD} = \dfrac{tanB + tanC}{1 - tanBtanC}$

Let we assume that

tanA = a

tanB = b

tanC = c

tanD = d

So, above can be rewritten as

$\rm :\longmapsto\:\dfrac{a - d}{1 + ad} = \dfrac{b + c}{1 - bc}$

$\rm :\longmapsto\:(a - d)(1 - bc) = (1 + ad)(b + c)$

$\rm :\longmapsto\:a - abc - d + dbc =b + c + abd + acd$

$\rm :\longmapsto\:a - b - c - abc = abd + acd + d - dbc$

$\rm :\longmapsto\:a - b - c - abc = d(ab + ac+ 1 - bc)$

$\rm :\implies\:d = \dfrac{a - b - c - abc}{1 + ab - bc + ca}$

So, on substituting the values of a, b, c and d, we get

$\rm :\implies\:tanD= \dfrac{tanA - tanB - tanC- tanAtanBtanC}{1 + tanAtanB - tanBtanC + tanCtanA}$

On substituting the value of D = A - B - C, we get

$\rm :\implies\:tan(A - B - C)= \dfrac{tanA - tanB - tanC- tanAtanBtanC}{1 + tanAtanB - tanBtanC + tanCtanA}$

Hence, derived

Additional Information :-

$\underbrace{\boxed{ \tt{tan(A + B + C)= \dfrac{tanA + tanB + tanC- tanAtanBtanC}{1 - tanAtanB - tanBtanC - tanCtanA}}}}$

$\underbrace{\boxed{ \tt{sin(A + B + C) = cosAcosBcosC(tanA + tanB + tanC - tanAtanBtanC}}}$

$\underbrace{\boxed{ \tt{cos(A + B + C) = cosAcosBcosC(1 - tanAtanB - tanBtanC - tanCtanA}}}$