Question

plz solve this please​

Answer 1

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

$\rm :\longmapsto\:A + B + C = \pi$

and

$\rm :\longmapsto\:cosA = cosBcosC$

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

$\rm :\longmapsto\:tanA = tanB + tanC$

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

Now, As it is given that

$\rm :\longmapsto\:A + B + C = \pi$

can be rewritten as

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

Apply sin on both sides, we get

$\rm :\longmapsto\:sin(B + C) = sin(\pi - A)$

$\rm :\longmapsto\:sinBcosC + sinCcosB = sinA$

Divide both sides by cosA, we get

$\rm :\longmapsto\:\dfrac{sinBcosC}{cosA} + \dfrac{sinCcosB}{cosA} = \dfrac{sinA}{cosA}$

$\rm :\longmapsto\:\dfrac{sinBcosC}{cosBcosC} + \dfrac{sinCcosB}{cosBcosC} =tanA$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: cosBcosC = cosA\bigg \}}$

$\bf\implies \:tanB + tanC = tanA$

$\large{{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\red{ \boxed{ \rm{ \: sin(A + B) = sinAcosB + sinBcosA}}}$

$\red{ \boxed{ \rm{ \: sin(A - B) = sinAcosB - sinBcosA}}}$

$\red{ \boxed{ \rm{ \: cos(A - B) = cosAcosB + sinBsinA}}}$

$\red{ \boxed{ \rm{ \: cos(A + B) = cosAcosB - sinBsinA}}}$

$\red{ \boxed{ \rm{ \: tan(A + B) = \frac{tanA + tanB}{1 - tanAtanB}}}}$

$\red{ \boxed{ \rm{ \: tan(A - B) = \frac{tanA - tanB}{1 + tanAtanB}}}}$