Math, asked by symashah000, 5 hours ago

plz solve this please​

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Answered by mathdude500
2

\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}}}}

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