Question

If A+B+C = r and cosA= cosB cosc
show that tan A= TanB + Tanc​

Answer 1

Step-by-step explanation:

A+B+C=π <＝> A = π - (B+C)

cos A = cos (π-(B+C)) = - cos (B+C) =-(cosB cosC -sinB sinC) = sinB sinC - cosB cosC

sinB sinC - cosB cosC = cosB cosC <＝> sinB sinC = 2 cosB cosC <＝> sinB sinC/cosB cosC =2

<＝> tanB tanC=2

tan A = tan (π-(B+C))=- tan (B+C)=

-(tanB+tanC)/(1-tanB tanC) = - (tanB+tanC)/(1–2)

= -(tanB+tanC)/(-1) = tanB + tanC (proven)

Answer 2

Answer:

Given

A+

B+

C=

π⟹

A=

π−

(B+

C)

And also given

cosA=

cosBcosC

⟹ cos(π− (B+ C))= cosBcosC

⟹ −cosBcosC+ sinBsinC= cosBcosC

⟹ sinBsinC= 2cosBcosC

⟹ tanBtanC= 2