Question

2.tanA .tan B .tan C=(SinA/cosB.cosc) +(sin B/cosC.cosA)+(sinC/cosA.cosB)​

Answer 1

Answer:

We have A−cosB=cosC

⇒sinA=cosB−cosC

⇒2sin2Acos2A=2sin2Acos(2B−C)

⇒cos2A=cos(2B−C sin2A=0]⇒2A=2B−C

⇒A=B−C

But A+B+C=π

Therefore, 2B=π

=B=2uponπ

​

Step-by-step explanation:

Answer 2

Answer:

We have A−cosB=cosC

⇒sinA=cosB−cosC

⇒2sin2Acos2A=2sin2Acos(2B−C)

⇒cos2A=cos(2B−C sin2A=0]⇒2A=2B−C

⇒A=B−C

But A+B+C=π

Therefore, 2B=π

=B=2uponπ