Question

if tanA =mtanB than prove that sin(A-B)/sin(A+B)=m-1/m+1.

Answer 1

Step-by-step explanation:

= > tanA =mtanB

= > tanA =mtanB

= > tanA/tanB = m

Using componendo dividendo:

= > (tanA+tanB)/(tanA-tanB) = (m+1)/(m-1)

Now,

= > tanA + tanB

= > (sinA/cosA) + (sinB/cosB)

= > ( sinAcosB + sinBcosA )/sinAcosB

= > sin(A+B) /sinAcosB

Similarly,

= > tanA - tanB

= > (sinA/cosA) - (sinB/cosB)

= > ( sinAcosB - sinBcosA )/sinAcosB

= > sin(A-B) /sinAcosB

Continue,

= > (tanA+tanB)/(tanA-tanB) = (m+1)/(m-1)

= > [sin(A+B)/sinAcosB]/[sin(A-B) /sinAcosB] = (m+1)/(m-1)

= > sin(A+B)/sin(A-B) = (m+1)/(m-1)

= > sin(A-B)/sin(A+B) = (m-1)/(m+1)

Proved.

Answer 2

Answer:

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