Question

If tan theta A=5/6 tan B= 1/11 then show that A+B=pai/4​

Answer 1

Question:

If tanA = 5/6 and tanB = 1/11 then show that

A + B = π/4 .

Note:

• tan(A+B) = (tanA + tanB)/(1 - tanA•tanB)

• tan(A-B) = (tanA - tanB)/(1 + tanA•tanB)

• tan0 = 0

• tan(π/6) = 1/√3

• tan(π/4) = 1

• tan(π/3) = √3

• tan(π/2) = ∞

Solution:

Given:

tanA = 5/6

tanB = 1/11

To prove:

A + B = π/4

We have;

tanA = 5/6

tanB = 1/11

Now;

=> tan(A+B) = (tanA + tanB)/(1 - tanA•tanB)

=> tan(A+B) = (5/6 + 1/11)/{1 - (5/6)•(1/11)}

=> tan(A+B) = (5/6 + 1/11)/(1 - 5/66)

=> tan(A+B) = {(55+6)/66}/{(66-5)/66}

=> tan(A+B) = (61/66)/(61/66)

=> tan(A+B) = 1

=> tan(A+B) = tan(π/4)

=> A + B = π/4

Hence proved.

Answer 2

Answer:

We have the following values:

tanA= 5/6.

tanB = 1/11.

Now,

tan(A+B) = (tanA + tanB)/(1 - tanA•tanB).

tan(A+B)= (5/6 + 1/11)/{1-(5/6)•(1/11)}.

tan(A+B)= (5/6 + 1/11)/{1 -5/66).

tan(A+B)={(55+6)/66}/{(66-5)/66}.

tan(A+B)=(61/66)/(61/66).

tan(A+B) = 1.

tan(A+B) = tan(pi/4)

hence A+B=PI/4

HENCE PROVED