Question

If tan(A+B) =√3 and tan (A-B) =1/√3 0°angle A+B 90°A B find A and B​

Answer 1

Step-by-step explanation:

tan(A+B)= √3 (tan60°=√3)

A+B= 60 °

A= 60°-B (1)

tan(A-B)= 1/√3 (tan30°= 1/√3)

A-B= 30°

From equation (1)

(60°-B)-B= 30°

60°-2B= 30°

-2B= 30°-60°

B= -30/-2= 15°

Therefore

A= 60°-B

= 60°-15

= 45

Therefore A= 45° and B = 15°

Answer 2

EXPLANATION.

⇒ tan(A + B) = √3 - - - - - (1).

⇒ tan(A - B) = 1/√3. - - - - - (2).

⇒ 0° ≤ A + B ≤ 90°.

As we know that,

We can write equation as,

⇒ tan(A + B) = tan60° - - - - - (1).

⇒ tan(A - B) = tan30° - - - - - (2).

Adding equation (1) & (2), we get.

⇒ A + B = 60° - - - - - (1).

⇒ A - B = 30° - - - - - (2).

We get,

⇒ 2A = 90°.

⇒ A = 45°.

Put the value of A = 45° in equation (1), we get.

⇒ A + B = 60°.

⇒ 45° + B = 60°.

⇒ B = 60° - 45°.

⇒ B = 15°.

Value of A = 45° & B = 15°.

                                                                                                                       

MORE INFORMATION.

Fundamental trigonometric identities.

(1) = sin²θ + cos²θ = 1.

(2) = 1 + tan²θ = sec²θ.

(3) = 1 + cot²θ = cosec²θ.