Question

If tan (A + B) = √3 and tan (A – B) = 1 /√ 3 where 0° < A + B ≤ 90°; A > B. find A and B. ALSO CALCULATE
tan A. sin(A+B) + cos A. tan(A-B)​

Answer 1

Answer:

Solution: tan (A + B) = √3

tan (A + B) = tan 60°

(A + B) = 60° ... (i)

tan (AB) = 1/√3

tan (AB) = tan 30°

(A - B) = 30°... (ii)

Adding (i) and (ii), we get

A+B+ A-B = 60° +30°

2A = 90°

A= 45°

Putting the value of A in equation (i)

45° + B = 60°

⇒ B= 60° -45°

⇒ B= 15°

Thus, A = 45° and B = 15 °.

Answer 2

Answer:

tan(A + B) = √3

Since, A + B ≤ 90°,

Therefore,

A ≤ 90° and B ≤ 90°

According to the table, the angle at which tan acquires the value √3, is at 60°.

Therefore,

tan(A + B) = √3 = tan 60°

tan(A + B) = tan 60°

A + B = 60°  ⇢   (i)

tan(A – B) = 1/√3

According to the table, the angle at which tan acquires the value 1/√3, is at 30°. Therefore,

tan(A – B) = 1/√3 = tan 30°

tan(A – B) = tan 30°

A – B = 30°  ⇢   (ii)

Adding equation (i) and (ii),  

2A = 90°

A = 45°

Putting the value of A in equation (i),  

45° + B = 60°

B = 60°- 45°

B = 15°

Explanation:

Therefore, the value of A and B that satisfy the given equation is 45° and 15°, respectively.