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Hint: Compare the given two trigonometric functions and use the value of tan${30^0}$ is $\dfrac{1}{{\sqrt 3 }}$ and period of tangent is ${180^0}$.

Given,

$\tan ({\text{A + B) = }}\dfrac{1}{{\sqrt 3 }}$………………………………………………………..(1)

$\tan ({\text{A - B) = }}\dfrac{1}{{\sqrt 3 }}$………………………………………………………….(2)

So, the period of tan is ${180^0}$. Neither A nor B can be near this angle because we have stated in the question that the sum of A and B must be less than 90$^0$.

So, from equation (1) and (2)

tan(A+B) = tan(A – B)

Equating the angles of tangents , we get

A + B = A – B

A + B – A + B = 0

2B = 0

B = 0$^0$

Putting the value of B in equation (1), we get

tan(A + 0) = $\dfrac{1}{{\sqrt 3 }}$

tanA = tan${30^0}$

A = ${30^0}$.

Hence,

A = ${30^0}$ and B = 0$^0$ is the answer.

Note: In these types of questions, the periodicity of the trigonometric function matters. Periodicity is the property of the function such that it repeats itself after a fixed interval of time called period of the function.

Given,

$\tan ({\text{A + B) = }}\dfrac{1}{{\sqrt 3 }}$………………………………………………………..(1)

$\tan ({\text{A - B) = }}\dfrac{1}{{\sqrt 3 }}$………………………………………………………….(2)

So, the period of tan is ${180^0}$. Neither A nor B can be near this angle because we have stated in the question that the sum of A and B must be less than 90$^0$.

So, from equation (1) and (2)

tan(A+B) = tan(A – B)

Equating the angles of tangents , we get

A + B = A – B

A + B – A + B = 0

2B = 0

B = 0$^0$

Putting the value of B in equation (1), we get

tan(A + 0) = $\dfrac{1}{{\sqrt 3 }}$

tanA = tan${30^0}$

A = ${30^0}$.

Hence,

A = ${30^0}$ and B = 0$^0$ is the answer.

Note: In these types of questions, the periodicity of the trigonometric function matters. Periodicity is the property of the function such that it repeats itself after a fixed interval of time called period of the function.

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