Question

If $\tan ({\text{A + B) = }}\dfrac{1}{{\sqrt 3 }}$ and $\tan ({\text{A - B) = }}\dfrac{1}{{\sqrt 3 }}$, ${0^0}$< A + B < ${90^0}$. Find A and B.

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.