If the value of trigonometric expression $\sin \left( {A + B} \right) = 1$and $\sin \left( {A - B} \right) = \dfrac{1}{{\sqrt 3 }}$, find the value of $\tan A + \tan B$.

Hint – In this question take the sin part of the given trigonometric relations to the right hand side such that inverse trigonometric relations can be formed, we will get two equations involving A and B. Then take cos both sides so as to evaluate the expression $\dfrac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}}$ as $\tan A + \tan B = \dfrac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}}$ after simple simplification.

Complete step-by-step answer:
Given data:
$\sin \left( {A + B} \right) = 1$ and $\sin \left( {A - B} \right) = \dfrac{1}{{\sqrt 3 }}$
So these equations is written as
$ \Rightarrow A + B = {\sin ^{ - 1}}1$
$ \Rightarrow A + B = {90^0}$, $\left[ {\because {{\sin }^{ - 1}}1 = {{90}^0}} \right]$
Now take cos on both sides we have,
$ \Rightarrow \cos \left( {A + B} \right) = \cos {90^0} = 0$....................... (1), $\left[ {\because \cos {{90}^0} = 0} \right]$
And $A - B = {\sin ^{ - 1}}\dfrac{1}{{\sqrt 3 }}$.............. (2)
Now take cos on both sides in equation (2) we have,
$ \Rightarrow \cos \left( {A - B} \right) = \cos \left( {{{\sin }^{ - 1}}\dfrac{1}{{\sqrt 3 }}} \right)$
Now let $x = {\sin ^{ - 1}}\dfrac{1}{{\sqrt 3 }}$
$ \Rightarrow \cos \left( {A - B} \right) = \cos \left( x \right)$........................ (3)
$ \Rightarrow \sin x = \dfrac{1}{{\sqrt 3 }}$
As we know sin is the ratio of perpendicular to hypotenuse,
Therefore perpendicular = 1 and hypotenuse = $\sqrt 3 $
So apply Pythagoras theorem and calculate the base,
$ \Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}$
$ \Rightarrow {\left( {\sqrt 3 } \right)^2} = {\left( 1 \right)^2} + {\left( {{\text{base}}} \right)^2}$
$ \Rightarrow {\left( {{\text{base}}} \right)^2} = 3 - 1 = 2$
$ \Rightarrow {\text{Base}} = \sqrt 2 $
So as we know that cos is the ratio of base to hypotenuse,
$ \Rightarrow \cos x = \dfrac{{\sqrt 2 }}{{\sqrt 3 }}$
$ \Rightarrow x = {\cos ^{ - 1}}\sqrt {\dfrac{2}{3}} $
Now from equation (3) we have,
$ \Rightarrow \cos \left( {A - B} \right) = \cos \left( {{{\cos }^{ - 1}}\sqrt {\dfrac{2}{3}} } \right)$
$ \Rightarrow \cos \left( {A - B} \right) = \sqrt {\dfrac{2}{3}} $....................... (4)
Now add equation (1) and (4) we have,
$ \Rightarrow \cos \left( {A + B} \right) + \cos \left( {A - B} \right) = 0 + \sqrt {\dfrac{2}{3}} $
Now as we know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ and $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ so use this property in above equation we have,
$ \Rightarrow \cos A\cos B - \sin A\sin B + \cos A\cos B + \sin A\sin B = 0 + \sqrt {\dfrac{2}{3}} $
$ \Rightarrow 2\cos A\cos B = \sqrt {\dfrac{2}{3}} $
$ \Rightarrow \cos A\cos B = \dfrac{1}{2}\sqrt {\dfrac{2}{3}} $......................... (5)
Now we have to find the value of $\tan A + \tan B$
$ \Rightarrow \tan A + \tan B = \dfrac{{\sin A}}{{\cos A}} + \dfrac{{\sin B}}{{\cos B}} = \dfrac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}}$
Now as we know that $\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)$
$ \Rightarrow \tan A + \tan B = \dfrac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}} = \dfrac{{\sin \left( {A + B} \right)}}{{\cos A\cos B}}$
Now substitute the values in this equation we have,
$ \Rightarrow \tan A + \tan B = \dfrac{{\sin \left( {A + B} \right)}}{{\cos A\cos B}} = \dfrac{1}{{\dfrac{1}{2}\sqrt {\dfrac{2}{3}} }} = \dfrac{{2\sqrt 3 }}{{\sqrt 2 }}$
Now multiply and divide by $\sqrt 2 $ we have,
$ \Rightarrow \tan A + \tan B = \dfrac{{2\sqrt 3 }}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = \dfrac{{2\sqrt 6 }}{2} = \sqrt 6 $
So this is the required answer.

Note – It is always advisable to remember basic trigonometric identities while solving problems of this type. One trigonometric ratio can easily be converted into another trigonometric ratio by using the phenomena of a right angled triangle, the same thus applies to inverse trigonometric conversions. The basics of tan that is $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and cot that is $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ is very helpful for trigonometric problems.