Question

If $A$ and $B$ are acute angles such that $\tan A=\dfrac{1}{3}$ , $\tan B=\dfrac{1}{2}$ and $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$, show that \[A+\text{ }B\text{ }=\text{ }45{}^\circ \].

Answer 1

Hint: Here we are given, $\tan A=\dfrac{1}{3}$ , $\tan B=\dfrac{1}{2}$and $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. So, take RHS and solve it by putting the values and equate it to LHS.

Complete step-by-step answer:
Now we are given that, $A$ and $B$ are acute angles such that $\tan A=\dfrac{1}{3}$ , $\tan B=\dfrac{1}{2}$and $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.
Now taking,
$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
So, taking RHS we get,
RHS$=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
RHS$=\dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{1-\left( \dfrac{1}{3} \right)\left( \dfrac{1}{2} \right)}$ ……………. (given $\tan A=\dfrac{1}{3}$ , $\tan B=\dfrac{1}{2}$ )
RHS$=\dfrac{\dfrac{5}{6}}{1-\left( \dfrac{1}{6} \right)}$
Simplifying we get,
RHS$=\dfrac{\dfrac{5}{6}}{\dfrac{5}{6}}$
RHS$=1$
Now we have RHS$=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=1$.
Now we are given LHS$=$RHS,
$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=1$
So,
$\tan (A+B)=1$
Now we know that, $\tan 45{}^\circ =1$
Hence, we get, $\tan (A+B)=\tan 45{}^\circ $
So now from both sides we get,
$(A+B)=45{}^\circ $
Hence proved.

Additional information:
Trigonometric ratio is defined as the values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle. The ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle. Trigonometric functions are also known as a Circular Functions can be simply defined as the functions of an angle of a triangle. It means that the relationship between the angles and sides of a triangle are given by these trig functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant.

Note: Tangent of an angle is defined as the ratio of the side opposite to that angle to the side adjacent to that angle. The tangent function is the ratio of the length of the opposite side to that of the adjacent side. It should be noted that the tan can also be represented in terms of sine and cos as their ratio.