Question

if \[\dfrac{x}{y} = \dfrac{{\cos A}}{{\cos B}}\] then \[\dfrac{{x\tan A + y\tan B}}{{x + y}} = \]
(A) \[\cot \dfrac{{A + B}}{2}\]
(B) \[\cot \dfrac{{A - B}}{2}\]
(C) \[\tan \dfrac{{A - B}}{2}\]
(D) \[\tan \dfrac{{A + B}}{2}\]

Answer 1

Hint:
According to the given question, firstly calculate x in terms of y. Then convert all the trigonometric functions in terms of sin and cos and put the calculated value of x in the \[\dfrac{{x\tan A + y\tan B}}{{x + y}}\] . Then simplify the equations using trigonometric formulas to get the required answer.

Complete step by step solution:
As, it is given that \[\dfrac{x}{y} = \dfrac{{\cos A}}{{\cos B}}\] and we have to find the value of \[\dfrac{{x\tan A + y\tan B}}{{x + y}}\] ---equation 1
So, we have \[\dfrac{x}{y} = \dfrac{{\cos A}}{{\cos B}}\]
Here, we will calculate x in terms of y we get,
\[x = \dfrac{{y\cos A}}{{\cos B}}\]
Now, we put the values of x and tan in terms of sin and cos in equation 1.
\[ \Rightarrow \dfrac{{\dfrac{{y\cos A}}{{\cos B}} \times \dfrac{{\sin A}}{{\cos A}} + \dfrac{{y\sin B}}{{\cos B}}}}{{\dfrac{{y\cos A}}{{\cos B}} + y}}\]
By cancelling \[\cos A\]we get,
\[ \Rightarrow \dfrac{{\dfrac{{y\sin A}}{{\cos B}} + \dfrac{{y\sin B}}{{\cos B}}}}{{\dfrac{{y\cos A}}{{\cos B}} + y}}\]
Now we will take the L.C.M of numerator and denominator separately.
\[ \Rightarrow \dfrac{{\dfrac{{y\sin A + y\sin B}}{{\cos B}}}}{{\dfrac{{y\cos A + y\cos B}}{{\cos B}}}}\]
On rearranging the above equation we get,
\[ \Rightarrow \dfrac{{y\sin A + y\sin B}}{{\cos B}} \times \dfrac{{\cos B}}{{y\cos A + y\cos B}}\]
Cancelling out \[\cos B\] from both numerator and denominator we get,
\[ \Rightarrow \dfrac{{y\sin A + y\sin B}}{{y\cos A + y\cos B}}\]
Taking out y common from both numerator and denominator separately.
\[ \Rightarrow \dfrac{{y\left( {\sin A + \sin B} \right)}}{{y\left( {\cos A + \cos B} \right)}}\]
Cancelling y from both numerator and denominator.
Thus, we get
\[ \Rightarrow \dfrac{{\left( {\sin A + \sin B} \right)}}{{\left( {\cos A + \cos B} \right)}}\] --- Equation 2
By using the trigonometric formulas for the above equation:
As, we know
Also,
Substituting all the formulas in equation 2 we get,

Cancelling 2 and \[\cos \left( {\dfrac{{A - B}}{2}} \right)\] from both numerator and denominator.
\[ \Rightarrow \dfrac{{\sin \left( {\dfrac{{A + B}}{2}} \right)}}{{\cos \left( {\dfrac{{A + B}}{2}} \right)}}\]
As we know, \[\tan A = \dfrac{{\sin A}}{{\cos A}}\]
Therefore, we get \[ \Rightarrow \tan \left( {\dfrac{{A + B}}{2}} \right)\]
Hence, \[\dfrac{{x\tan A + y\tan B}}{{x + y}} = \tan \left( {\dfrac{{A + B}}{2}} \right)\]

So, option (D) \[\tan \dfrac{{A + B}}{2}\] is correct.

Note:
To solve these types of questions, you must remember the basic and the half angle formulas of trigonometric functions. Change the values according to the given statement and see the options so that it will give you a basic idea to reach to the solution.