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Question: \(\dfrac{x}{y} = \dfrac{{\cos A}}{{\cos B}}\)then \(\dfrac{{x\tan A + y\tan B}}{{x + y}} = \) A) \...

xy=cos⁡Acos⁡B\dfrac{x}{y} = \dfrac{{\cos A}}{{\cos B}}then xtan⁡A+ytan⁡Bx+y=\dfrac{{x\tan A + y\tan B}}{{x + y}} =
A) cot⁡A+B2\cot \dfrac{{A + B}}{2}
B) cot⁡A−B2\cot \dfrac{{A - B}}{2}
C) tan⁡A+B2\tan \dfrac{{A + B}}{2}
D) tan⁡A−B2\tan \dfrac{{A - B}}{2}

Explanation

Solution

We have given the value of xy=cos⁡Acos⁡B\dfrac{x}{y} = \dfrac{{\cos A}}{{\cos B}}, we have to find xtan⁡A+ytan⁡Bx+y\dfrac{{x\tan A + y\tan B}}{{x + y}}
For this firstly, we have to find the value of x then we have to put the value of x in the trigonometric expression. After that we have to simplify the expression. The simplification is done by converting the tan function into sin and cos function. Once a tan function is written in the sin and cos function, we will cancel out the common factor and apply a trigonometric formula. This will lead to the result.

Complete step by step solution:
We have given that xy=cos⁡Acos⁡B\dfrac{x}{y} = \dfrac{{\cos A}}{{\cos B}}………………..(i)
We have to find xtan⁡A+ytan⁡Bx+y\dfrac{{x\tan A + y\tan B}}{{x + y}}…………(ii)
Now firstly we have to find value of y from (i)
y=xcos⁡Bcos⁡Ay = \dfrac{{x\cos B}}{{\cos A}}now put this value of y in the equation (ii)
xtan⁡A+ytan⁡Bx+y=xtan⁡A+(xcos⁡Bcos⁡A)tan⁡Bx+xcos⁡Bcos⁡A\dfrac{{x\tan A + y\tan B}}{{x + y}} = \dfrac{{x\tan A + \left( {\dfrac{{x\cos B}}{{\cos A}}} \right)\tan B}}{{x + x\dfrac{{\cos B}}{{\cos A}}}}……………….(iii)
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
Therefore tan⁡A=sin⁡Acos⁡A,tan⁡B=sin⁡Bcos⁡B\tan A = \dfrac{{\sin A}}{{\cos A}},\tan B = \dfrac{{\sin B}}{{\cos B}}putting these in equation (iii)
xtan⁡A+ytan⁡Bx+y=xsin⁡Acos⁡A+(xcos⁡Bcos⁡A)sin⁡Bcos⁡Bx+xcos⁡Bcos⁡A\dfrac{{x\tan A + y\tan B}}{{x + y}} = \dfrac{{x\dfrac{{\sin A}}{{\cos A}} + \left( {\dfrac{{x\cos B}}{{\cos A}}} \right)\dfrac{{\sin B}}{{\cos B}}}}{{x + x\dfrac{{\cos B}}{{\cos A}}}}
Take x common from numerator and denominator and cancel them with each other. We get;
xtan⁡A+ytan⁡Bx+y=sin⁡Acos⁡A+(cos⁡Bcos⁡A)sin⁡Bcos⁡B1+cos⁡Bcos⁡A\dfrac{{x\tan A + y\tan B}}{{x + y}} = \dfrac{{\dfrac{{\sin A}}{{\cos A}} + \left( {\dfrac{{\cos B}}{{\cos A}}} \right)\dfrac{{\sin B}}{{\cos B}}}}{{1 + \dfrac{{\cos B}}{{\cos A}}}}
=sin⁡A+sin⁡Bcos⁡Acos⁡A+cos⁡Bcos⁡A=sin⁡A+sin⁡Bcos⁡A+cos⁡B= \dfrac{{\dfrac{{\sin A + \sin B}}{{\cos A}}}}{{\dfrac{{\cos A + \cos B}}{{\cos A}}}} = \dfrac{{\sin A + \sin B}}{{\cos A + \cos B}}…………………..(iv)
Now using the trigonometric formula sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}and cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}
Putting these values in equation (iv)
xtan⁡A+ytan⁡Bx+y=2sin⁡A+B2cos⁡A−B22cos⁡A+B2cos⁡A−B2\dfrac{{x\tan A + y\tan B}}{{x + y}} = \dfrac{{2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}}}{{2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}}}
=sin⁡(A+B2)cos⁡(A−B2)=tan⁡A+B2= \dfrac{{\sin \left( {\dfrac{{A + B}}{2}} \right)}}{{\cos \left( {\dfrac{{A - B}}{2}} \right)}} = \tan \dfrac{{A + B}}{2}

So, xtan⁡A+ytan⁡Bx+y=tan⁡A+B2\dfrac{{x\tan A + y\tan B}}{{x + y}} = \tan \dfrac{{A + B}}{2}. Therefore option D is correct.

Note:
Trigonometry is the branch of mathematics that studies the relationship between side lengths and angle of triangle. Trigonometry has six functions which are sin, cos, tan, cosec, sec, and cot. Trigonometric functions are the real functions which relate an angle right angle triangle to the ratio of two sides of a triangle.