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Question: If \[\tan \theta = \dfrac{x}{y}\] then find \[\dfrac{{x\sin \theta + y\cos \theta }}{{x\sin \theta -...

If tan⁡θ=xy\tan \theta = \dfrac{x}{y} then find xsin⁡θ+ycos⁡θxsin⁡θ−ycos⁡θ\dfrac{{x\sin \theta + y\cos \theta }}{{x\sin \theta - y\cos \theta }}

Explanation

Solution

Trigonometry is a study of triangles which is defined by three basic functions called Sine, Cosine and Tangent. It helps to find angles in the right-angled triangle and also used to find missing sides of a triangle. In a right-angled triangle the side opposite to the angle is called oppositeopposite , the slanted arm of the right angle is called hypotenusehypotenuse and the other arm of the right-angle is called adjacentadjacent .
The basic trigonometry functions are calculated by sin⁡=opphyp\sin = \dfrac{{opp}}{{hyp}} , cos⁡=adjhyp\cos = \dfrac{{adj}}{{hyp}} and tan⁡=oppadj\tan = \dfrac{{opp}}{{adj}} .
Thus, tangent function can be defined as, tan⁡=sin⁡cos⁡\tan = \dfrac{{\sin }}{{\cos }} .

Complete step by step answer:
It is given that tan⁡θ=xy\tan \theta = \dfrac{x}{y} then we have to find the value of xsin⁡θ+ycos⁡θxsin⁡θ−ycos⁡θ\dfrac{{x\sin \theta + y\cos \theta }}{{x\sin \theta - y\cos \theta }} .
Consider the given expression xsin⁡θ+ycos⁡θxsin⁡θ−ycos⁡θ\dfrac{{x\sin \theta + y\cos \theta }}{{x\sin \theta - y\cos \theta }} , let us divide and multiply this expression by cos⁡θ\cos \theta .
xsin⁡θ+ycos⁡θxsin⁡θ−ycos⁡θ=xsin⁡θcos⁡θ+ycos⁡θcos⁡θxsin⁡θcos⁡θ−ycos⁡θcos⁡θ\dfrac{{x\sin \theta + y\cos \theta }}{{x\sin \theta - y\cos \theta }} = \dfrac{{\dfrac{{x\sin \theta }}{{\cos \theta }} + \dfrac{{y\cos \theta }}{{\cos \theta }}}}{{\dfrac{{x\sin \theta }}{{\cos \theta }} - \dfrac{{y\cos \theta }}{{\cos \theta }}}}
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} , applying this to the above expression we get,
=xtan⁡θ+yxtan⁡θ−y= \dfrac{{x\tan \theta + y}}{{x\tan \theta - y}}
It is given in the question that, tan⁡θ=xy\tan \theta = \dfrac{x}{y} . So let us apply this to the above expression,
=x(xy)+yx(xy)−y= \dfrac{{x\left( {\dfrac{x}{y}} \right) + y}}{{x\left( {\dfrac{x}{y}} \right) - y}}
Simplifying the above expression, we get
=(x2y)+y(x2y)−y= \dfrac{{\left( {\dfrac{{{x^2}}}{y}} \right) + y}}{{\left( {\dfrac{{{x^2}}}{y}} \right) - y}}
On further simplification we get,
=x2+y2yx2−y2y= \dfrac{{\dfrac{{{x^2} + {y^2}}}{y}}}{{\dfrac{{{x^2} - {y^2}}}{y}}}
Making few more simplification we get,
=x2+y2x2−y2= \dfrac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}}
We cannot simplify it further, since the above expression contains algebraic identities. If we try to apply the identity it will become an elaborated expression so we will stop simplifying this further.
Thus, xsin⁡θ+ycos⁡θxsin⁡θ−ycos⁡θ=x2+y2x2−y2\dfrac{{x\sin \theta + y\cos \theta }}{{x\sin \theta - y\cos \theta }} = \dfrac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}} .

Note: This question can also be solved by comparing the given value of tangent function that is, tan⁡θ=sin⁡θcos⁡θ=xy⇒sin⁡θ=x,cos⁡θ=y\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{x}{y} \Rightarrow \sin \theta = x,\cos \theta = y .By substituting this in the given expression and simplifying it will also yield a solution in terms of trigonometry solution. Since the given hypothesis is in the terms of x&yx\& y we have solved this question in this method to yield the answer in terms of x&yx\& y .