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Question: How do you simplify \(\tan (x + y)\) to trigonometric functions of x and y....

How do you simplify tan⁡(x+y)\tan (x + y) to trigonometric functions of x and y.

Explanation

Solution

In order to solve this question we need to have the knowledge of trigonometric identities which will give us the relation.

Step by step solution:
For solving this we need to know the formula of:
sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta
And also we need to know:
cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
Now we will see the formula:
tan⁡(α+β)=sin⁡(α+β)cos⁡(α+β)\tan (\alpha + \beta ) = \dfrac{{\sin (\alpha + \beta )}}{{\cos (\alpha + \beta )}}
Then we will put the expanded form of sin⁡(α+β)\sin (\alpha + \beta ) and cos⁡(α+β)\cos (\alpha + \beta ) in this formula:
tan⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡βcos⁡αcos⁡β−sin⁡αsin⁡β\tan (\alpha + \beta ) = \dfrac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta - \sin \alpha \sin \beta }}
Now we will multiplying and dividing the L.H.S by cos⁡αcos⁡β\cos \alpha \cos \beta ;
tan⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡βcos⁡αcos⁡βcos⁡αcos⁡β−sin⁡αsin⁡βcos⁡αcos⁡β\tan (\alpha + \beta ) = \dfrac{{\dfrac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}{{\dfrac{{\cos \alpha \cos \beta - \sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}
On further solving we will get;
tan⁡(α+β)=sin⁡αcasα+sin⁡βcos⁡βcos⁡αcos⁡βcos⁡αcos⁡β−sin⁡αsin⁡βcos⁡αcos⁡β\tan (\alpha + \beta ) = \dfrac{{\dfrac{{\sin \alpha }}{{cas\alpha }} + \dfrac{{\sin \beta }}{{\cos \beta }}}}{{\dfrac{{\cos \alpha \cos \beta }}{{\cos \alpha \cos \beta }} - \dfrac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}
Now on further solving it we get:
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}
We have reached almost now we will be replacing α\alpha by x and β\beta by y.
tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}
So this will be our final answer.
But we should keep in mind that we always should remember these identities.

Note: Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation.
Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angled triangle.
The six basic trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent. All these trigonometric ratios are defined using the sides of the right triangle, such as an adjacent side, opposite side, and hypotenuse side. All the fundamental trigonometric identities are derived from the six trigonometric ratios