How do you simplify $\tan (x + y)$ to trigonometric functions of x and y.
Answer
628.2k+ views
Hint: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 (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $
And also we need to know:
$\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $
Now we will see the formula:
$\tan (\alpha + \beta ) = \dfrac{{\sin (\alpha + \beta )}}{{\cos (\alpha + \beta )}}$
Then we will put the expanded form of $\sin (\alpha + \beta )$ and $\cos (\alpha + \beta )$ in this formula:
$\tan (\alpha + \beta ) = \dfrac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta - \sin \alpha \sin \beta }}$
Now we will multiplying and dividing the L.H.S by $\cos \alpha \cos \beta $ ;
$\tan (\alpha + \beta ) = \dfrac{{\dfrac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}{{\dfrac{{\cos \alpha \cos \beta - \sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}$
On further solving we will get;
$\tan (\alpha + \beta ) = \dfrac{{\dfrac{{\sin \alpha }}{{cas\alpha }} + \dfrac{{\sin \beta }}{{\cos \beta }}}}{{\dfrac{{\cos \alpha \cos \beta }}{{\cos \alpha \cos \beta }} - \dfrac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}$
Now on further solving it we get:
$\tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}$
We have reached almost now we will be replacing $\alpha $ by x and $\beta $ by 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
Step by step solution:
For solving this we need to know the formula of:
$\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $
And also we need to know:
$\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $
Now we will see the formula:
$\tan (\alpha + \beta ) = \dfrac{{\sin (\alpha + \beta )}}{{\cos (\alpha + \beta )}}$
Then we will put the expanded form of $\sin (\alpha + \beta )$ and $\cos (\alpha + \beta )$ in this formula:
$\tan (\alpha + \beta ) = \dfrac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta - \sin \alpha \sin \beta }}$
Now we will multiplying and dividing the L.H.S by $\cos \alpha \cos \beta $ ;
$\tan (\alpha + \beta ) = \dfrac{{\dfrac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}{{\dfrac{{\cos \alpha \cos \beta - \sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}$
On further solving we will get;
$\tan (\alpha + \beta ) = \dfrac{{\dfrac{{\sin \alpha }}{{cas\alpha }} + \dfrac{{\sin \beta }}{{\cos \beta }}}}{{\dfrac{{\cos \alpha \cos \beta }}{{\cos \alpha \cos \beta }} - \dfrac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}$
Now on further solving it we get:
$\tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}$
We have reached almost now we will be replacing $\alpha $ by x and $\beta $ by 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
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