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The trigonometric ratios cosecant, secant, and cotangent are reciprocals of sine, cosine, and tangent, respectively.

$cscθ=sinθ1 $

$secθ=cosθ1 $

$cotθ=tanθ1 $

Consider a right triangle with the three sides labeled with respect to an acute angle $θ.$

Next, the sine, cosine, tangent, cosecant, secant, and cotangent ratios are written.$sinθ=hypopp cosθ=hypadj tanθ=adjopp cscθ=opphyp secθ=adjhyp cotθ=oppadj $

The reciprocal of the sine ratio will now be calculated.
$sinθ=hypopp $

Solve for $sinθ1 $

MultEqn

$LHS⋅sinθ1 =RHS⋅sinθ1 $

$1=hypopp (sinθ1 )$

MultEqn

$LHS⋅opphyp =RHS⋅opphyp $

$opphyp =sinθ1 $

RearrangeEqn

Rearrange equation

$sinθ1 =opphyp $

$⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧ sinθ1 =opphyp .cscθ=opphyp ⇓cscθ=sinθ1 $

By following a similar procedure, the other two identities for secant and cotangent can be proven.
The tangent of an angle $θ$ can be expressed as the ratio of the sine of $θ$ to the cosine of $θ.$

$tanθ=cosθsinθ $

Similarly, the cotangent of $θ$ can be expressed as the ratio of the cosine of $θ$ to the sine of $θ.$

$cotθ=sinθcosθ $

Two proofs will be written for this identity, one using a right triangle and the other using a unit circle.

In a right triangle, the tangent of an angle $θ$ is defined as the ratio of the length of the opposite side $k$ to the length of the adjacent side $ℓ.$

At the same time, the sine and cosine of $θ$ can be written as follows.$sinθ=mk cosθ=mℓ $

By manipulating the right-hand side of the equation $tanθ=ℓk ,$ the tangent can be expressed as the sine over the cosine of $θ.$
$tanθ=ℓk $

ReduceFrac

$ba =b/ma/m $

$tanθ=ℓ/mk/m $

SubstituteII

$k/m=sin(θ)$, $ℓ/m=cos(θ)$

$tanθ=cosθsinθ $

Consider a unit circle and an angle $θ$ in standard position.

It is known that the point of intersection $P$ of the terminal side of the angle and the unit circle has coordinates $(cosθ,sinθ).$

Draw a right triangle using the origin and $P(cosθ,sinθ)$ as two of its vertices. The length of the hypotenuse is $1$ and the lengths of the legs are $sinθ$ and $cosθ.$

As shown previously, the tangent of a right triangle is defined as the ratio of the length of the opposite side — in this case, $sinθ$ — to the length of the adjacent side, which here is $cosθ.$$tanθ=cosθsinθ ✓ $

Two more proofs will be written for this identity, one of them using just a right triangle and the other using a unit circle.

In a right triangle, the cotangent of an angle $θ$ is defined as the ratio of the length of the adjacent side $ℓ$ to the length of the opposite side $k.$

Additionally, the sine and cosine of $θ$ can be written as follows.$sinθ=mk cosθ=mℓ $

By manipulating the right-hand side of the equation $cotθ=kℓ ,$ the cotangent can be expressed as the cosine over the sine of $θ.$
$cotθ=kℓ $

ReduceFrac

$ba =b/ma/m $

$cotθ=k/mℓ/m $

SubstituteII

$ℓ/m=cos(θ)$, $k/m=sin(θ)$

$cotθ=sinθcosθ $

$tanθ=cosθsinθ $

By manipulating the above equation, it can be shown that the cotangent of $θ$ is the ratio of the cosine of $θ$ to the sine of $θ.$
$tanθ=cosθsinθ $

MultEqn

$LHS⋅sinθcosθ =RHS⋅sinθcosθ $

$tanθ(sinθcosθ )=1$

DivEqn

$LHS/tanθ=RHS/tanθ$

$sinθcosθ =tanθ1 $

$cot(θ)=tan(θ)1 $

$sinθcosθ =cotθ$

RearrangeEqn

Rearrange equation

$cotθ=sinθcosθ ✓$

For any angle $θ,$ the following trigonometric identities hold true.

$sin_{2}θ+cos_{2}θ=1$

$1+tan_{2}θ=sec_{2}θ$

$1+cot_{2}θ=csc_{2}θ$

Consider a right triangle with a hypotenuse of $1.$

Since $cosθ$ represents a side length, it is **not** $0.$ Therefore, by diving both sides of the above equation by $cos_{2}θ,$ the second identity can be obtained.
The second identity was obtained.
Since $sinθ$ represents a side length, it is **not** $0.$ Therefore, by dividing both sides of $sin_{2}θ+cos_{2}θ=1$ by $sin_{2}θ,$ the third identity can be proven.
Finally, the third identity was obtained.

By recalling the sine and cosine ratios, the lengths of the opposite and adjacent sides to $∠θ$ can be expressed in terms of the angle.

Definition | Substitute | Simplify | |
---|---|---|---|

$sinθ$ | $HypotenuseLength ofoppositeside to∠θ $ | $1opp $ | $opp$ |

$cosθ$ | $HypotenuseLength ofadjacentside to∠θ $ | $1adj $ | $adj$ |

It can be seen that if the hypotenuse of a right triangle is $1,$ the sine of an acute angle is equal to the length of its opposite side. Similarly, the cosine of the angle is equal to the length of its adjacent side.

By the Pythagorean Theorem, the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Therefore, for the above triangle, the sum of the squares of $sinθ$ and $cosθ$ is equal to the square of $1.$

$sin_{2}θ+cos_{2}θ⇓sin_{2}θ+cos_{2}θ =1_{2}=1 $

$sin_{2}θ+cos_{2}θ=1$

DivEqn

$LHS/cos_{2}θ=RHS/cos_{2}θ$

$cos_{2}θsin_{2}θ+cos_{2}θ =cos_{2}θ1 $

Simplify

WriteSumFrac

Write as a sum of fractions

$cos_{2}θsin_{2}θ +cos_{2}θcos_{2}θ =cos_{2}θ1 $

QuotOne

$aa =1$

$cos_{2}θsin_{2}θ +1=cos_{2}θ1 $

WritePow

Write as a power

$cos_{2}θsin_{2}θ +1=cos_{2}θ1_{2} $

$b_{m}a_{m} =(ba )_{m}$

$(cosθsinθ )_{2}+1=(cosθ1 )_{2}$

$cosθsinθ =tanθ$

$tan_{2}θ+1=(cosθ1 )_{2}$

$cosθ1 =secθ$

$tan_{2}θ+1=sec_{2}θ$

CommutativePropAdd

Commutative Property of Addition

$1+tan_{2}θ=sec_{2}θ$

$1+tan_{2}θ=sec_{2}θ$

$sin_{2}θ+cos_{2}θ=1$

DivEqn

$LHS/sin_{2}θ=RHS/sin_{2}θ$

$sin_{2}θsin_{2}θ+cos_{2}θ =sin_{2}θ1 $

Simplify

WriteSumFrac

Write as a sum of fractions

$sin_{2}θsin_{2}θ +sin_{2}θcos_{2}θ =sin_{2}θ1 $

QuotOne

$aa =1$

$1+sin_{2}θcos_{2}θ =sin_{2}θ1 $

WritePow

Write as a power

$1+sin_{2}θcos_{2}θ =sin_{2}θ1_{2} $

$b_{m}a_{m} =(ba )_{m}$

$1+(sinθcosθ )_{2}=(sinθ1 )_{2}$

$sinθcosθ =cotθ$

$1+cot_{2}θ=(sinθ1 )_{2}$

$sinθ1 =cscθ$

$1+cot_{2}θ=csc_{2}θ$

$1+cot_{2}θ=csc_{2}θ$

The first identity can be shown using the unit circle and the Pythagorean Theorem. Consider a point $(x,y)$ on the unit circle in the first quadrant, corresponding to the angle $θ.$ A right triangle can be constructed with $θ.$

By the Pythagorean Theorem, the sum of the squares of $x$ and $y$ equals $1.$$x_{2}+y_{2}=1 $

In fact, this is true not only for points in the first quadrant, but for $cos_{2}θ+sin_{2}θ=1$

The function $y=sin(x)$ has odd symmetry, and $y=cos(x)$ has even symmetry, which can be seen from their graphs. As a result, the corresponding identities hold true.

$sin(-θ)=-sinθcos(-θ)=cosθtan(-θ)=-tanθ$

The identities will be proven using a unit circle. Consider an arbitrary $∠θ$ on a unit circle. Let $A$ be the point that the angle forms on the circle.

Recall that the values on the $x-$axis are represented by cosine and the value on the $y-$axis are represented by sine. Therefore, the coordinates of $A$ are $(cosθ,sinθ).$

Next, the point $A$ can be reflected over the $x-$axis.
Since $A_{′}$ is the same distance from the $y-$axis as $A,$ its $x-$coordinate is also $cosθ.$ Note that a reflection is a congruent transformation, so $A$ and $A_{′}$ are equidistant from the $x-$axis too. However, $A_{′}$ is below the $x-$axis, so its $y-$coordinate is a negative $sinθ.$

Additionally, after the reflection, the angle between $A_{′},$ the origin, and the $x-$axis is $-θ.$ This means that the lengths of the newly created horizontal and vertical segments are $cos(-θ)$ and $sin(-θ).$ Therefore, the coordinates of $A_{′}$ can also be written as $(cos(-θ),sin(-θ)).$

This way there are two pairs of coordinates of the same point $A_{′}.$ This allows to conclude the following.$A_{′}(cos(-θ),sin(-θ))cos(-θ)sin(-θ) andA_{′}(cosθ,-sinθ)⇓=cosθ=-sinθ $

By expressing $tan(-θ)$ using sine and cosine, this identity can be shown.

$tan(-θ)$

$tan(θ)=cos(θ)sin(θ) $

$cos(-θ)sin(-θ) $

$sin(-θ)=-sin(θ)$

$cos(-θ)-sin(θ) $

$cos(-θ)=cos(θ)$

$cos(θ)-sin(θ) $

MoveNegNumToFrac

Put minus sign in front of fraction

$-cos(θ)sin(θ) $

$cos(θ)sin(θ) =tan(θ)$

$-tan(θ)$

For any angle $θ,$ the following trigonometric identities hold true.

$sin(2π −θ)=cosθ$

$cos(2π −θ)=sinθ$

$tan(2π −θ)=cotθ$

Consider a right triangle. The measure of its right angle is $90_{∘}$ or $2π $ radians. Let $θ$ be the radian measure of one of the acute angles. Since the sum of two acute angles in a right triangle is $2π ,$ the measure of the third acute angle must be $2π −θ.$

Let also $a,$ $b,$ and $c$ represent the side lengths of the triangle. In this case, cosine of $θ$ can be expressed as the ratio of the lengths of the angle's adjacent side and the hypotenuse.$cosθ=cb $

At the same time, sine of the opposite angle can be expressed as the ratio of the lengths of the angle's opposite side and the hypotenuse.
$sin(2π −θ)=cb $

Since the right-hand sides of the equations are equal, by the Transitive Property of Equality, the left-hand sides are also equal. $sin(2π −θ)=cosθ$

This identity is true for all angles, not just those that make it possible to construct a right triangle. Using similar reasoning, the corresponding identities for cosine and tangent can be proven.

The given identities can be proven using the Angle Sum and Difference Identities for any angle measure. Consider the identity for the sine of the difference between two angles.
Therefore, the Cofunction Identity for sine was obtained.

$sin(x−y)=sinxcosy−cosxsiny $

This identity will be applied to the left-hand side of the first identity.
$sin(2π −θ)$

$sin(α−β)=sin(α)cos(β)−cos(α)sin(β)$

$sin(2π )cos(θ)−cos(2π )sin(θ)$

$1⋅cos(θ)−cos(2π )sin(θ)$

MultByOne

$a⋅1=a$

$cos(θ)−cos(2π )sin(θ)$

$cos(θ)−0⋅sin(θ)$

ZeroPropMult

Zero Property of Multiplication

$cos(θ)−0$

IdPropAdd

Identity Property of Addition

$cos(θ)$

$sin(2π −θ)=cos(θ)$

To evaluate trigonometric functions of the sum of two angles, the following identities can be applied.

$sin(x+y)cos(x+y)tan(x+y) =sinxcosy+cosxsiny=cosxcosy−sinxsiny=1−tanxtanytanx+tany $

There are also similar identities for the difference of two angles.

$sin(x−y)cos(x−y)tan(x−y) =sinxcosy−cosxsiny=cosxcosy+sinxsiny=1+tanxtanytanx−tany $

Let $△AFD$ be a right triangle with hypotenuse $1$ and an acute angle with measure $x+y.$

By definition, the sine of an angle is the ratio between the lengths of the opposite side and the hypotenuse.$sin(x+y)=1DF ⇓DF=sin(x+y) $

The idea now is to rewrite $DF$ in terms of $sinx,$ $siny,$ $cosx,$ and $cosy.$ To do it, draw a ray so that $∠A$ is divided into two angles with measures $x$ and $y.$ Let $C$ be a point on this ray such that $△ACD$ and $△ABC$ are right triangles.
Consider $△ACD.$ By calculating the sine and cosine of $x,$ the legs of this triangle can be rewritten.
$sinx=1DC ⇒DC=sinxcosx=1AC ⇒AC=cosx $

Now consider $△ABC.$ Knowing that $AC=cosx,$ the sine of $y$ can be used to write $BC$ in terms of $x$ and $y.$
$siny=ACBC $

Solve for $BC$

Substitute

$AC=cosx$

$siny=cosxBC $

MultEqn

$LHS⋅cosx=RHS⋅cosx$

$cosxsiny=BC$

RearrangeEqn

Rearrange equation

$BC=cosxsiny$

By the Third Angle Theorem, it is known that $∠GAF≅∠GDC.$ Therefore, $m∠GDC=y.$

Since the purpose is to rewrite $DF,$ plot a point $E$ on $DF$ such that $EC∥AB.$ This way a rectangle $ECBF$ is formed. The opposite sides of a rectangle have the same length, so $EF$ and $CB$ are equal. Also, $CE⊥DF$ makes $△CED$ a right triangle.

Consequently, $EF=cosxsiny$ and $DE$ can be written in terms of $sinx$ and $cosy$ using the cosine ratio.$cosy=sinxDE ⇓DE=sinxcosy $

Finally, by the Segment Addition Postulate, $DF$ is equal to the sum of $DE$ and $EF.$ All these lengths have been rewritten in terms of the sine and cosine of $x$ and $y.$
$DF=DE+EF⇓sin(x+y)=sinxcosy+cosxsiny $

This concludes the proof of the first identity. The other identities can be proven using similar reasoning.
Consider the following process for calculating the exact value of $sin120_{∘}.$

- To be able to use the angle sum identities, the angle $120_{∘}$ needs to be rewritten as the sum of two angles for which the sine and cosine are known. For example, $120_{∘}$ can be rewritten as $90_{∘}+30_{∘}.$
- Use the first formula for the angle sum.
- Based on the trigonometric ratios of common angles, it is known that $sin90_{∘}=1,$ $sin30_{∘}=21 ,$ $cos90_{∘}=0,$ and $cos30_{∘}=23 .$

$sin120_{∘}$

Rewrite

Rewrite $120_{∘}$ as $90_{∘}+30_{∘}$

$sin(90_{∘}+30_{∘})$

$sin(x+y)=sinxcosy+cosxsiny$

$sin90_{∘}⋅cos30_{∘}+cos90_{∘}⋅sin30_{∘}$

SubstituteValues

Substitute values

$1⋅23 +0⋅21 $

Simplify

OneMult

$1⋅a=a$

$23 +0⋅21 $

ZeroPropMult

Zero Property of Multiplication

$23 +0$

IdPropAdd

Identity Property of Addition

$23 $

The double-angle identities materialize when two angles with the same measure are substituted into the angle sum identities.

$Sine:Cosine:Tangent: sin2θ=2sinθcosθcos2θ=cos_{2}θ−sin_{2}θcos2θ=2cos_{2}θ−1cos2θ=1−2sin_{2}θtan2θ=1−tan_{2}θ2tanθ $

Start by writing the Angle Sum Identity for sine and cosine.
### Sine Identity

### Cosine Identities

Now, recall that, by the Pythagorean Identity, the sine square plus the cosine square of the same angle equals $1.$ From this identity, two different equations can be set.
That way, the second identity for the cosine has been obtained. To obtain the third cosine identity, substitute Equation (II) into the first identity for the cosine.
### Tangent Identity

To prove the tangent identity, start by rewriting $tan2θ$ in terms of sine and cosine.

$sin(x+y)=sinxcosy+cosxsinycos(x+y)=cosxcosy−sinxsiny $

Let $x=θ$ and $y=θ.$ With this, $x+y$ becomes $2θ.$ Then, these two formulas can be rewritten in terms of $θ.$
$sin2θ=sinθcosθ+cosθsinθcos2θ=cosθcosθ−sinθsinθ $

Approaching the first equation, the Commutative Property of Multiplication can be applied to the second term of its right-hand side. Then, by adding the terms on the right-hand side of this equation, the formula for $sin2θ$ is obtained.

$sin2θ=sinθcosθ+sinθcosθ⇓sin2θ=2sinθcosθ✓ $

Approaching the second equation, the Product of Powers Property can be used to rewrite its right-hand side. By doing this, the first identity for the cosine of the double of an angle is obtained.

$cos2θ=cosθcosθ−sinθsinθ⇓cos2θ=cos_{2}θ−sin_{2}θ✓ $

$sin_{2}θ+cos_{2}θ=1⇓{sin_{2}θ=1−cos_{2}θcos_{2}θ=1−sin_{2}θ (I)(II) $

Next, substitute Equation (I) into the first identity for the cosine.
$cos2θ=cos_{2}θ−sin_{2}θ$

Substitute $1−cos_{2}θ$ for $sin_{2}θ$ and simplify

Substitute

$sin_{2}θ=1−cos_{2}θ$

$cos2θ=cos_{2}θ−(1−cos_{2}θ)$

Distr

Distribute $-1$

$cos2θ=cos_{2}θ−1+cos_{2}θ$

AddTerms

Add terms

$cos2θ=2cos_{2}θ−1✓$

$cos2θ=cos_{2}θ−sin_{2}θ$

Substitute $1−sin_{2}θ$ for $cos_{2}θ$ and simplify

$cos2θ=1−2sin_{2}θ✓$

$tan2θ=cos2θsin2θ $

Next, substitute the first sine identity in the numerator and the first cosine identity in the denominator.
$tan2θ=cos_{2}θ−sin_{2}θ2sinθcosθ $

Then, divide the numerator and denominator by $cos_{2}θ.$
$tan2θ=cos_{2}θcos_{2}θ−sin_{2}θ cos_{2}θ2sinθcosθ $

Finally, simplifying the right-hand side the tangent identity will be obtained.
$tan2θ=cos_{2}θcos_{2}θ−sin_{2}θ cos_{2}θ2sinθcosθ $

Simplify right-hand side

WriteDiffFrac

Write as a difference of fractions

$tan2θ=cos_{2}θcos_{2}θ −cos_{2}θsin_{2}θ cos_{2}θ2sinθcosθ $

CrossCommonFac

Cross out common factors

$tan2θ=cos_{2}θcos_{2}θ −cos_{2}θsin_{2}θ cos_{2}θ2sinθcosθ $

CancelCommonFac

Cancel out common factors

$tan2θ=1−cos_{2}θsin_{2}θ cosθ2sinθ $

$b_{m}a_{m} =(ba )_{m}$

$tan2θ=1−(cosθsinθ )_{2}cosθ2sinθ $

MovePartNumLeft

$ca⋅b =a⋅cb $

$tan2θ=1−(cosθsinθ )_{2}2⋅cosθsinθ $

$cos(θ)sin(θ) =tan(θ)$

$tan2θ=1−tan_{2}θ2tanθ ✓$

To calculate the exact value of $cos120_{∘},$ these steps can be followed.

- To be able to use the double-angle identities, the angle $120_{∘}$ needs to be rewritten as $2$ multiplied by another angle. Therefore, rewrite $120_{∘}$ as $2⋅60_{∘}.$
- Use the second formula for the cosine of twice an angle.
- Based on the trigonometric ratios of common angles, it is known that $cos60_{∘}=21 .$

$cos120_{∘}$

Rewrite

Rewrite $120_{∘}$ as $2⋅60_{∘}$

$cos(2⋅60_{∘})$

$cos2θ=2cos_{2}θ−1$

$2cos_{2}(60_{∘})−1$

$cos60_{∘}=21 $

$2(21 )_{2}−1$

$-21 $

The half-angle identities are special cases of angle difference identities. To evaluate trigonometric functions of half an angle, the following identities can be applied.

$sin2θ cos2θ tan2θ =±21−cosθ =±21+cosθ =±1+cosθ1−cosθ $

The sign of each formula is determined by the quadrant where the angle $2θ $ lies.

These identities are useful when finding the exact value of the sine, cosine, or tangent at a given angle.

First, write two of the Double-Angle Identities for cosine.
### Sine Identity

Start by solving the first identity written above for $sinx.$
Next, substitute $2θ $ for $x$ to obtain the half-angle identity for the sine.
### Cosine Identity

Start by solving the second identity written at the beginning for $cosx.$
Next, substitute $2θ $ for $x$ to obtain the half-angle identity for the cosine.
### Tangent Identity

To derive the tangent identity, start by recalling the definition of the tangent ratio.

$cos2xcos2x =1−2sin_{2}x=2cos_{2}x−1 $

$cos2x=1−2sin_{2}x$

Solve for $sinx$

SubEqn

$LHS−1=RHS−1$

$cos2x−1=-2sin_{2}x$

DivEqn

$LHS/(-2)=RHS/(-2)$

$-2cos2x−1 =sin_{2}x$

RearrangeEqn

Rearrange equation

$sin_{2}x=-2cos2x−1 $

MoveNegDenomToFrac

Put minus sign in front of fraction

$sin_{2}x=-2cos2x−1 $

DistrNegSignSwap

$-(b−a)=a−b$

$sin_{2}x=21−cos2x $

SqrtEqn

$LHS =RHS $

$sinx=±21−cos2x $

$sin2θ sin2θ =±21−cos(2⋅2θ ) ⇓=±21−cosθ ✓ $

$cos2x=2cos_{2}x−1$

Solve for $cosx$

AddEqn

$LHS+1=RHS+1$

$1+cos2x=2cos_{2}x$

DivEqn

$LHS/2=RHS/2$

$21+cos2x =cos_{2}x$

RearrangeEqn

Rearrange equation

$cos_{2}x=21+cos2x $

SqrtEqn

$LHS =RHS $

$cosx=±21+cos2x $

$cos2θ cos2θ =±21+cos(2⋅2θ ) ⇓=±21+cosθ ✓ $

$tanx=cosxsinx $

Next, substitute $2θ $ for $x.$
$tan2θ =cos(2θ )sin(2θ ) $

Finally, substitute the half-identities for the sine and cosine into the equation above and simplify the right-hand side.
$tan2θ =cos(2θ )sin(2θ ) $

SubstituteII

$sin2θ =±21−cosθ $, $cos2θ =±21+cosθ $

$tan2θ =±21+cosθ ±21−cosθ $

Simplify right-hand side

DivSqrt

$b a =ba $

$tan2θ =±21+cosθ 21−cosθ $

DivFracByFracD

$c/da/b =ba ⋅cd $

$tan2θ =±21−cosθ ⋅1+cosθ2 $

CrossCommonFac

Cross out common factors

$tan2θ =±2 1−cosθ ⋅1+cosθ2 $

CancelCommonFac

Cancel out common factors

$tan2θ =±11−cosθ ⋅1+cosθ1 $

MultFrac

Multiply fractions

$tan2θ =±1+cosθ1−cosθ ✓$

Consider the calculation of the exact value of $cos15_{∘}.$

- To be able to use the half-angle identities, the angle $15_{∘}$ needs to be rewritten as a certain angle divided by $2.$ Therefore, rewrite $15_{∘}$ as $230_{∘} .$
- Based on the trigonometric ratios of common angles, it is known that $cos30_{∘}=23 .$
- According to the diagram of the quadrants, an angle that measures $15_{∘}$ is in the first quadrant. Therefore, the cosine ratio is positive.

$cos15_{∘}$

Rewrite

Rewrite $15_{∘}$ as $230_{∘} $

$cos230_{∘} $

$cos2θ =21+cosθ $

$21+cos30_{∘} $

$cos30_{∘}=23 $

$21+23 $

Simplify

OneToFrac

Rewrite $1$ as $22 $

$222 +23 $

AddFrac

Add fractions

$222+3 $

MoveNumRight

$ba =b1 ⋅a$

$21 (22+3 ) $

MultFrac

Multiply fractions

$42+3 $