Opposite Of Tangent

In Trigonometry, we have six important trigonometric functions. The main three trigonometric functions equations are sine, cosine, tangent. The opposite three trigonometric functions are cosecant, secant, cotangent.

So, Opposite of tangent is Cotangent.

Cotangent is shortly called as ctg.

Cotangent is also referred as reciprocal of tangent trigonometric function.

The cotangent of angle ? in a right-angle triangle is the ratio of adjacent side and opposite side.

Cotangent is used to find unknown angle and unknown side.

The notation of Cotangent is cot ? (or) ctg ?.

Let us see some important formulas of opposite of tangent.
Formulas of Opposite of Tangent:

Formula for cotangent in a right triangle:

Cot ? = `("adjacent side")/("opposite side") `

Pythagorean identities for ctg:

1 + cot2 ? = csc2 ?

We can also rewritten the above one.

Cot2 ? = csc2 ? – 1

Derivative of ctg:

`d/dx` cot x   = - csc2 x

Integral of ctg:

`int`cot x dx = ln(sin x) + C

The cotangent value table for standard angles is shown below.

Angle/Function


   0°


30° (or) `pi/6`


45° (or) `pi/4`


60° (or) `pi/3`


90° (or) `pi/2`

    Cot ?


   8


  `sqrt(3)`


     1


  `1/sqrt(3)`


  0



Negative argument of ctg:

Cot(-?) = - cot(?)

Double angle of ctg:

Cot(2?) =` (cot^2 theta-1)/(2cot theta)` = `1/2` (cot ? – tan ?)

Triple angle of ctg:

Cot(3?) = `(cot^3 theta-3 cot theta)/(3 cot^2 theta-1)`

Half angle of ctg:

Cot(`theta/2`) = cot ? + csc ?

Cot(`theta/2`) = `(sin theta)/(1-cos theta)`

Cot(`theta/2`) = `sqrt((1+cos theta)/(1-cos theta))`

Relation between cotangent and inverse cotangent:

Cot(cot-1(?)) = ?

Simple Relation between cotangent and other trigonometric functions:

Cot(?) = tan(`pi/2` – ?)

Cot(?) = `(cos theta)/(sin theta)`

Let us see sample problem of opposite of tangent.
Example Problem - Opposite of Tangent:

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Example Problem 1:

Find the angle using cotangent if adjacent side is 5 cm and opposite side is 12 cm in a right triangle. (note: ? be angle)

Solution:

As we know that,

Cot ? = `("adjacent side")/("opposite side") `

cot ? = `5/12`

cot ? = 0.4167

Take inverse cotangent both sides. we get,

? = cot -1(0.4167)        

? = 67.4°

Therefore, angle of right triangle is 67.4°.

Example Problem 2:

Find the value of cot 45° using sin 45° = `1/sqrt(2)` and cos 45° = `1/sqrt(2)` .

Solution:

As we know that,

Cot ? = `(cos theta)/(sin theta)`

So, cot 45° = `cos 45/sin 45`

cot 45° = `(1/sqrt(2))/(1/sqrt(2))`

Cot 45° = 1.

Solving Complex Rational Expressions

Rational Expressions: Rational expressions are defined as one of the basis of mathematics. All the rational are having polynomials terms in both of the numerator function and the denominator function. The polynomials are present along with the variables.

 Complex rational expressions: Complex rational expressions are having the polynomials with the fraction format. The expressions should have atleast one fraction with the polynomials.
Explanations for Solving Complex Rational Expressions

There are many steps are followed for solving the complex rational expressions. They are defined as follows,

Step 1: Write the given rational expressions.

Step 2: Bring the denominator terms to the multiplication format by taking inverse.

Step 3: Then in the next step, we have to solve the obtained result.
Example Problem for Solving Complex Rational Expressions

Problem 1: Solve the given rational expressions, `((4a)/3)/((3a)/2)` .

Solution:

Step 1: Write the given complex rational expressions,

`((4a)/3)/((3a)/2)`

Step 2: Bring the denominator terms to the numerator by taking the inverse, we get,

`((4a)/3)` `xx` `(2/(3a))`

Step 3: In the next step, we have to simplify the obtained terms,we get,

`(8a)/(9a)`

Step 4: By eliminating the like terms, we get,

`8/9`

This is the obtained result for solving the complex rational expressions.

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Problem 2: Solve the given rational expressions, `((6a)/3)/((9a)/4)` .

Solution:

Step 1: Write the given complex rational expressions,

`((6a)/3)/((9a)/4)`

Step 2: Bring the denominator terms to the numerator by taking the inverse, we get,

`((6a)/3)` `xx` `(4/(9a))`

Step 3: In the next step, we have to simplify the obtained terms,we get,

`(8a)/(9a)`

Step 4: By eliminating the like terms, we get,

`8/9`

This is the obtained result for solving the complex rational expressions.

Problem 3: Solve the given rational expressions, `((8a)/2)/((4a)/2)` .

Solution:

Step 1: Write the given complex rational expressions,

`((8a)/2)/((4a)/2)`

Step 2: Bring the denominator terms to the numerator by taking the inverse, we get,

`((8a)/2)` `xx` `(2/(4a))`

Step 3: In the next step, we have to simplify the obtained terms,we get,

= 2

This is the obtained result for solving the complex rational expressions.
Practice Problem for Solving Complex Rational Expressions

Problem 1: Solve the given rational expressions, `((5a)/10)/((25a)/5)` .

Answer: `1/10`

Problem 1: Solve the given rational expressions, `((5a)/4)/((15a)/8)` .

Answer: `2/3`

Introduction to preparation for differentiation strategies

The preparation process for differentiation strategies represents the process of differentiation under the polar coordinates, variables in equations. The differential equations may be present in the ordinary differential equations with different functions like algebraic functions, exponential functions, etc.. In this article we deal with the differential equations with the variables to differentiate for the strategy in the differentiation.

Preparation for Differentiation Strategies with Examples

Preparation for the differentiation strategies on the differential equations  `[x+y]^2 dy/dx = 25`  
Solution:

The equation given as   `[x+y]^2 dy/dx = 25`      

Put x + y = z

Differentiating with respect to 'x' we get,

  `1 + dy/dx` =  `dz/dx`    

  `=>`   ` dy/dx` =  `dz/dx -1`            

The given equation becomes

  `z^2[dz/dx - 1] = 25`  

  `=>`  `dz/dx -1` = `25/z^2`                          

  `=>`  `dz/dx` = `1 + 25/z^2`                          

  `=>`  `dz/dx` = `[z^2+25]/z^2`                          

  `=>`  `dz` = `[z^2+25]/z^2 dx`                          

  `=>`  `dz``z^2/[z^2+25]` =   `dx`                            

Integrating we have

  `=>` `int z^2/[z^2+25]dz` =   `int dx`                            

  `=>` `int [z^2+25 - 25]/[z^2+25]dz` =   `int dx`                            

  `=>` `int [[z^2+25]/[z^2+25] - 25/[z^2+25]]dz` =   `int dx`                  

  `=>` `int [1 - 25/[z^2+25]]dz` =   `int dx`                  

  `=>` `int dz - int 25/[z^2+25]dz` =   `int dx`                            

  `=>` `z - 25 1/5 tan^-1z/5` =   `x + c`                            

Put x + y = z

  `=>` `x+y - 5. tan^-1[[x+y]/5]` =   `x + c`                

Reduce 'x' on both sides,

  `=>` `y - 5. tan^-1[[x+y]/5]` =   ` c`   is the required solution.                

Problems for the Preparation for Differentiation Strategies

Homework or practice problems on the preparation for differentiation strategies:

Solve `x dy = ( y + 4x^5 e^[x^4]) dx`  
Solution:

The answer is `y/x = e^[x^4] + c`    

Solve    `[x^2 -y] dy + [y^2-x]dy =0`  
Solution:

The answer is    `x^3 + y^3 = 3xy`