Trigonometrical Identity

Definition of trigonometrical identity:

An equation which is true for all values of the variable involved is called an identity. An equation which involves trigonometric ratios of an angle and is true for all the values of the angle is called trigonometrical identities.

When the solutions of any trigonometric ratio problems represent the same expression in the L.H.S. and R.H.S. and the relation is satisfied for all the values of θ then such relation is called a trigonometrical identity.

Mutual relations among the trigonometrical ratios are generally used to establish the equality of such trigonometrical identities.


To solve different types of trignometrical identity follow the formula:

sin θ ∙ csc θ = 1   csc θ = 1/sin θ

cos θ ∙ sec θ = 1 sec θ = 1/cos θ 

 tan θ ∙ cot θ = 1  cot θ = 1/tan θ 

 tan θ = sin θ/cos θ                 

 cot θ = cos θ/sin θ

sin2 θ implies (sin θ)2
similarly, tan3 θ means (tan θ)3 etc.

sin2 θ + cos2 θ = 1

cos2 θ = 1 - sin2 θ
sin2 θ = 1 - cos2 θ

sec2 θ = 1 + tan2 θ
sec2 θ - tan2 θ = 1
tan2 θ = sec2 θ - 1

csc2 θ = 1 + cot2 θ
csc2 θ - 1 = cot2 θ
csc2 θ - cot2 θ = 1

The trigonometrical ratios of a positive acute angle θ are always non-negative and

(i) sin θ and cos θ can never be greater than 1;

(ii) sec θ and csc θ can never be less than 1;

(iii) tan θ and cot θ can have any value.


Worked-out problems on trigonometric identity:

1. Proof the identity:

tan2 θ – (1/cos2 θ) + 1 = 0

Solution:

L.H.S = tan2 θ – (1/cos2 θ) + 1

= tan2 θ - sec2 θ + 1 [since, 1/cos θ = sec θ]

= tan2 θ – (1 + tan2 θ) +1 [since, sec2 θ = 1 + tan2 θ]

= tan2 θ – 1 – tan2 θ + 1

= 0 = R.H.S. Proved


2. Verify that:

1/(sin θ + cos θ) + 1/(sin θ - cos θ) = 2 sin θ/(1 – 2 cos2 θ)

Solution:

L.H.S = 1/(sin θ + cos θ) + 1/(sin θ - cos θ)

= [(sin θ - cos θ) + (sin θ + cos θ)]/(sin θ + cos θ)(sin θ - cos θ)

= [sin θ - cos θ + sin θ + cos θ]/(sin2 θ - cos2 θ)

= 2 sin θ/[(1 - cos2 θ) - cos2 θ] [since, sin2 θ = 1 - cos2 θ]

= 2 sin θ/[1 - cos2 θ - cos2 θ]

= 2 sin θ/[1 – 2 cos2 θ] = R.H.S. Proved


3. Prove that:

sec2 θ + csc2 θ = sec2 θ ∙ csc2 θ

Solution:

L.H.S. = sec2 θ + csc2 θ

= 1/cos2 θ + 1/sin2 θ [since, sec θ = 1/cos θ and csc θ = 1/sin θ]

= (sin2 θ + cos2 θ)/(cos2 θ sin2 θ)

= 1/cos2 θ ∙ sin2 θ [since, sin2 θ + cos2 θ = 1]

= 1/cos2 θ ∙ 1/sin2 θ

= sec2 θ ∙ csc2 θ = R.H.S. Proved




More examples on trigonometrical identity are explained below. To proof the identities step-by-step follow the above trig formulas.

4. Prove the identity:

cos θ/(1 + sin θ) = (1 + cos θ - sin θ)/(1 + cos θ + sin θ)

Solution:

R. H. S. = (1 + cos θ - sin θ)/(1 + cos θ + sin θ)

= {(1 + cos θ - sin θ) (1 + cos θ + sin θ)}/{(1+ cos θ + sin θ) (1 + cos θ + sin θ)} [multiplying both numerator and denominator by (1 + cos θ + sin θ)]

= {(1 + cos θ)2 - sin2 θ}/(1 + cos θ + sin θ)2

= (1 + cos2 θ + 2 cos θ - sin2 θ)/{(1 + cos θ)2 + 2 ∙ (1 + cos θ) sin θ + sin2 θ}

= (cos2 θ + 2 cos θ + 1 - sin2 θ)/{1 + cos2 θ + 2 cos θ + 2 ∙ (1 + cos θ) ∙ sin θ + sin2 θ}

= (cos2 θ + 2 cos θ + cos2 θ)/{2 + 2 cos θ + 2 ∙ (1 + cos θ) ∙ sin θ} [since, sin2 θ + cos2 θ = 1 and 1 - sin2 θ = cos2 θ]

= {2 cos θ (1 + cos θ)}/{2 (1 + cos θ)(1 + sin θ)}

= cos θ/(1 + sin θ) = L.H.S. Proved


5. Verify the trigonometrical identity:

(cot θ + csc θ – 1)/(cot θ - csc θ + 1) = (1 + cos θ)/sin θ

L.H.S. = (cot θ + csc θ – 1)/(cot θ - csc θ + 1)

= {cot θ + csc θ - (csc2 θ - cot2 θ)}/(cot θ - csc θ + 1)

[csc2 θ = 1 + cot2 θ ⇒ csc2 θ - cot2 θ = 1]

= {(cot θ + csc θ) - (csc θ + cot θ) (csc θ - cot θ)}/(cot θ - csc θ + 1)

= {(cot θ + csc θ) (1 - csc θ + cot θ)}/ (1 - csc θ + cot θ)

= cot θ + csc θ

= (cos θ/sin θ) + (1/sin θ)

= (1 + cos θ)/sin θ = R.H.S. Proved

Trigonometric Functions


10th Grade Math

From Trigonometrical Identity to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Share this page: What’s this?

Recent Articles

  1. Worksheet on Triangle | Homework on Triangle | Different types|Answers

    Jun 21, 24 02:19 AM

    Find the Number of Triangles
    In the worksheet on triangle we will solve 12 different types of questions. 1. Take three non - collinear points L, M, N. Join LM, MN and NL. What figure do you get? Name: (a)The side opposite to ∠L…

    Read More

  2. Worksheet on Circle |Homework on Circle |Questions on Circle |Problems

    Jun 21, 24 01:59 AM

    Circle
    In worksheet on circle we will solve 10 different types of question in circle. 1. The following figure shows a circle with centre O and some line segments drawn in it. Classify the line segments as ra…

    Read More

  3. Circle Math | Parts of a Circle | Terms Related to the Circle | Symbol

    Jun 21, 24 01:30 AM

    Circle using a Compass
    In circle math the terms related to the circle are discussed here. A circle is such a closed curve whose every point is equidistant from a fixed point called its centre. The symbol of circle is O. We…

    Read More

  4. Circle | Interior and Exterior of a Circle | Radius|Problems on Circle

    Jun 21, 24 01:00 AM

    Semi-circular Region
    A circle is the set of all those point in a plane whose distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is known

    Read More

  5. Quadrilateral Worksheet |Different Types of Questions in Quadrilateral

    Jun 19, 24 09:49 AM

    In math practice test on quadrilateral worksheet we will practice different types of questions in quadrilateral. Students can practice the questions of quadrilateral worksheet before the examinations

    Read More