Trigonometric Identities

XYZ is a right-angled triangle in which ∠XZY = 90° and ∠XYZ = θ, which in an acute angle.

We know,

sin θ = \(\frac{\textrm{Opposite}}{\textrm{Hypotenuse}}\) = \(\frac{o}{h}\);

cos θ = \(\frac{\textrm{Adjacent}}{\textrm{Hypotenuse}}\) = \(\frac{a}{h}\);

tan θ = \(\frac{\textrm{Opposite}}{\textrm{Adjacent}}\) = \(\frac{o}{a}\);

csc θ = \(\frac{\textrm{Hypotenuse}}{\textrm{Opposite}}\) = \(\frac{h}{o}\);

sec θ = \(\frac{\textrm{Hypotenuse}}{\textrm{Adjacent}}\) = \(\frac{h}{a}\);

cot θ = \(\frac{\textrm{Adjacent}}{\textrm{Opposite}}\) = \(\frac{a}{o}\).

Let’s multiply sin θ and csc θ

sin θ ∙ csc θ = \(\frac{o}{h}\) × \(\frac{h}{o}\) = 1

Therefore, csc θ = \(\frac{1}{sin θ}\)

Now, multiply cos θ and sec θ

cos θ ∙ sec θ = \(\frac{a}{h}\) × \(\frac{h}{a}\) = 1

Therefore, sec θ = \(\frac{1}{cos θ}\)

Again, multiply tan θ and cot θ

tan θ ∙ cot θ = \(\frac{o}{a}\) × \(\frac{a}{a}\) = 1

Therefore, cot θ = \(\frac{1}{tan θ}\)

Let’s divide sin θ by cos θ

sin θ ÷ cos θ = \(\frac{sin θ}{cos θ}\) = \(\frac{\frac{o}{h}}{\frac{a}{h}}\) = \(\frac{o}{a}\) = tan θ.

Therefore, tan θ = \(\frac{sin θ}{cos θ}\).

Similarly, divide cos θ by sin θ

cos θ ÷ sin θ = \(\frac{cos θ}{sin θ}\) = \(\frac{\frac{a}{h}}{\frac{o}{h}}\) = \(\frac{a}{o}\) = cot θ.

Therefore, cot θ = \(\frac{cos θ}{sin θ}\).

If a relation of equality between two expressions involving trigonometric ratios of an angle θ holds true for all values of θ then the equality is called a trigonometric identity. But it holds true only for some values of θ, the equality gives a trigonometric equation. There are a number of fundamental trigonometric identities.

I.  sin\(^{2}\) θ + cos\(^{2}\) θ = 1

We have, sin θ = \(\frac{o}{h}\) and cos θ = \(\frac{a}{h}\).

Therefore, sin\(^{2}\) θ + cos\(^{2}\) θ = \(\frac{o^{2}}{h^{2}}\) + \(\frac{a^{2}}{h^{2}}\).

                                       = \(\frac{o^{2} + a^{2}}{h^{2}}\)

                                       = \(\frac{h^{2}}{h^{2}}\); [Since, in the right-angled ∆XYZ, o\(^{2}\) + a\(^{2}\) = h\(^{2}\) (by Pythagoras’ Theorem)]

                                        = 1.

Therefore, sin\(^{2}\) θ + cos\(^{2}\) θ = 1.

Consequently, 1 - sin\(^{2}\) θ = cos\(^{2}\) θ and 1 - cos\(^{2}\) θ = sin\(^{2}\) θ.

II. sec\(^{2}\) θ - tan\(^{2}\) θ = 1

We have, sec θ = \(\frac{h}{a}\) and tan θ = \(\frac{o}{a}\).

Therefore, sec\(^{2}\) θ - tan\(^{2}\) θ = \(\frac{h^{2}}{a^{2}}\) - \(\frac{o^{2}}{a^{2}}\).

                                      = \(\frac{h^{2} - o^{2}}{a^{2}}\)

                                      = \(\frac{a^{2}}{a^{2}}\); [Since, in the right-angled ∆XYZ, o\(^{2}\) + a\(^{2}\) = h\(^{2}\) ⟹ h\(^{2}\) - o\(^{2}\) = a\(^{2}\) (by Pythagoras’ Theorem)]

                                      = 1.

Therefore, sec\(^{2}\) θ - tan\(^{2}\) θ = 1.

Consequently, 1 + tan\(^{2}\) θ = sec\(^{2}\) θ and sec\(^{2}\) θ - 1 = tan\(^{2}\) θ.

III. csc\(^{2}\) θ - cot\(^{2}\) θ = 1

We have, csc θ = \(\frac{h}{o}\) and cot θ = \(\frac{a}{o}\).

Therefore, csc\(^{2}\) θ - cot\(^{2}\) θ = \(\frac{h^{2}}{o^{2}}\) - \(\frac{a^{2}}{o^{2}}\).

                                       = \(\frac{h^{2} - a^{2}}{o^{2}}\)

                                       = \(\frac{o^{2}}{o^{2}}\); [Since, in the right-angled ∆XYZ, o\(^{2}\) + a\(^{2}\) = h\(^{2}\) ⟹ h\(^{2}\) - a\(^{2}\) = o\(^{2}\) (by Pythagoras’ Theorem)]

                                       = 1.

Therefore, csc\(^{2}\) θ - cot\(^{2}\) θ = 1.

Consequently, 1 + cot\(^{2}\) θ = csc\(^{2}\) θ and csc\(^{2}\) θ - 1 = cot\(^{2}\) θ.

These three trigonometric identities are also called Pythagorean identities.

Note: The equalities csc θ = \(\frac{1}{sin θ}\), sec θ = \(\frac{1}{cos θ}\), cot θ = \(\frac{1}{tan θ}\), tan θ = \(\frac{sin θ}{cos θ}\) and cot θ = \(\frac{cos θ}{sin θ}\) are holds for all values of θ. Therefore, these equalities are trigonometric identities.

Thus, we have the following trigonometric identities.

Table of Trigonometric Identities

1. csc θ = \(\frac{1}{sin θ}\),

2. sec θ = \(\frac{1}{cos θ}\),

3. cot θ = \(\frac{1}{tan θ}\),

4. tan θ = \(\frac{sin θ}{cos θ}\),

5. cot θ = \(\frac{cos θ}{sin θ}\)

6. sin\(^{2}\) θ + cos\(^{2}\) θ = 1; 1 - sin\(^{2}\) θ = cos\(^{2}\) θ, 1 - cos\(^{2}\) θ = sin\(^{2}\) θ.

7. sec\(^{2}\) θ - tan\(^{2}\) θ = 1; 1 + tan\(^{2}\) θ = sec\(^{2}\) θ; sec\(^{2}\) θ - 1 = tan\(^{2}\) θ.

8. csc\(^{2}\) θ - cot\(^{2}\) θ = 1

9. 1 + cot\(^{2}\) θ = csc\(^{2}\) θ,

10. csc\(^{2}\) θ - 1 = cot\(^{2}\) θ.

Solved Examples on Trigonometric Identities:

1. Prove that sin\(^{4}\) θ + cos\(^{4}\)  θ + 2 sin\(^{2}\) θ ∙ cos\(^{2}\)  θ = 1

Solution:

LHS = sin\(^{4}\) θ + cos\(^{4}\)  θ + 2 sin\(^{2}\) θ ∙ cos\(^{2}\)  θ

       = (sin\(^{2}\) θ)\(^{2}\) + (cos\(^{2}\)  θ)\(^{2}\) + 2 sin\(^{2}\) θ ∙ cos\(^{2}\)  θ

       = (sin\(^{2}\) θ + cos\(^{2}\) θ)\(^{2}\)

       = 1\(^{2}\); [Since, sin\(^{2}\) θ + cos\(^{2}\) θ = 1]

       = 1 = RHS. (Proved).

How to Solve Trigonometric Identities Proving Problems?

2. Show that sec θ - cos θ = sin θ tan θ

Solution:

LHS =  sec θ - cos θ

      = \(\frac{1}{cos θ}\) - cos θ

      = \(\frac{1 - cos^{2} θ}{cos θ}\)

      = \(\frac{sin^{2} θ}{cos θ}\);  [Since, sin\(^{2}\) θ + cos\(^{2}\) θ = 1]

      = sin θ ∙ \(\frac{sin θ}{cos θ}\)

      = sin θ ∙ tan θ = RHS. (Proved)

3. Prove that 1 - \(\frac{cos^{2} A}{1 + sin A}\) = sin A

Solution:

LHS = 1 - \(\frac{cos^{2} A}{1 + sin A}\)

       = 1 - \(\frac{1 - sin^{2} A}{1 + sin A}\); [cos\(^{2}\) θ = 1 - sin\(^{2}\) θ]

       = 1 - \(\frac{(1 + sin A)(1 – sin A)}{1 + sin A}\)

       = 1 – (1 – sin A)

       = 1 – 1 + sin A

       = sin A = RHS. (Proved).

Verifying Trigonometric Identities & Equations

4. Prove that \(\frac{sin A}{1 + cos A}\) + \(\frac{sin A}{1 - cos A}\) = 2 csc A.

Solution:

LHS = \(\frac{sin A}{1 + cos A}\) + \(\frac{sin A}{1 - cos A}\)

       = \(\frac{sin A(1 – cos A) + sin A(1 + cos A)}{(1 + cos A)(1 – cos A)}\)

       = \(\frac{sin A – sin A ∙ cos A + sin A + sin A ∙ cos A)}{1 - cos^{2} A}\)

       = \(\frac{2 sin A}{sin^{2} A}\); [Since, 1 - cos\(^{2}\) A = sin\(^{2}\) A]

       = \(\frac{2}{sin A}\)

       = 2 ∙ \(\frac{1}{sin A}\); [Since, csc A = \(\frac{1}{sin A}\)]

       = RHS. (Proved).

Proving Trigonometric Identities Practice Problems Online

5. Prove that \(\frac{sin A}{1 + cos A}\) = \(\frac{1 – cos A}{sin A}\).

Solution:

LHS = \(\frac{sin A}{1 + cos A}\)

       = \(\frac{sin A}{1 + cos A}\) ∙ \(\frac{1 - cos A}{1 - cos A}\)

       = \(\frac{sin A(1 - cos A)}{1 - cos^{2} A}\)

       = \(\frac{sin A(1 - cos A)}{sin^{2} A}\); [Since 1 - cos\(^{2}\) θ = sin\(^{2}\) θ]

       = \(\frac{1 - cos A}{sin A}\) = RHS. (Proved).

Trigonometry Problems and Questions with Solutions

6. Prove that \(\sqrt{\frac{1 + sin θ}{1 - sin θ}}\) = sec θ +tan θ.

Solution:

LHS = \(\sqrt{\frac{1 + sin θ}{1 - sin θ}}\)

       = \(\sqrt{\frac{(1 + sin θ) (1 + sin θ)}{(1 - sin θ)(1 + sin θ)}}\)

       = \(\sqrt{\frac{(1 + sin θ)^{2}}{1 – sin^{2} θ}}\)

       = \(\sqrt{\frac{(1 + sin θ)^{2}}{cos^{2} θ}}\); [Since 1 - sin\(^{2}\) θ = cos\(^{2}\) θ]

       = \(\frac{1 + sin θ}{cos θ}\)

       = \(\frac{1}{cos θ}\) + \(\frac{sin θ}{cos θ}\)

       = secθ + tan θ = RHS. (Proved).

Trigonometric identities problems for class 10

7. Prove that tan\(^{2}\) A – tan\(^{2}\) B = \(\frac{sin^{2} A - sin^{2} B}{cos^{2} A ∙ cos^{2} B}\) = sec\(^{2}\) A - sec\(^{2}\) B

Solution:

LHS = tan\(^{2}\) A – tan\(^{2}\) B

       = \(\frac{sin^{2} A}{cos^{2} A}\) - \(\frac{ sin^{2} B}{cos^{2} B}\)

       = \(\frac{sin^{2} A ∙ cos^{2} B - sin^{2} B ∙ cos^{2} A}{cos^{2} A ∙ cos^{2} B}\)

       = \(\frac{sin^{2} A (1 - sin^{2} B) - sin^{2} B(1 - sin^{2} A)}{cos^{2} A ∙ cos^{2} B}\); [Since, cos² B = 1 - sin² B and cos² A = 1 - sin² A]

       = \(\frac{sin^{2} A  - sin^{2} A  sin^{2} B - sin^{2} B + sin^{2} B sin^{2} A}{cos^{2} A ∙ cos^{2} B}\);

        = \(\frac{sin^{2} A - sin^{2} B}{cos^{2} A ∙ cos^{2} B}\) = MHS

        = \(\frac{(1 - cos^{2} A) - (1 - cos^{2} B)}{cos^{2} A ∙ cos^{2} B}\); [Since, sin² A = 1 - cos² A and sin² B = 1 - cos² B]

        = \(\frac{cos^{2} B - cos^{2} A}{cos^{2} A ∙ cos^{2} B}\)

        = \(\frac{cos^{2} B}{cos^{2} A ∙ cos^{2} B}\) - \(\frac{cos^{2} A}{cos^{2} A ∙ cos^{2} B}\)

        = \(\frac{1}{cos^{2} A}\) - \(\frac{1}{cos^{2} B}\)

        = sec\(^{2}\) A – sec\(^{2}\) B = RHS. (Proved)

Again, 

LHS = tan\(^{2}\) A – tan\(^{2}\) B

       = (sec\(^{2}\) A – 1) – (sec\(^{2}\) B – 1); [Since, tan\(^{2}\) θ = sec\(^{2}\) θ – 1 ]

       = sec\(^{2}\) A – 1 – (sec\(^{2}\) B +1

       = sec\(^{2}\) A – sec\(^{2}\) B = RHS. (Proved).

Solving Trigonometric Equations using Trigonometric Identities

8. Prove that tan\(^{2}\) θ - sin\(^{2}\) θ = tan\(^{2}\) θ ∙ sin\(^{2}\) θ

Solution:

LHS = tan\(^{2}\) θ - sin\(^{2}\) θ

       = tan\(^{2}\) θ - \(\frac{sin^{2} θ}{cos^{2} θ}\) ∙ cos\(^{2}\)

       = tan\(^{2}\) θ - tan\(^{2}\) θ ∙ cos\(^{2}\)

       = tan\(^{2}\) θ(1 - cos\(^{2}\))

       = tan\(^{2}\) θ ∙ sin\(^{2}\) θ = RHS. (Proved).

List of trigonometric identities Problems

9. Prove that 1 + \(\frac{cot^{2} θ}{1 + csc θ}\) = csc θ

Solution:

LHS = 1 + \(\frac{cot^{2} θ}{1 + csc θ}\)

       = 1 + \(\frac{csc^{2} θ - 1}{1 + csc θ}\)

       = 1 + \(\frac{(csc θ + 1)(csc θ – 1)}{1 + csc θ}\)

       = 1 + (csc θ – 1)

       = 1 + csc θ – 1

       = csc θ = RHS. (Proved).

10. Prove that \(\frac{sec θ - )}{sec θ + 1}\) = (cot θ – csc θ)\(^{2}\).

Solution:

LHS = \(\frac{sec θ - 1}{sec θ + 1}\)

       = \(\frac{sec θ - 1}{sec θ + 1}\) ∙ \(\frac{sec θ - 1}{sec θ - 1}\)

       = \(\frac{(sec θ - 1)^{2}}{sec^{2} θ - 1}\)

       = \(\frac{(sec θ - 1)^{2}}{tan^{2} θ}\); [Since sec\(^{2}\) θ – 1 = tan\(^{2}\) θ]

       = \((\frac{sec θ - 1}{tan θ})^{2}\)

       = \((\frac{\frac{1}{cos θ} - 1}{\frac{sin θ }{cos θ} })^{2}\)

       = \((\frac{1 – cos θ}{cos θ} ∙ \frac{cos θ}{sin θ})^{2}\)

       = \((\frac{1 – cos θ}{sin θ})^{2}\)

       = \((\frac{1}{sin θ} - \frac{cos θ}{sin θ})^{2}\)

       = (csc θ – cot θ)\(^{2}\)

       = (cot θ – csc θ)\(^{2}\) = RHS. (Proved).

Proving Trigonometric Identities

11. Prove that \(\frac{sin A}{cot A + csc A}\) = 2 + \(\frac{sin A}{cot A - csc A }\)

Solution:

LHS = \(\frac{sin A}{cot A + csc A}\)

       = \(\frac{sin A(cot A - csc A)}{(cot A + csc A)(cot A - csc A)}\)

       = \(\frac{sin A ∙ cot A – sin A ∙ csc A}{cot^{2} A – csc^{2} A}\)

       = \(\frac{sin A ∙ \frac{cos A}{sin A} – 1}{-1}\); [since, sin θ csc θ = 1, csc\(^{2}\) θ – cot\(^{2}\)  θ = 1]

       = \(\frac{cos A - 1}{-1}\)

       = 1 – cos A

RHS = 2 + \(\frac{sin A}{cot A - csc A}\)

        = 2 + \(\frac{sin A}{ cot A - csc A}\) ∙ \(\frac{(cot A + csc A)}{(cot A + csc A)}\)

        = 2 + \(\frac{sin A(cot A + csc A)}{cot^{2} A – csc^{2} A}\)

        = 2 + \(\frac{sin A(\frac{cos A}{sin A} + \frac{1}{sin A})}{-1}\)

        = 2 + \(\frac{cos A + 1}{-1}\)

        = 2 – (cos A + 1)

        = 2 – cos A – 1

        = 1 – cos A

Therefore, LHS = RHS. (Proved).

Alternative Method

The identity is established if we can prove that

\(\frac{sin A}{cot A + csc A}\) - \(\frac{sin A}{cot A - csc A }\) = 2

Now, here  

LHS = \(\frac{sin A}{cot A + csc A}\) - \(\frac{sin A}{cot A - csc A }\)

       = \(\frac{sin A(cot A - csc A) – sin A(cot A + csc A)}{( cot A + csc A)(cot A - csc A)}\)

       = \(\frac{sin A ∙ cot A – sin A ∙ csc A – sin A ∙ cot A – sin A ∙ csc A}{(cot A + csc A)(cot A - csc A)}\)

       = \(\frac{–2 sin A ∙ csc A}{cot^{2} A – csc^{2} A}\)

       = \(\frac{–2}{-1 }\); [Since sin θ csc θ = 1, csc\(^{2}\) θ – cot\(^{2}\)  θ = 1]

       = 2 = RHS. (Proved)

Problems on Evaluation using Trigonometric Identities:

12. If sin θ + csc θ = 2, find the value of sin\(^{10}\) θ + csc\(^{11}\) θ

Solution:

Given that, sin θ + csc θ = 2 ……………. (i)

⟹ sin θ + \(\frac{1}{ sin θ}\) = 2

⟹ \(\frac{ sin^{2} θ + 1}{sin θ }\) = 2

⟹ sin\(^{2}\) θ + 1= 2 sin θ

⟹ sin\(^{2}\) θ - 2 sin θ + 1 = 0

⟹ (sin θ - 1)\(^{2}\) = 0

⟹ sin θ - 1 = 0

⟹ sin θ = 1

⟹ csc θ = \(\frac{1}{ sin θ}\) = \(\frac{1}{ 1}\) = 1

Therefore, csc θ = 1.

Now, sin\(^{10}\) θ + csc\(^{11}\) θ

= 1\(^{10}\) + 1\(^{11}\)

= 1 + 1

= 2.

13. If sec θ – tan θ = \(\frac{1}{3}\), find the value of sec A and tan A.

Solution:

Given, sec θ – tan θ = \(\frac{1}{3}\) ………… (i)

We know that the Pythagorean identity, sec\(^{2}\) θ - tan\(^{2}\) θ = 1.

                   ⟹ (sec θ + tan θ)(sec θ - tan θ) = 1

                   ⟹ (sec θ + tan θ) ∙ \(\frac{1}{3}\) = 1, [Given sec θ – tan θ = \(\frac{1}{3}\)]

                   ⟹ sec θ + tan θ = 3 ……….. (ii)

Now adding (i) and (ii) we get

                       2 sec θ = \(\frac{1}{3}\) + 3

                   ⟹ 2 sec θ = \(\frac{10}{3}\)

                   ⟹ sec θ = \(\frac{10}{3}\) ∙ \(\frac{1}{2}\)

                   ⟹ sec θ = \(\frac{5}{3}\)

Therefore, sec θ = \(\frac{5}{3}\)

Putting the value of sec θ = \(\frac{5}{3}\) in (ii), we get

\(\frac{5}{3}\) + tan θ = 3

⟹ tan θ = 3 - \(\frac{5}{3}\)

⟹ tan θ = \(\frac{4}{3}\)

Therefore, tan θ = \(\frac{4}{3}\).

14. If tan A + sec A = \(\frac{2}{\sqrt{3}}\), find the value of sin A

Solution:

Given that, tan A + sec A = \(\frac{2}{\sqrt{3}}\) ……………….. (i)

We know the Pythagorean trigonometric identity,

       sec\(^{2}\) A - tan\(^{2}\) A = 1.

⟹ (sec A + tan A)(sec A – tan A) = 1

⟹ \(\frac{2}{\sqrt{3}}\)(sec A – tan A) = 1; [given , tan A + sec A = \(\frac{2}{\sqrt{3}}\)]

⟹ sec A – tan A = \(\frac{\sqrt{3}}{2}\) ……………….. (ii)

Solving these two equations we get,

     2 sec A = \(\frac{2}{\sqrt{3}}\) + \(\frac{\sqrt{3}}{2}\)

 ⟹ 2 sec A = \(\frac{4 + 3}{2\sqrt{3}}\)

 ⟹ 2 sec A = \(\frac{7}{2\sqrt{3}}\)

⟹ sec A = \(\frac{7}{4\sqrt{3}}\) ……………….. (iii)

Putting the value of sec A = \(\frac{7}{4\sqrt{3}}\) in (i) we get,

     tan A + \(\frac{7}{4\sqrt{3}}\) = \(\frac{2}{\sqrt{3}}\)

⟹ tan A = \(\frac{2}{\sqrt{3}}\) - \(\frac{7}{4\sqrt{3}}\)

⟹ tan A = \(\frac{8 - 7}{4\sqrt{3}}\)

⟹ tan A = \(\frac{1}{4\sqrt{3}}\)……………….. (iv)

Now dividing (iv) by (iii) we get,

\(\frac{tan A}{sec A}\) = \(\frac{1}{7}\)

⟹ sin A = \(\frac{1}{7}\)

Therefore, sin A = \(\frac{1}{7}\).

Problems on establishing conditional results

15. If cos A + sin A = \(\sqrt{2}\) cos A, prove that cos A - sin A = \(\sqrt{2}\) sin A.

Solution:

Let cos A - sin A = k. Now,

(cos A + sin A)\(^{2}\) + (cos A - sin A)\(^{2}\) = cos\(^{2}\) A + sin\(^{2}\) A + 2 cos A ∙ sin A + cos\(^{2}\) A + sin\(^{2}\) A - 2 cos A ∙ sin A

                                            = 1 + 2 cos A sin A + 1 - 2 cos A sin A

                                            = 2

Therefore, (\(\sqrt{2}\) cos A)\(^{2}\) + k\(^{2}\) = 2

⟹ k\(^{2}\) = 2 - (\(\sqrt{2}\) cos A)\(^{2}\) 

⟹ k\(^{2}\) = 2 - 2cos\(^{2}\) A

⟹ k\(^{2}\) = 2(1 - cos\(^{2}\) A)

⟹ k\(^{2}\) = 2sin\(^{2}\) A; [Since, sin\(^{2}\) θ + cos\(^{2}\) θ = 1]

⟹ k = \(\sqrt{2}\) sin A

⟹ cos A - sin A = \(\sqrt{2}\) sin A. (Proved).

10th Grade Math

From Trigonometric Identities to HOME PAGE