# 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).

Solve the problem using trigonometric identities

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).

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