Reciprocal identities in trigonometry express the inverse relationships between pairs of trigonometric functions. These identities are fundamental in simplifying expressions, solving equations, and understanding the relationships between different trigonometric functions.

In this article, we will discuss reciprocal identities and practice questions related to them.

What are Reciprocal Identities?

Reciprocal identities in trigonometry are relationships that express trigonometric functions in terms of the reciprocals of other trigonometric functions. These identities are useful for simplifying expressions and solving trigonometric equations.

Reciprocal Identities in trigonometry are:

• sin θ = 1 / csc θ
• cos θ = 1 / sec θ
• tan θ = 1 / cot θ
•  csc θ = 1 / sin θ
• sec θ = 1 / cos θ
• cot θ = 1 / tan θ

Key Points:

• These identities hold for all angles θ where the functions are defined.
• They’re called “reciprocal” because each pair of functions is the reciprocal of the other.

Derivation of Reciprocal Identities

These identities come from the definitions of the trigonometric functions in a right-angled triangle:

• sin θ = opposite/hypotenuse
• csc θ = hypotenuse/opposite

Thus, sin θ and csc θ are reciprocals, and similarly for the other pairs are derived.

Formulas of Reciprocal Identities

The formulas of reciprocal identities are given below:

Sine and Cosecant Relation:

• sin θ = 1 / csc θ
• csc θ = 1 / sin θ

Cosine and Secant Relation:

• cos θ = 1 / sec θ
• sec θ = 1 / cos θ

Tangent and Cotangent Relation:

• tan θ = 1 / cot θ
• cot θ = 1 / tan θ

These formulas express the reciprocal relationships between each pair of trigonometric functions. They hold true for all angles θ where the functions are defined.

It’s important to note that these identities are valid except when the denominator equals zero. For example, sin θ = 1 / csc θ is undefined when csc θ = 0, which occurs when sin θ = 0.

Reciprocal Identities Uses

Some key uses of reciprocal identites with examples are added below:

• Simplifying Trigonometric Expressions:

Example: Simplify (sin θ + csc θ) / (1 + sin θ csc θ)

Solution:

Using csc θ = 1 / sin θ

we get: (sin θ + 1/sin θ) / (1 + 1)

= (sin² θ + 1) / (2 sin θ)

= (1 + cos² θ) / (2 sin θ)

• Solving Trigonometric Equations:

Example: Solve csc x = 2 for 0° ≤ x ≤ 360°

Solution:

Using sin x = 1 / csc x

we get: sin x = 1/2

x = 30° or 150°

• Proving Other Trigonometric Identities:

Example: Prove that tan θ cot θ = 1

Proof:

tan θ.cot θ

= tan θ (1 / tan θ)

= 1

• Converting Complex Expressions:

Example: Convert sec² θ – 1 to an expression involving tan

Solution:

sec² θ = 1 + tan² θ

sec² θ – 1 = (1 + tan² θ) – 1

= tan² θ

• Calculus Applications: In calculus, these identities are useful for differentiating and integrating trigonometric functions.

Example: find derivative of csc x

Solution:

= d/dx (csc x)

= -csc x cot x (Using Trigonometric Formula)

This is derived using the reciprocal identity and the chain rule.

• Simplifying Rational Expressions:

Example: Simplify (1 – cos θ) / sin θ

Solution:

Multiply numerator and denominator by csc θ:

((1 – cos θ) / sin θ) × (csc θ / csc θ)

= (csc θ – cos θ csc θ) / sin θ × csc θ

= (csc θ – cos θ csc θ) / 1

= csc θ – cot θ

• Inverse Trigonometric Functions: These identities are useful when working with inverse trig functions.

Example: If y = arcsin x

y = sin^-1(x)

sin y = x

csc y = 1/x

Reciprocal Identities Practice Test

Question 1: Simplify (sin θ + csc θ)².

Solution:

(sin θ + csc θ)² = sin² θ + 2sin θ csc θ + csc² θ

= sin² θ + 2(sin θ)(1/sin θ) + (1/sin θ)²

= sin² θ + 2 + (1/sin² θ)

= (sin⁴ θ + 2sin² θ + 1) / sin² θ

= (sin² θ + 1)² / sin² θ

= (csc² θ)² / sin² θ(since csc² θ = 1 + cot² θ = (sin² θ + cos² θ) / sin² θ = 1/sin² θ)

= csc² θ csc² θ = csc⁴ θ

Question 2: If sec θ = -5/3, find the value of cos θ.

Solution:

Using the reciprocal identity cos θ = 1 / sec θ

cos θ = 1 / (-5/3) = -3/5

Question 3: Prove that (tan θ + cot θ)² = sec² θ + csc² θ.

Solution:

LHS = (tan θ + cot θ)² = tan² θ + 2 + cot² θ

RHS = sec² θ + csc² θ

We know that:

sec² θ = 1 + tan² θ

csc² θ = 1 + cot² θ

Therefore, RHS = (1 + tan² θ) + (1 + cot² θ) = 2 + tan² θ + cot² θ = LHS

Question 4: If sin θ = 3/5, find csc θ.

Solution:

Using the reciprocal identity csc θ = 1 / sin θ

csc θ = 1 / (3/5) = 5/3

Question 5: Simplify (sec θ – tan θ)(sec θ + tan θ).

Solution:

(sec θ – tan θ)(sec θ + tan θ) = sec² θ – tan² θ

We know that sec² θ = 1 + tan² θ

Therefore, sec² θ – tan² θ = (1 + tan² θ) – tan² θ = 1

Question 6: If cot θ = 4/3, find tan θ.

Solution:

Using the reciprocal identity tan θ = 1 / cot θ

tan θ = 1 / (4/3) = 3/4

Question 7: Prove that sin θ csc θ + cos θ sec θ = 2.

Solution:

sin θ csc θ + cos θ sec θ

= sin θ (1/sin θ) + cos θ (1/cos θ) (using reciprocal identities)

= 1 + 1 = 2

Question 8: If sec θ = 13/5, find tan θ.

Solution:

We know that sec² θ = 1 + tan² θ

(13/5)² = 1 + tan² θ

169/25 = 1 + tan² θ

tan² θ = 169/25 – 1 = 144/25

tan θ = ±12/5 (but since sec θ is positive, tan θ must be positive)

Therefore, tan θ = 12/5

Question 9: Simplify (csc θ – cot θ) / (1 – cos θ).

Solution:

(csc θ – cot θ) / (1 – cos θ)

= ((1/sin θ) – (cos θ/sin θ)) / (1 – cos θ)

= (1 – cos θ) / (sin θ (1 – cos θ))

= 1 / sin θ = csc θ

Question 10: Prove that (sec θ + tan θ)(sec θ – tan θ) = 1.

Solution:

(sec θ + tan θ)(sec θ – tan θ)

= sec² θ – tan² θ

= (1 + tan² θ) – tan² θ(using sec² θ = 1 + tan² θ)

= 1

Practice Questions Reciprocal Identities

Q1. Simplify: (sin θ csc θ + cos θ sec θ – 1) / (tan θ + cot θ)

Q2. If csc θ = -5/3, find the value of sin θ.

Q3. Prove the identity: tan² θ + 1 = sec² θ

Q4. If sec θ = 7/4 and θ is in the second quadrant, find the value of cos θ and tan θ.

Q5. Simplify: (1 – sin θ)(1 + csc θ)

Q6. Prove that: (csc θ + cot θ)(csc θ – cot θ) = 1

Q7. If tan θ = 3/4, find the value of sec θ.

Q8. Simplify: (sec² θ – tan² θ) / (csc² θ – cot² θ)

Q9. Prove the identity: sin θ / (1 – cos θ) = (1 + cos θ) / sin θ

Q10. If cot θ = 5/12 and θ is in the third quadrant, find the values of sin θ and cos θ.

Application of Reciprocal Identities

Some application of reciprocal identites includes:

• Physics and Engineering: In fields like physics and engineering, these identities help simplify complex formulas involving trigonometric functions.

• Graphing: Understanding reciprocal relationships helps in graphing trigonometric functions and their reciprocals.

• Periodic Properties: Reciprocal identities help in understanding the periodic nature of trigonometric functions and their reciprocals.

Also Check,

• Reciprocal of Trigonometric Ratios
• Trigonometric identities
• Trigonometry Table | Trigonometric Ratios and Formulas

Frequently Asked Questions

What are the Basic Reciprocal Identities?

Basic reciprocal identites are:

• sin θ = 1/csc θ
• cos θ = 1/sec θ
• tan θ = 1/cot θ

and their inverses.

Why are they Called “Reciprocal” Identities?

They are called reciprocal identites because each pair of functions is the reciprocal of the other.

When are Reciprocal Identities Undefined?

When the denominator equals zero, e.g., sin θ = 1/csc θ is undefined when csc θ = 0.

What are Applications of Reciprocal Identities?

Simplifying expressions, solving equations, proving other identities, and applications in calculus and physics.