Question 1

If tan θ = 5/12, find sin θ (where θ is acute).

Accepted Answer

Step 1: tan θ = opposite/adjacent = 5/12. Step 2: Using Pythagoras, hypotenuse = √(5² + 12²) = √(25 + 144) = √169 = 13. Step 3: sin θ = opposite/hypotenuse = 5/13.

Question 2

Simplify: tan θ × cot θ

Accepted Answer

Step 1: Recall that cot θ = 1/tan θ. Step 2: tan θ × cot θ = tan θ × (1/tan θ) = 1. Step 3: This is a fundamental trigonometric identity.

Question 3

If cos θ = 7/25, find tan θ (where θ is acute).

Accepted Answer

Step 1: cos θ = adjacent/hypotenuse = 7/25. Step 2: Using Pythagoras, opposite = √(25² - 7²) = √(625 - 49) = √576 = 24. Step 3: tan θ = opposite/adjacent = 24/7.

Question 4

Simplify: (1 - sin²θ) / cos θ

Accepted Answer

Step 1: Apply the Pythagorean identity: 1 - sin²θ = cos²θ. Step 2: Substitute: cos²θ / cos θ. Step 3: Simplify: cos²θ / cos θ = cos θ.

Question 5

Simplify: (sin⁴ A - cos⁴ A) / (sin² A - cos² A)

Accepted Answer

Step 1: Recognize that sin⁴ A - cos⁴ A is a difference of squares: (sin² A)² - (cos² A)² = (sin² A - cos² A)(sin² A + cos² A). Step 2: Substitute into the fraction: [(sin² A - cos² A)(sin² A + cos² A)] / (sin² A - cos² A). Step 3: Cancel (sin² A - cos² A) from numerator and denominator. Step 4: Use sin² A + cos² A = 1. Therefore, the answer is 1.

Question 6

Prove that (1 - sin² θ) sec² θ = 1. What is the value of the left side when θ = 60°?

Accepted Answer

Step 1: Simplify the left side using the identity 1 - sin² θ = cos² θ. Step 2: (1 - sin² θ) sec² θ = cos² θ · sec² θ = cos² θ · (1/cos² θ) = 1. Step 3: The identity is proven; the left side equals 1 for all values of θ (including θ = 60°).

Question 7

Simplify: (sin⁴ θ − cos⁴ θ)/(sin² θ − cos² θ) + 2cos² θ.

Accepted Answer

Step 1: Factor the numerator using difference of squares: sin⁴ θ − cos⁴ θ = (sin² θ − cos² θ)(sin² θ + cos² θ). Step 2: Simplify the fraction: (sin⁴ θ − cos⁴ θ)/(sin² θ − cos² θ) = (sin² θ − cos² θ)(sin² θ + cos² θ)/(sin² θ − cos² θ) = sin² θ + cos² θ = 1. Step 3: Add 2cos² θ: 1 + 2cos² θ.

Question 8

If sin θ − cos θ = 1/3, find the value of sin³ θ − cos³ θ.

Accepted Answer

Step 1: Use the identity a³ − b³ = (a − b)(a² + ab + b²). So sin³ θ − cos³ θ = (sin θ − cos θ)(sin² θ + sin θ cos θ + cos² θ). Step 2: Substitute sin θ − cos θ = 1/3 and sin² θ + cos² θ = 1: sin³ θ − cos³ θ = (1/3)(1 + sin θ cos θ). Step 3: Find sin θ cos θ by squaring sin θ − cos θ = 1/3. We get (sin θ − cos θ)² = 1/9, so sin² θ − 2sin θ cos θ + cos² θ = 1/9. Step 4: Substitute sin² θ + cos² θ = 1: 1 − 2sin θ cos θ = 1/9, so 2sin θ cos θ = 8/9, thus sin θ cos θ = 4/9. Step 5: Substitute back: sin³ θ − cos³ θ = (1/3)(1 + 4/9) = (1/3)(13/9) = 13/27.

Question 9

Simplify: (sin²θ + cos²θ) / sin θ × cos θ

Accepted Answer

Step 1: Apply the Pythagorean identity sin²θ + cos²θ = 1. Step 2: Substitute: 1 / (sin θ × cos θ). Step 3: This equals 1/(sin θ cos θ), which cannot be simplified further without specific values. However, recognizing the identity gives us the simplified form: 1/(sin θ cos θ) or equivalently sec θ csc θ.

IBPS Clerk Trig Identities

IBPS Clerk Trig Identities — Past Exam Questions

Trig Identities— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Trig Identities — Practice Questions

60-Second Revision — Trig Identities