Question 1

Simplify: tan θ × cot θ + sin²θ

Accepted Answer

Step 1: Recall that tan θ × cot θ = 1 (since tan θ = sin θ/cos θ and cot θ = cos θ/sin θ). Step 2: The expression becomes 1 + sin²θ. Step 3: This cannot be simplified further without additional information, so the answer is 1 + sin²θ.

Question 2

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 Pythagorean theorem, opposite = √(25² - 7²) = √(625 - 49) = √576 = 24. Step 3: tan θ = opposite/adjacent = 24/7.

Question 3

Simplify: (1 - sin²θ) / cos θ

Accepted Answer

Step 1: Apply the Pythagorean identity: 1 - sin²θ = cos²θ. Step 2: The expression becomes cos²θ / cos θ. Step 3: Simplify by canceling one factor of cos θ: cos²θ / cos θ = cos θ.

Question 4

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 Pythagorean theorem, hypotenuse = √(5² + 12²) = √(25 + 144) = √169 = 13. Step 3: sin θ = opposite/hypotenuse = 5/13.

Question 5

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

Accepted Answer

Step 1: Numerator sin⁴ α - cos⁴ α is a difference of squares: (sin² α - cos² α)(sin² α + cos² α). Step 2: Using sin² α + cos² α = 1, numerator becomes (sin² α - cos² α) × 1 = sin² α - cos² α. Step 3: Divide by denominator (sin² α - cos² α). Step 4: Result = 1.

Question 6

Simplify: (1 - cos² θ) sec² θ

Accepted Answer

Step 1: Use the identity 1 - cos² θ = sin² θ. Step 2: Expression becomes sin² θ sec² θ. Step 3: sec θ = 1/cos θ, so sec² θ = 1/cos² θ. Step 4: sin² θ × (1/cos² θ) = sin² θ / cos² θ = tan² θ.

Question 7

If sin θ + cos θ = √2, find sin θ cos θ.

Accepted Answer

Step 1: Let sin θ + cos θ = √2. Square both sides: (sin θ + cos θ)² = 2. Step 2: Expand: sin² θ + 2 sin θ cos θ + cos² θ = 2. Step 3: Use sin² θ + cos² θ = 1: 1 + 2 sin θ cos θ = 2. Step 4: 2 sin θ cos θ = 1, so sin θ cos θ = 1/2.

Question 8

If tan θ + cot θ = 4, find the value of tan² θ + cot² θ.

Accepted Answer

Step 1: Let tan θ + cot θ = 4. Square both sides: (tan θ + cot θ)² = 16. Step 2: Expand: tan² θ + 2 tan θ cot θ + cot² θ = 16. Step 3: Since tan θ cot θ = 1, we have tan² θ + 2(1) + cot² θ = 16. Step 4: tan² θ + cot² θ = 16 - 2 = 14.

Question 9

If sin θ + sin²θ = 1, then find cos²θ + cos⁴θ.

Accepted Answer

Step 1: From sin θ + sin²θ = 1, rearrange to get sin θ = 1 − sin²θ = cos²θ. Step 2: Square both sides: sin²θ = cos⁴θ. Step 3: Since sin²θ + cos²θ = 1, substitute sin²θ = cos⁴θ: cos⁴θ + cos²θ = 1. Step 4: Therefore cos²θ + cos⁴θ = 1.

Question 10

If cos²θ − sin²θ = 3/5, then find cos 2θ.

Accepted Answer

Step 1: Recall the double angle identity cos 2θ = cos²θ − sin²θ. Step 2: The given equation directly states cos²θ − sin²θ = 3/5. Step 3: Therefore cos 2θ = 3/5.

Question 11

If sin θ − cos θ = 1/2, then find the value of sin θ · cos θ.

Accepted Answer

Step 1: Square both sides of sin θ − cos θ = 1/2. We get (sin θ − cos θ)² = 1/4. Step 2: Expand: sin²θ − 2sin θ cos θ + cos²θ = 1/4. Step 3: Use sin²θ + cos²θ = 1, so 1 − 2sin θ cos θ = 1/4. Step 4: Rearrange: −2sin θ cos θ = 1/4 − 1 = −3/4. Step 5: Therefore sin θ cos θ = 3/8.

Question 12

If sec θ − tan θ = 2/5, then find sec θ + tan θ.

Accepted Answer

Step 1: Use the identity sec²θ − tan²θ = 1, which factors as (sec θ − tan θ)(sec θ + tan θ) = 1. Step 2: Substitute sec θ − tan θ = 2/5: (2/5)(sec θ + tan θ) = 1. Step 3: Solve for sec θ + tan θ: sec θ + tan θ = 1 ÷ (2/5) = 5/2.

Question 13

If tan α + cot α = 4, then find tan²α + cot²α.

Accepted Answer

Step 1: Square both sides of tan α + cot α = 4. We get (tan α + cot α)² = 16. Step 2: Expand: tan²α + 2tan α cot α + cot²α = 16. Step 3: Since tan α · cot α = 1, we have tan²α + 2(1) + cot²α = 16. Step 4: Therefore tan²α + cot²α = 16 − 2 = 14.

SSC CPO Trig Identities

Trig Identities— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Trig Identities — Practice Questions

60-Second Revision — Trig Identities