Question 1

Simplify: sec²θ - tan²θ

Accepted Answer

Step 1: Recall the Pythagorean identity sec²θ - tan²θ = 1. Step 2: This is a standard trigonometric identity that directly simplifies to 1.

Question 2

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

Accepted Answer

Step 1: Recall that tan θ = opposite/adjacent = 5/12. Step 2: Use the Pythagorean theorem to find the hypotenuse: h² = 5² + 12² = 25 + 144 = 169, so h = 13. Step 3: sin θ = opposite/hypotenuse = 5/13.

Question 3

Simplify: (1 - sin²θ) / cos θ

Accepted Answer

Step 1: Use the identity 1 - sin²θ = cos²θ. Step 2: Substitute: cos²θ / cos θ. Step 3: Simplify by cancelling one cos θ: cos θ.

Question 4

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

Accepted Answer

Step 1: Numerator sin⁴α − cos⁴α is a difference of squares: (sin²α − cos²α)(sin²α + cos²α). Step 2: Using sin²α + cos²α = 1, numerator becomes (sin²α − cos²α) × 1 = sin²α − cos²α. Step 3: Denominator is sin²α − cos²α. Step 4: Dividing: (sin²α − cos²α)/(sin²α − cos²α) = 1.

Question 5

Simplify: (cos² θ − sin² θ) / (1 − 2 sin² θ) and identify the result.

Accepted Answer

Step 1: Numerator cos²θ − sin²θ is the double angle formula cos 2θ. Step 2: Denominator 1 − 2 sin²θ is also the double angle formula cos 2θ (using 1 − 2 sin²θ = cos 2θ). Step 3: (cos 2θ) / (cos 2θ) = 1.

Question 6

Prove that (1 − sin² θ) sec² θ = 1. What is the simplified form of the left side before equating to 1?

Accepted Answer

Step 1: (1 − sin²θ) sec²θ. Step 2: Using 1 − sin²θ = cos²θ, expression becomes cos²θ sec²θ. Step 3: sec θ = 1/cos θ, so sec²θ = 1/cos²θ. Step 4: cos²θ × (1/cos²θ) = 1. The simplified form before the final step is cos²θ sec²θ.

Question 7

If sin(θ) + cosec(θ) = 2, then find the value of sin²(θ) + cosec²(θ).

Accepted Answer

Step 1: Given sin(θ) + cosec(θ) = 2. Step 2: Square both sides: [sin(θ) + cosec(θ)]² = 4. Step 3: Expand: sin²(θ) + 2·sin(θ)·cosec(θ) + cosec²(θ) = 4. Step 4: Since sin(θ)·cosec(θ) = 1, we have: sin²(θ) + 2(1) + cosec²(θ) = 4. Step 5: Therefore sin²(θ) + cosec²(θ) = 4 - 2 = 2.

Question 8

If sin(x) + sin²(x) = 1, then find cos²(x) + cos⁴(x).

Accepted Answer

Step 1: Given sin(x) + sin²(x) = 1. Step 2: Rearrange: sin(x) = 1 - sin²(x) = cos²(x). Step 3: Since sin²(x) + cos²(x) = 1, we have sin²(x) = 1 - cos²(x). Step 4: From sin(x) = cos²(x), square both sides: sin²(x) = cos⁴(x). Step 5: Substitute into sin²(x) = 1 - cos²(x): cos⁴(x) = 1 - cos²(x). Step 6: Therefore cos²(x) + cos⁴(x) = cos²(x) + (1 - cos²(x)) = 1.

Question 9

If 3sin(θ) - 4cos(θ) = 5, then find 3cos(θ) + 4sin(θ).

Accepted Answer

Step 1: Given 3sin(θ) - 4cos(θ) = 5. Step 2: Let 3cos(θ) + 4sin(θ) = k (what we seek). Step 3: Square the given equation: (3sin(θ) - 4cos(θ))² = 25, which gives 9sin²(θ) - 24sin(θ)cos(θ) + 16cos²(θ) = 25. Step 4: Square the expression we seek: (3cos(θ) + 4sin(θ))² = k², which gives 9cos²(θ) + 24sin(θ)cos(θ) + 16sin²(θ) = k². Step 5: Add the two squared equations: 9(sin²(θ) + cos²(θ)) + 16(sin²(θ) + cos²(θ)) + (-24 + 24)sin(θ)cos(θ) = 25 + k². Step 6: Simplify: 9 + 16 = 25 + k², so 25 = 25 + k², thus k² = 0, so k = 0.

Question 10

If tan(A) + tan(B) = 6 and tan(A)·tan(B) = 2, then find tan(A + B).

Accepted Answer

Step 1: Use the addition formula: tan(A + B) = [tan(A) + tan(B)] / [1 - tan(A)·tan(B)]. Step 2: Substitute the given values: tan(A + B) = 6 / [1 - 2]. Step 3: Simplify: tan(A + B) = 6 / (-1) = -6.

Question 11

If cos(θ) - sec(θ) = -2, then find cos²(θ) + sec²(θ).

Accepted Answer

Step 1: Given cos(θ) - sec(θ) = -2. Step 2: Square both sides: [cos(θ) - sec(θ)]² = 4. Step 3: Expand: cos²(θ) - 2·cos(θ)·sec(θ) + sec²(θ) = 4. Step 4: Since cos(θ)·sec(θ) = 1, we have: cos²(θ) - 2(1) + sec²(θ) = 4. Step 5: Therefore cos²(θ) + sec²(θ) = 4 + 2 = 6.

Question 12

If tan(α) + cot(α) = 4, then find tan²(α) + cot²(α).

Accepted Answer

Step 1: Given tan(α) + cot(α) = 4. Step 2: Square both sides: [tan(α) + cot(α)]² = 16. Step 3: Expand: tan²(α) + 2·tan(α)·cot(α) + cot²(α) = 16. Step 4: Since tan(α)·cot(α) = 1, we have: tan²(α) + 2(1) + cot²(α) = 16. Step 5: Therefore tan²(α) + cot²(α) = 16 - 2 = 14.

IBPS RRB PO Trig Identities

IBPS RRB PO Trig Identities — Past Exam Questions

Trig Identities— Rules & Concept

Key Points to Remember

Exam-Specific Tips

60-Second Revision — Trig Identities