Exercise 8.3 answers: Introduction to Trigonometry

Open NCERT p. 131Matches NCERT’s answer

1. tan A is the reciprocal of cot A, so tan A = 1/cot A.
2. From the identity cosec²A = 1 + cot²A, cosec A = √(1 + cot²A). Take the positive root, because A is acute and cosec A is positive.
3. sin A = 1/cosec A = 1/√(1 + cot²A).
4. cot A = cos A/sin A, so cos A = cot A × sin A = cot A/√(1 + cot²A).
5. sec A = 1/cos A = √(1 + cot²A)/cot A.

Answersin A = 1/√(1 + cot²A), sec A = √(1 + cot²A)/cot A, tan A = 1/cot A

Watch this explained “One ratio, and then all six”, 8:04 into Two more identities from the same equation, and the angles they hold for

Open NCERT p. 131Matches NCERT’s answer

1. cos A is simply the reciprocal of sec A: cos A = 1/sec A.
2. sec²A = 1 + tan²A, so tan A = √(sec²A − 1) (positive root, since A is acute).
3. sin A = tan A × cos A = √(sec²A−1)/sec A.
4. cosec A = 1/sin A = sec A/√(sec²A−1).
5. cot A = 1/tan A = 1/√(sec²A−1).

Answercos A = 1/sec A, sin A = √(sec²A−1)/sec A, tan A = √(sec²A−1), cosec A = sec A/√(sec²A−1), cot A = 1/√(sec²A−1)

Watch this explained “One ratio, and then all six”, 8:04 into Two more identities from the same equation, and the angles they hold for

Open NCERT p. 131Matches NCERT’s answer

(i) 9 sec² A – 9 tan² A =

1. From 1 + tan²A = sec²A, sec²A − tan²A = 1.
2. So 9 sec²A − 9 tan²A = 9(sec²A − tan²A) = 9 × 1 = 9.

Answer(B) 9

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =

1. Write tan θ = sin θ/cos θ, sec θ = 1/cos θ, cot θ = cos θ/sin θ and cosec θ = 1/sin θ.
2. First bracket: 1 + sin θ/cos θ + 1/cos θ = (cos θ + sin θ + 1)/cos θ.
3. Second bracket: 1 + cos θ/sin θ − 1/sin θ = (sin θ + cos θ − 1)/sin θ.
4. Multiply the tops: (sin θ + cos θ + 1)(sin θ + cos θ − 1) = (sin θ + cos θ)² − 1 = sin²θ + 2 sin θ cos θ + cos²θ − 1 = 1 + 2 sin θ cos θ − 1 = 2 sin θ cos θ.
5. So the product = 2 sin θ cos θ/(cos θ sin θ) = 2.

Answer(C) 2

(iii) (sec A + tan A) (1 – sin A) =

1. sec A + tan A = 1/cos A + sin A/cos A = (1 + sin A)/cos A.
2. So (sec A + tan A)(1 − sin A) = (1 + sin A)(1 − sin A)/cos A = (1 − sin²A)/cos A.
3. 1 − sin²A = cos²A, so this is cos²A/cos A = cos A.

Answer(D) cos A

(iv) (1 + tan² A)/(1 + cot² A) =

1. 1 + tan²A = sec²A and 1 + cot²A = cosec²A.
2. So the expression = sec²A/cosec²A = (1/cos²A) ÷ (1/sin²A) = sin²A/cos²A = tan²A.

Answer(D) tan² A

Watch this explained “First collapse: convert to sines and cosines”, 10:28 into Two more identities from the same equation, and the angles they hold for

Open NCERT p. 131One way to think about it

(i)

1. To prove: (cosec θ – cot θ)² = (1 – cos θ)/(1 + cos θ).
2. Write cosecθ = 1/sinθ and cotθ = cosθ/sinθ, so cosecθ−cotθ = (1−cosθ)/sinθ.
3. Squaring: (cosecθ−cotθ)² = (1−cosθ)²/sin²θ.
4. sin²θ = 1−cos²θ = (1−cosθ)(1+cosθ), so the fraction becomes (1−cosθ)²/[(1−cosθ)(1+cosθ)] = (1−cosθ)/(1+cosθ).

In shortBoth sides simplify to (1−cosθ)/(1+cosθ), so the identity holds.

(ii)

1. To prove: cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A.
2. Combine the two fractions over the common denominator cosA(1+sinA): [cos²A + (1+sinA)²] / [cosA(1+sinA)].
3. Expand (1+sinA)² = 1+2sinA+sin²A, so the top becomes cos²A+sin²A+1+2sinA = 1+1+2sinA = 2(1+sinA).
4. So the expression is 2(1+sinA)/[cosA(1+sinA)] = 2/cosA = 2 secA.

In shortLHS simplifies to 2 sec A, which is the RHS.

(iii)

1. To prove: tan θ/(1 – cot θ) + cot θ/(1 – tan θ) = 1 + sec θ cosec θ.
2. Write tanθ = sinθ/cosθ and cotθ = cosθ/sinθ.
3. The first term becomes sin²θ/[cosθ(sinθ−cosθ)] and the second becomes −cos²θ/[sinθ(sinθ−cosθ)].
4. Adding them over the common factor (sinθ−cosθ) gives (sin³θ−cos³θ) / [sinθcosθ(sinθ−cosθ)].
5. sin³θ−cos³θ factorises to (sinθ−cosθ)(sin²θ+sinθcosθ+cos²θ) = (sinθ−cosθ)(1+sinθcosθ).
6. Cancelling (sinθ−cosθ) leaves (1+sinθcosθ)/(sinθcosθ) = 1/(sinθcosθ) + 1 = secθ cosecθ + 1.

In shortLHS simplifies to 1 + secθ cosecθ, which is the RHS.

(iv)

1. To prove: (1 + sec A)/sec A = sin² A/(1 – cos A).
2. Simplify the LHS: (1+secA)/secA = 1/secA + 1 = cosA + 1.
3. Simplify the RHS: sin²A = 1−cos²A = (1−cosA)(1+cosA), so sin²A/(1−cosA) = 1+cosA.
4. Both sides equal 1+cosA.

In shortBoth sides equal 1 + cos A, so the identity holds.

(v)

1. To prove: (cos A – sin A + 1)/(cos A + sin A – 1) = cosec A + cot A, using the identity cosec² A = 1 + cot² A.
2. Divide every term of the top and bottom by sinA: top becomes cotA+cosecA−1, bottom becomes cotA−cosecA+1.
3. In the bottom, replace the +1 with cosec²A−cot²A (from cosec²A = 1+cot²A).
4. The bottom becomes cotA−cosecA+cosec²A−cot²A, which factorises to (cosecA−cotA)(cosecA+cotA−1).
5. The top, cotA+cosecA−1, is exactly the second factor, so it cancels, leaving 1/(cosecA−cotA).
6. Multiplying top and bottom by (cosecA+cotA) turns the bottom into cosec²A−cot²A = 1, leaving cosecA+cotA.

In shortLHS simplifies to cosec A + cot A, which is the RHS.

(vi)

1. To prove: √((1 + sin A)/(1 – sin A)) = sec A + tan A.
2. Multiply inside the square root, top and bottom, by (1+sinA): (1+sinA)² / [(1−sinA)(1+sinA)] = (1+sinA)²/(1−sin²A) = (1+sinA)²/cos²A.
3. Taking the square root (everything positive, since A is acute) gives (1+sinA)/cosA.
4. Split the fraction: 1/cosA + sinA/cosA = secA + tanA.

In shortLHS simplifies to sec A + tan A, which is the RHS.

(vii)

1. To prove: (sin θ – 2 sin³ θ)/(2 cos³ θ – cos θ) = tan θ.
2. Factor sinθ from the top: sinθ(1−2sin²θ). Factor cosθ from the bottom: cosθ(2cos²θ−1).
3. Using sin²θ = 1−cos²θ, rewrite 1−2sin²θ = 1−2(1−cos²θ) = 2cos²θ−1, the same bracket as on the bottom.
4. The bracket (2cos²θ−1) cancels, leaving sinθ/cosθ = tanθ.

In shortLHS simplifies to tan θ, which is the RHS.

(viii)

1. To prove: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.
2. Expand both squares: sin²A+2sinA·cosecA+cosec²A + cos²A+2cosA·secA+sec²A.
3. sinA·cosecA = 1 and cosA·secA = 1, so the two middle terms give 2+2 = 4.
4. sin²A+cos²A = 1, so the total so far is 1+4+cosec²A+sec²A = 5+cosec²A+sec²A.
5. Using cosec²A = 1+cot²A and sec²A = 1+tan²A, this becomes 5+1+cot²A+1+tan²A = 7+tan²A+cot²A.

In shortLHS simplifies to 7 + tan²A + cot²A, which is the RHS.

(ix)

1. To prove: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A).
2. Simplify the LHS: cosecA−sinA = 1/sinA − sinA = (1−sin²A)/sinA = cos²A/sinA. Similarly secA−cosA = sin²A/cosA.
3. Multiplying these: (cos²A/sinA)(sin²A/cosA) = sinA cosA.
4. Simplify the RHS: tanA+cotA = sinA/cosA + cosA/sinA = (sin²A+cos²A)/(sinAcosA) = 1/(sinAcosA), so 1/(tanA+cotA) = sinAcosA.
5. Both sides equal sinA cosA.

In shortBoth sides equal sin A cos A, so the identity holds.

(x)

1. To prove: (1 + tan² A)/(1 + cot² A) = ((1 – tan A)/(1 – cot A))² = tan² A.
2. 1+tan²A = sec²A and 1+cot²A = cosec²A, so the first expression is sec²A/cosec²A = sin²A/cos²A = tan²A.
3. For the middle expression, write cotA = 1/tanA, so 1−cotA = (tanA−1)/tanA.
4. (1−tanA)/(1−cotA) = (1−tanA)×tanA/(tanA−1) = −tanA, since (1−tanA) = −(tanA−1).
5. Squaring gives (−tanA)² = tan²A, the same as the first expression.

In shortBoth the first expression and the squared middle expression equal tan²A.

Watch this explained “First collapse: convert to sines and cosines”, 10:28 into Two more identities from the same equation, and the angles they hold for