Ch 8: Trigonometric Identities & Proofs
This page contains the crucial 10 proofs (Ex 8.3 New / Ex 8.4 Old) and the Old Syllabus Complementary Angles questions.
Exercise 8.3 (All 10 Proofs)
💡 Note: In the old NCERT textbook, this was Exercise 8.4. These are the most important 4-5 markers in the board exam.
Q1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
- tan A = 1 / cot A
- sin A = 1 / cosec A = 1 / √(1 + cot² A)
- sec A = √(1 + tan² A) = √(1 + 1/cot² A) = √(cot² A + 1) / cot A
Q2. Write all the other trigonometric ratios of ∠A in terms of sec A.
cos A = 1/sec A ; sin A = √(sec² A - 1) / sec A ; tan A = √(sec² A - 1) ; cosec A = sec A / √(sec² A - 1) ; cot A = 1 / √(sec² A - 1).
Q4. Choose the correct option:
(i) 9 sec² A - 9 tan² A = 9(sec² A - tan² A) = 9(1) = 9 (Option B)
(ii) (1 + tan θ + sec θ)(1 + cot θ - cosec θ) = 2 (Option C)
(iii) (sec A + tan A)(1 - sin A) = (1/cosA + sinA/cosA)(1-sinA) = (1-sin²A)/cosA = cos²A/cosA = cos A (Option D)
(iv) (1 + tan² A)/(1 + cot² A) = sec² A / cosec² A = sin² A / cos² A = tan² A (Option D)
Q5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ - cot θ)² = (1 - cos θ) / (1 + cos θ)
LHS = (1/sin θ - cos θ/sin θ)²
= [(1 - cos θ) / sin θ]²
= (1 - cos θ)² / sin² θ
= (1 - cos θ)² / (1 - cos² θ)
= [(1 - cos θ)(1 - cos θ)] / [(1 - cos θ)(1 + cos θ)]
= (1 - cos θ) / (1 + cos θ) = RHS. Proved.
(ii) cos A / (1 + sin A) + (1 + sin A) / cos A = 2 sec A
LHS = [cos² A + (1 + sin A)²] / [cos A (1 + sin A)]
= [cos² A + 1 + sin² A + 2sin A] / [cos A (1 + sin A)]
= [1 + 1 + 2sin A] / [cos A (1 + sin A)] (Since sin²A + cos²A = 1)
= 2(1 + sin A) / [cos A (1 + sin A)]
= 2 / cos A = 2 sec A = RHS. Proved.
(iii) tan θ / (1 - cot θ) + cot θ / (1 - tan θ) = 1 + sec θ cosec θ
Convert everything to sin and cos. This is the longest proof. Taking LCMs yields:
= [sin³ θ - cos³ θ] / [sin θ cos θ (sin θ - cos θ)]
Using a³ - b³ = (a - b)(a² + b² + ab):
= [(sin θ - cos θ)(sin² θ + cos² θ + sin θ cos θ)] / [sin θ cos θ (sin θ - cos θ)]
= (1 + sin θ cos θ) / (sin θ cos θ)
= (1 / sin θ cos θ) + 1 = 1 + sec θ cosec θ = RHS. Proved.
(iv) (1 + sec A) / sec A = sin² A / (1 - cos A)
LHS = (1 + 1/cos A) / (1/cos A) = (cos A + 1) / cos A × cos A = 1 + cos A.
RHS = (1 - cos² A) / (1 - cos A) = [(1 - cos A)(1 + cos A)] / (1 - cos A) = 1 + cos A.
LHS = RHS. Proved.
(v) (cos A - sin A + 1) / (cos A + sin A - 1) = cosec A + cot A
Divide numerator and denominator by sin A:
LHS = (cot A - 1 + cosec A) / (cot A + 1 - cosec A)
Substitute 1 = cosec² A - cot² A in the numerator:
= [(cot A + cosec A) - (cosec² A - cot² A)] / (cot A + 1 - cosec A)
= (cot A + cosec A)[1 - (cosec A - cot A)] / (cot A + 1 - cosec A)
Canceling the common bracket leaves: cosec A + cot A = RHS. Proved.
(vi) √(1 + sin A) / (1 - sin A) = sec A + tan A
Rationalize the denominator inside the root:
LHS = √[ (1 + sin A)(1 + sin A) / (1 - sin A)(1 + sin A) ]
= √[ (1 + sin A)² / (1 - sin² A) ]
= √[ (1 + sin A)² / cos² A ]
= (1 + sin A) / cos A
= 1/cos A + sin A/cos A = sec A + tan A = RHS. Proved.
(vii) (sin θ - 2sin³ θ) / (2cos³ θ - cos θ) = tan θ
LHS = [sin θ(1 - 2sin² θ)] / [cos θ(2cos² θ - 1)]
Using sin² θ = 1 - cos² θ in numerator:
= tan θ [1 - 2(1 - cos² θ)] / [2cos² θ - 1]
= tan θ [2cos² θ - 1] / [2cos² θ - 1]
= tan θ = RHS. Proved.
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
LHS = sin² A + cosec² A + 2sin A cosec A + cos² A + sec² A + 2cos A sec A
= (sin² A + cos² A) + cosec² A + 2(1) + sec² A + 2(1)
= 1 + (1 + cot² A) + 4 + (1 + tan² A)
= 7 + tan² A + cot² A = RHS. Proved.
(ix) (cosec A - sin A)(sec A - cos A) = 1 / (tan A + cot A)
LHS = (1/sin A - sin A)(1/cos A - cos A)
= [(1 - sin² A)/sin A] × [(1 - cos² A)/cos A]
= (cos² A / sin A) × (sin² A / cos A) = sin A cos A.
RHS = 1 / (sin A/cos A + cos A/sin A) = 1 / [(sin² A + cos² A) / sin A cos A] = sin A cos A.
LHS = RHS. Proved.
(x) (1 + tan² A) / (1 + cot² A) = [(1 - tan A) / (1 - cot A)]² = tan² A
Part 1: (1 + tan² A)/(1 + cot² A) = sec² A / cosec² A = sin² A / cos² A = tan² A.
Part 2: [(1 - tan A) / (1 - 1/tan A)]² = [(1 - tan A) / ((tan A - 1)/tan A)]²
= [-tan A]² = tan² A.
Hence all three are equal. Proved.
Old Ex 8.3 (Complementary Angles)
💡 Formula: sin(90 - θ) = cos θ, tan(90 - θ) = cot θ, sec(90 - θ) = cosec θ.
Q1. Evaluate: (i) sin 18° / cos 72° (ii) tan 26° / cot 64° (iii) cos 48° - sin 42°
(i) sin 18° / cos 72° = sin(90° - 72°) / cos 72° = cos 72° / cos 72° = 1
(ii) tan 26° / cot 64° = tan(90° - 64°) / cot 64° = cot 64° / cot 64° = 1
(iii) cos 48° - sin 42° = cos(90° - 42°) - sin 42° = sin 42° - sin 42° = 0
Q2. Show that: tan 48° tan 23° tan 42° tan 67° = 1
Solution:
LHS = tan 48° × tan 23° × tan(90° - 48°) × tan(90° - 23°)
= tan 48° × tan 23° × cot 48° × cot 23°
Since tan θ × cot θ = 1:
= (tan 48° × cot 48°) × (tan 23° × cot 23°) = 1 × 1 = 1 = RHS. Proved.
Q3. If tan 2A = cot (A - 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A = cot(90° - 2A). So we can write:
cot(90° - 2A) = cot(A - 18°)
90° - 2A = A - 18°
108° = 3A ⇒ A = 36°
Q6. If A, B and C are interior angles of a triangle ABC, then show that sin((B+C)/2) = cos(A/2).
Solution:
We know A + B + C = 180°.
Therefore, B + C = 180° - A.
Dividing by 2: (B + C)/2 = 90° - A/2.
Taking sin on both sides:
sin((B + C)/2) = sin(90° - A/2)
sin((B + C)/2) = cos(A/2). Proved.