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Trigonometric Identities

Key Concepts

Solved Examples: 50

Practice Questions: 3

Solved Examples

Question 1: If tan θ = 3/4, find the value of (tan θ + cot θ)/(tan θ × cot θ)

Using the reciprocal identity, cot θ = 4/3
Numerator
= 3/4 + 4/3
= (9 + 16)/12
= 25/12
Denominator
= 3/4 × 4/3
= 1
Therefore,
= 25/12

Question 2: If cosec θ = 25/24, find the value of (1 + sin θ)/(1 − sin θ)

Using the reciprocal identity,
sin θ = 24/25
Therefore,
= (1 + 24/25)/(1 − 24/25)
= (49/25)/(1/25)
= 49

Question 3: If sec θ = 5/3 and tan θ = 4/3, find the value of (1/sec θ) + (1/tan θ)

Using reciprocal identities,
1/sec θ = cos θ = 3/5
1/tan θ = cot θ = 3/4
Therefore,
= 3/5 + 3/4
= (12 + 15)/20
= 27/20

Question 4: If sin θ = 5/13 and cos θ = 12/13, find the value of: tan θ + cot θ

tan θ = (5/13)/(12/13)
= 5/12
cot θ = (12/13)/(5/13)
= 12/5
Therefore,
tan θ + cot θ
= 5/12 + 12/5
= (25 + 144)/60
= 169/60

Question 5: If sin θ = 3/5 and θ is an acute angle, find cos θ.

Using the identity,
sin² θ + cos² θ = 1
cos² θ = 1 − (3/5)²
= 1 − 9/25
= 16/25
Since θ is acute,
cos θ = 4/5

Question 6: If tan θ = 3/4, find sec θ.

Using the identity,
1 + tan² θ = sec² θ
sec² θ
= 1 + (3/4)²
= 1 + 9/16
= 25/16
Therefore,
sec θ = 5/4

Question 7: Simplify (sec² θ − tan² θ) + (cosec² θ − cot² θ)

Using the identities,
sec² θ − tan² θ = 1
cosec² θ − cot² θ = 1
Therefore,
= 1 + 1
= 2

Question 8: Simplify (1 + cot² θ)(1 − sin² θ)

Using the identities,
1 + cot² θ = cosec² θ
1 − sin² θ = cos² θ
Therefore,
= cosec² θ × cos² θ
= (1/sin² θ) × cos² θ
= cot² θ

Question 9: Simplify (cosec² θ − cot² θ)(sec² θ − tan² θ)

Using the identities,
cosec² θ − cot² θ = 1
sec² θ − tan² θ = 1
Therefore,
= 1 × 1
= 1

Question 10: Simplify (sec² θ + cosec² θ) − (tan² θ + cot² θ)

Using the identities,
sec² θ = 1 + tan² θ
cosec² θ = 1 + cot² θ
Therefore,
= (1 + tan² θ) + (1 + cot² θ) − tan² θ − cot² θ
= 2

Question 11: Verify that (sec² θ − 1)(cosec² θ − 1) = 1

LHS
= (sec² θ − 1)(cosec² θ − 1)
Using,
sec² θ − 1 = tan² θ
cosec² θ − 1 = cot² θ
Therefore,
= tan² θ × cot² θ
= (tan θ × cot θ)²
= 1²
= 1
= RHS

Question 12: Verify that (1 − cos² θ)(1 + tan² θ) = tan² θ

LHS
= (1 − cos² θ)(1 + tan² θ)
Using,
1 − cos² θ = sin² θ
1 + tan² θ = sec² θ
Therefore,
= sin² θ × sec² θ
= sin² θ/cos² θ
= tan² θ
= RHS
Hence verified.

Question 13: If cos θ = 12/13 and θ is an acute angle, find the value of cosec θ − cot θ

Using,
sin² θ = 1 − cos² θ
= 1 − (12/13)²
= 25/169
∴ sin θ = 5/13
Hence,
cosec θ = 13/5
cot θ = 12/5
Therefore,
cosec θ − cot θ
= 13/5 − 12/5
= 1/5

Question 14: If cosec θ = 25/24 and θ is an acute angle, find the value of sec θ − tan θ

sin θ = 24/25
Using, cos² θ = 1 − (24/25)²
= 49/625
∴ cos θ = 7/25
Hence,
sec θ = 25/7
tan θ = (24/25)/(7/25)
= 24/7
Therefore,
sec θ − tan θ
= 25/7 − 24/7
= 1/7

Question 15: If cos θ = 4/5 and θ is an acute angle, find the value of (1 − sin² θ)sec² θ

Using,
1 − sin² θ = cos² θ
Therefore,
= cos² θ × sec² θ
= 1

Question 16: If A is an acute angle and 13 sin A = 5, evaluate (5 sin A − 2 cos A) / tan A

Given, 13 sin A = 5
∴ sin A = 5/13
Using the identity,
sin² A + cos² A = 1
cos² A = 1 − (5/13)²
= 1 − 25/169
= 144/169
Since A is acute,
cos A = 12/13
Also, tan A = sin A/cos A
= (5/13)/(12/13)
= 5/12
Now,
(5 sin A − 2 cos A)/tan A
= [5(5/13) − 2(12/13)]/(5/12)
= [(25 − 24)/13]/(5/12)
= (1/13) × (12/5)
= 12/65
∴ The value is 12/65.

Question 17: If A is an acute angle and tan A = 1/√5, find the value of (cosec² A − sec² A) / (cosec² A × sec² A).

Given, tan A = 1/√5
Using the identity,
1 + tan² A = sec² A
sec² A = 1 + (1/√5)² = 1 + 1/5 = 6/5
Also, 1 + cot² A = cosec² A
Since, cot A = √5
cosec² A = 1 + (√5)² = 6
∴ (cosec² A − sec² A) / (cosec² A × sec² A)
= (6 − 6/5)/(6 × 6/5)
= (24/5)/(36/5)
= 24/36
= 2/3
∴ The value is 2/3.

Question 18: Evaluate (sin² 63° + sin² 27°) / (cos² 17° + cos² 73°).

Using the complementary-angle relationships,
sin 63° = sin(90° − 27°) = cos 27° and
cos 73° = cos(90° − 17°) = sin 17°
∴ sin² 63° + sin² 27°
= cos² 27° + sin² 27°
= 1
Also, cos² 17° + cos² 73°
= cos² 17° + sin² 17°
= 1
Hence,
(sin² 63° + sin² 27°) / (cos² 17° + cos² 73°)
= 1/1
= 1
∴ The value is 1.

Question 19: Without using trigonometric tables, evaluate cosec² 57° − tan² 33° + cos 44° × cosec 46° − √2 cos 45° − tan² 60°.

Using complementary angles,
cosec 57° = sec(90° − 57°) = sec 33°
∴ cosec² 57° = sec² 33°
Also, tan 33° = cot(90° − 33°) = cot 57°
Using the identity,
sec² θ − tan² θ = 1 we get
cosec² 57° − tan² 33°
= sec² 33° − tan² 33°
= 1
Next, cos 44° = sin(90° − 44°) = sin 46°
∴ cos 44° × cosec 46°
= sin 46° × 1/sin 46°
= 1
Also, √2 cos 45°
= √2 × √2/2
= 1
And, tan² 60° = 3
Hence,
cosec² 57° − tan² 33° + cos 44° × cosec 46° − √2 cos 45° − tan² 60°.
= 1 + 1 − 1 − 3
= −2
∴ The value is −2.

Question 20: Evaluate (tan 20° / cosec 70°)² + (cot 20° / sec 70°)² + 2 tan 15° tan 45° tan 75°.

Using complementary angles,
cosec 70° = sec 20° and
sec 70° = cosec 20°
Therefore, (tan 20° / cosec 70°)²
= (tan 20° / sec 20°)²
= sin² 20°
Also, (cot 20° / sec 70°)²
= (cot 20° / cosec 20°)²
= cos² 20°
Hence, (tan 20° / cosec 70°)² + (cot 20° / sec 70°)²
= sin² 20° + cos² 20°
= 1
Now, 2 tan 15° tan 45° tan 75°
= 2 tan 15° × 1 × tan 75°
Since, tan 75° = cot 15°
we get = 2 tan 15° × cot 15°
= 2
Original expression = 1 + 2 = 3
∴ The value is 3.

Question 21: Prove that (sin A + cos A)(tan A + cot A) = sec A + cosec A.

LHS = (sin A + cos A)(tan A + cot A)
= (sin A + cos A)[sin A / cos A + cos A / sin A]
= (sin A + cos A)[(sin² A + cos² A) / (sin A cos A)]
Since sin² A + cos² A = 1
LHS = (sin A + cos A) / (sin A cos A)
= sin A / (sin A cos A) + cos A / (sin A cos A)
= 1 / cos A + 1 / sin A
= sec A + cosec A
= RHS
∴ LHS = RHS.

Question 22: Prove that (sin A × tan A) / (1 − cos A) = 1 + sec A.

LHS = (sin A × tan A) / (1 − cos A)
Using, tan A = sin A / cos A
= sin² A / [cos A(1 − cos A)]
Using, sin² A = 1 − cos² A
= (1 − cos² A) / [cos A(1 − cos A)]
Since, 1 − cos² A = (1 − cos A)(1 + cos A)
= [(1 − cos A)(1 + cos A)] / [cos A(1 − cos A)]
= (1 + cos A) / cos A
= 1/cos A + 1
= sec A + 1
= RHS
∴ LHS = RHS.

Question 23: Prove that sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A.

LHS = sin A(1 + sin A/cos A) + cos A(1 + cos A/sin A)
= sin A[(cos A + sin A)/cos A] + cos A[(sin A + cos A)/sin A]
= (sin A + cos A)[sin A/cos A + cos A/sin A]
= (sin A + cos A)[(sin² A + cos² A)/(sin A cos A)]
Since sin² A + cos² A = 1
= (sin A + cos A)/(sin A cos A)
= 1/cos A + 1/sin A
= sec A + cosec A
= RHS
∴ LHS = RHS.

Question 24: Prove that (1 + cot A − cosec A)(1 + tan A + sec A) = 2.

LHS = (1 + cos A/sin A − 1/sin A)(1 + sin A/cos A + 1/cos A)
= [(sin A + cos A − 1)/sin A] × [(cos A + sin A + 1)/cos A]
= [(sin A + cos A)² − 1]/(sin A cos A)
Since sin² A + cos² A = 1
= [sin² A + cos² A + 2 sin A cos A − 1]/(sin A cos A)
= 2 sin A cos A/(sin A cos A)
= 2
∴ LHS = 2.

Question 25: Prove that tan A/(1 − cot A) + cot A/(1 − tan A) = 1 + tan A + cot A.

LHS = tan A/(1 − cot A) + cot A/(1 − tan A)
= tan A/(1 − 1/tan A) + cot A/(1 − tan A)
= tan² A/(tan A − 1) + cot A/(1 − tan A)
= tan² A/(tan A − 1) − cot A/(tan A − 1)
= (tan² A − cot A)/(tan A − 1)
= (tan² A − 1/tan A)/(tan A − 1)
= (tan³ A − 1)/[tan A(tan A − 1)]
= [(tan A − 1)(tan² A + tan A + 1)]/[tan A(tan A − 1)]
= (tan² A + tan A + 1)/tan A
= tan A + 1 + cot A
= 1 + tan A + cot A
= RHS
∴ LHS = RHS.

Question 26: Prove that (tan A + sec A − 1)/(tan A − sec A + 1) = (1 + sin A)/cos A.

Since sec² A − tan² A = 1
(tan A + sec A − 1)/(tan A − sec A + 1) = (1 + sin A)/cos A.
= [(tan A + sec A) − (sec² A − tan² A)]/(tan A − sec A + 1)
= [(tan A + sec A) − (sec A + tan A)(sec A − tan A)]/(tan A − sec A + 1)
= (tan A + sec A)[1 − (sec A − tan A)]/(tan A − sec A + 1)
= (tan A + sec A)(tan A − sec A + 1)/(tan A − sec A + 1)
= tan A + sec A
= sin A/cos A + 1/cos A
= (1 + sin A)/cos A
= RHS
∴ LHS = RHS.

Question 27: Prove that sin⁴ A + cos⁴ A = 1 − 2 sin² A cos² A.

LHS = sin⁴ A + cos⁴ A
= (sin² A)² + (cos² A)²
Since a² + b² = (a + b)² − 2ab
= (sin² A + cos² A)² − 2 sin² A cos² A
Again sin² A + cos² A = 1
= 1² − 2 sin² A cos² A
= 1 − 2 sin² A cos² A
= RHS
∴ LHS = RHS.

Question 28: Prove that (1 + tan² A)/(1 + cot² A) = [(1 − tan A)/(1 − cot A)]² = tan² A.

First LH term = (1 + tan² A)/(1 + cot² A)
Since 1 + tan² A = sec² A and
1 + cot² A = cosec² A
So LH term = sec² A/cosec² A
= (1/cos² A)/(1/sin² A)
= sin² A/cos² A
= tan² A = RH term
Again Middle term
= [(1 − tan A)/(1 − cot A)]²
= [(1 − tan A)/(1 − 1/tan A)]²
= [(1 − tan A)/((tan A − 1)/tan A)]²
= [tan A(1 − tan A)/(tan A − 1)]²
Since, 1 − tan A = −(tan A − 1)
= (−tan A)²
= tan² A = RH term
Hence,
LH term = Middle term = RH term
∴ Proved.

Question 29: Given that A + B = 90°, prove that √(cos A cosec B − cos A sin B) = sin A.

Since A + B = 90°,
B = 90° − A
∴ sin B = sin(90° − A) = cos A and
cosec B = 1/sin B = 1/cos A
Now,
LHS = √(cos A cosec B − cos A sin B)
= √(cos A × 1/cos A − cos A × cos A)
= √(1 − cos² A)
Since 1 − cos² A = sin² A,
= √(sin² A)
= sin A
= RHS
∴ Proved.

Question 30: If 1 + sin² A = 3 sin A cos A, prove that tan A = 1 or 1/2.

Given, 1 + sin² A = 3 sin A cos A
Dividing both sides by cos² A,
1/cos² A + sin² A/cos² A = 3 sin A/cos A
or, sec² A + tan² A = 3 tan A
Since sec² A = 1 + tan² A,
1 + tan² A + tan² A = 3 tan A
or, 2 tan² A − 3 tan A + 1 = 0
Factoring,
(2 tan A − 1)(tan A − 1) = 0
So, 2 tan A − 1 = 0 or tan A − 1 = 0
or, tan A = 1/2 or tan A = 1
∴ tan A = 1 or 1/2.

Question 31: If p sin A + q cos A = r, prove that p cos A − q sin A = ± √(p² + q² − r²).

Given, p sin A + q cos A = r
Squaring both sides,
(p sin A + q cos A)² = r²
or, p² sin² A + q² cos² A + 2pq sin A cos A = r²
Now consider,
(p cos A − q sin A)²
= p² cos² A + q² sin² A − 2pq sin A cos A
Adding the two expressions,
(p sin A + q cos A)² + (p cos A − q sin A)²
= p² sin² A + q² cos² A + 2pq sin A cos A
p² cos² A + q² sin² A − 2pq sin A cos A
= p²(sin² A + cos² A) + q²(sin² A + cos² A)
= p² + q²
∴ r² + (p cos A − q sin A)² = p² + q²
or, (p cos A − q sin A)² = p² + q² − r²
Taking square roots,
p cos A − q sin A = ± √(p² + q² − r²)
∴ Proved.

Question 32: If sin A + 2 cos A = 1, prove that cos A − 2 sin A = 2.

Given sin A + 2 cos A = 1
or, 2 cos A = 1 − sin A
or, 2 cos A(1 + sin A) = (1 − sin A)(1 + sin A)
or, 2 cos A(1 + sin A) = 1 − sin² A
or, 2 cos A(1 + sin A) = cos² A
or, 2(1 + sin A) = cos A
or, cos A − 2 sin A = 2
∴ Proved.

Question 33: If tan A + sec A = l, prove that sec A = (l² + 1)/(2l).

Given, tan A + sec A = l
Squaring both sides,
(tan A + sec A)² = l²
or, tan² A + sec² A + 2 tan A sec A = l²
Since sec² A − tan² A = 1,
l² = 2 sec² A − 1 + 2 tan A sec A
or, l² + 1 = 2 sec A(sec A + tan A)
Since sec A + tan A = l,
l² + 1 = 2l sec A
or, sec A = (l² + 1)/(2l)
∴ Proved.

Question 34: If sin A + cos A = p and sec A + cosec A = q, prove that p = q(p² − 1)/2.

Given, sin A + cos A = p
Squaring, p² = sin² A + cos² A + 2 sin A cos A
or, p² − 1 = 2 sin A cos A
Also, sec A + cosec A
= 1/cos A + 1/sin A
= (sin A + cos A)/(sin A cos A)
= p/[(p² − 1)/2]
= 2p/(p² − 1)
∴ q = 2p/(p² − 1)
or, p = q(p² − 1)/2
∴ Proved.

Question 35: If A is an acute angle, tan A + sin A = m and tan A − sin A = n, prove that m² − n² = 4√(mn).

Given, m = tan A + sin A and
n = tan A − sin A
Now, m² − n²
= (m + n)(m − n)
= (2 tan A)(2 sin A)
= 4 sin A tan A
Also, mn
= (tan A + sin A)(tan A − sin A)
= tan² A − sin² A
= sin² A(sec² A − 1)
= sin² A tan² A
∴ √(mn) = sin A tan A
Hence, m² − n² = 4√(mn)
∴ Proved.

Question 36: Prove that cos A/(1 − tan A) − sin² A/(cos A − sin A) = sin A + cos A.

LHS = cos A/[1 − (sin A/cos A)] − sin² A/(cos A − sin A)
= cos A/[(cos A − sin A)/cos A] − sin² A/(cos A − sin A)
= cos² A/(cos A − sin A) − sin² A/(cos A − sin A)
= (cos² A − sin² A)/(cos A − sin A)
= [(cos A − sin A)(cos A + sin A)]/(cos A − sin A)
= sin A + cos A
= RHS
∴ Proved.

Question 37: Prove that 2 sec² A − sec⁴ A − 2 cosec² A + cosec⁴ A = cot⁴ A − tan⁴ A.

LHS = (2 sec² A − sec⁴ A) − (2 cosec² A − cosec⁴ A)
= sec² A(2 − sec² A) − cosec² A(2 − cosec² A)
Since sec² A = 1 + tan² A and cosec² A = 1 + cot² A,
= (1 + tan² A)(1 − tan² A) − (1 + cot² A)(1 − cot² A)
= 1 − tan⁴ A − (1 − cot⁴ A)
= cot⁴ A − tan⁴ A
= RHS
∴ Proved.

Question 38: Prove that (1 + cos A − sin² A)/[sin A(1 + cos A)] = cot A.

LHS = [1 + cos A − sin² A]/[sin A(1 + cos A)]
Since 1 − sin² A = cos² A,
= [cos² A + cos A]/[sin A(1 + cos A)]
= [cos A(1 + cos A)]/[sin A(1 + cos A)]
= cos A/sin A
= cot A
= RHS
∴ Proved.

Question 39: Prove that (tan³ A − 1)/(tan A − 1) = sec² A + tan A.

LHS = (tan³ A − 1)/(tan A − 1)
Using a³ − b³ = (a − b)(a² + ab + b²),
= [(tan A − 1)(tan² A + tan A + 1)]/(tan A − 1)
= tan² A + tan A + 1
= sec² A + tan A
= RHS
∴ Proved.

Question 40: Prove that √[(1 − cos A)/(1 + cos A)] = sin A/(1 + cos A).

LHS = √[((1 − cos A)(1 + cos A))/(1 + cos A)²]
= √[(1 − cos² A)/(1 + cos A)²]
= √[sin² A/(1 + cos A)²]
= sin A/(1 + cos A)
= RHS
∴ Proved.

Question 41: Prove that √[(sec A − 1)/(sec A + 1)] + √[(sec A + 1)/(sec A − 1)] = 2 cosec A.

LHS = √[(sec A − 1)/(sec A + 1)] + √[(sec A + 1)/(sec A − 1)]
Taking the two terms together,
= [√(sec A − 1)/√(sec A + 1)] + [√(sec A + 1)/√(sec A − 1)]
= [(sec A − 1) + (sec A + 1)]/√[(sec A − 1)(sec A + 1)]
= 2 sec A/√(sec² A − 1)
Since sec² A − 1 = tan² A,
= 2 sec A/tan A
= 2 cosec A
= RHS
∴ Proved.

Question 42: Prove that (cos A cot A)/(1 − sin A) = 1 + cosec A.

LHS = cos A × (cos A/sin A)/(1 − sin A)
= cos² A/[sin A(1 − sin A)]
Since cos² A = 1 − sin² A,
= (1 − sin² A)/[sin A(1 − sin A)]
= [(1 − sin A)(1 + sin A)]/[sin A(1 − sin A)]
= (1 + sin A)/sin A
= 1 + cosec A
= RHS
∴ Proved.

Question 43: Prove that sec⁴ A − tan⁴ A = 1 + 2 tan² A.

LHS = (sec² A − tan² A)(sec² A + tan² A)
Since sec² A − tan² A = 1,
= sec² A + tan² A
= 1 + tan² A + tan² A
= 1 + 2 tan² A
= RHS
∴ Proved.

Question 44: Prove that sin A/(cot A + cosec A) = 2 + sin A/(cot A − cosec A).

LHS = sin A/[(cos A + 1)/sin A]
= sin² A/(1 + cos A)
= 1 − cos A
Now, RHS
= 2 + sin A/[(cos A − 1)/sin A]
= 2 + sin² A/(cos A − 1)
= 2 − sin² A/(1 − cos A)
= 2 − (1 + cos A)
= 1 − cos A
∴ LHS = RHS Proved.

Question 45: Prove that (cosec A − sin A)(sec A − cos A) sec² A = tan A.

LHS = [(1/sin A) − sin A][(1/cos A) − cos A] sec² A
= [(1 − sin² A)/sin A][(1 − cos² A)/cos A] sec² A
= [cos² A/sin A][sin² A/cos A] sec² A
= sin A cos A × 1/cos² A
= sin A/cos A
= tan A
= RHS
∴ Proved.

Question 46: If sin A + cos A = √2 sin(90° − A), show that cot A = √2 + 1.

Given, sin A + cos A = √2 sin(90° − A)
Since sin(90° − A) = cos A,
sin A + cos A = √2 cos A
or, sin A = (√2 − 1) cos A
or, tan A = √2 − 1
Hence,
cot A = 1/(√2 − 1)
= (√2 + 1)/(2 − 1)
= √2 + 1
∴ Proved.

Question 47: If 7 sin² A + 3 cos² A = 4, 0° ≤ A ≤ 90°, find the value of A.

Given, 7 sin² A + 3 cos² A = 4
Using sin² A = 1 − cos² A,
7(1 − cos² A) + 3 cos² A = 4
or, 7 − 4 cos² A = 4
or, cos² A = 3/4
Since 0° ≤ A ≤ 90°,
cos A = √3/2
∴ A = 30°

Question 48: If x = h + a cos A and y = k + a sin A, prove that (x − h)² + (y − k)² = a².

Given, x = h + a cos A and y = k + a sin A
or, x − h = a cos A and y − k = a sin A
Squaring,
(x − h)² = a² cos² A and (y − k)² = a² sin² A
Adding, (x − h)² + (y − k)²
= a²(cos² A + sin² A)
= a²
∴ Proved.

Question 48: If x = h + a cos A and y = k + a sin A, prove that (x − h)² + (y − k)² = a².

Question 49: Evaluate: [sin² 22° + sin² 68°]/[cos² 22° + cos² 68°] + sin² 63° + sin 63° sin 27°.

We have,
[(sin² 22° + sin² 68°) / (cos² 22° + cos² 68°)] + sin² 63° + cos 63° sin 27°
We know that sin A = cos (90° − A) and
cos A = sin (90° − A)
Therefore,
[(sin² 22° + sin² (90° − 22°)) / (cos² 22° + cos² (90° − 22°))] + sin² 63° + cos 63° sin (90° − 63°)
= [(sin² 22° + cos² 22°) / (cos² 22° + sin² 22°)] + sin² 63° + cos 63° cos 63°
= 1/1 + sin² 63° + cos² 63°
= 1 + 1
= 2
∴ The value of the expression is 2.

Question 50: Prove that sec⁴ A(1 − sin⁴ A) − 2tan² A = 1

L.H.S. = sec⁴ A(1 − sin⁴ A) − 2tan² A
Using the identity a² − b² = (a − b)(a + b),
= sec⁴ A(1 − sin² A)(1 + sin² A) − 2tan² A
Since 1 − sin² A = cos² A,
= sec⁴ A · cos² A(1 + sin² A) − 2tan² A
Since sec² A = 1/cos² A,
= sec² A(1 + sin² A) − 2tan² A
= sec² A + sec² A sin² A − 2tan² A
Since sec² A sin² A = tan² A,
= sec² A + tan² A − 2tan² A
= sec² A − tan² A
= 1
∴ L.H.S. = R.H.S.
Hence proved.

Question 51: Prove that tan² A/(tan² A − 1) + cosec² A/(sec² A − cosec² A) = 1/(sin² A − cos² A).

We have,
tan² A/(tan² A − 1) + cosec² A/(sec² A − cosec² A)
Using tan A = sin A/cos A, cosec A = 1/sin A and sec A = 1/cos A,
= [sin² A/cos² A] / [(sin² A/cos² A) − 1] + [1/sin² A] / [(1/cos² A) − (1/sin² A)]
= [sin² A/cos² A] / [(sin² A − cos² A)/cos² A] + [1/sin² A] / [(sin² A − cos² A)/(sin² A cos² A)]
= sin² A/(sin² A − cos² A) + cos² A/(sin² A − cos² A)
= (sin² A + cos² A)/(sin² A − cos² A)
Using the identity sin² A + cos² A = 1,
= 1/(sin² A − cos² A)
∴ L.H.S. = R.H.S.

Question 51: Prove that tan² A/(tan² A − 1) + cosec² A/(sec² A − cosec² A) = 1/(sin² A − cos² A).

Question 52: Prove that 1/(sin A + cos A + 1) + 1/(sin A + cos A − 1) = sec A + cosec A.

We have, 1/(sin A + cos A + 1) + 1/(sin A + cos A − 1)
Taking the L.C.M.,
= [(sin A + cos A − 1) + (sin A + cos A + 1)] / [(sin A + cos A + 1)(sin A + cos A − 1)]
= [2(sin A + cos A)] / [(sin A + cos A)² − 1]
Using (sin A + cos A)² = sin² A + cos² A + 2 sin A cos A,
= [2(sin A + cos A)] / [sin² A + cos² A + 2 sin A cos A − 1]
Using the identity sin² A + cos² A = 1,
= [2(sin A + cos A)] / [2 sin A cos A]
= (sin A + cos A)/(sin A cos A)
= sin A/(sin A cos A) + cos A/(sin A cos A)
= 1/cos A + 1/sin A
= sec A + cosec A
∴ L.H.S. = R.H.S.
Hence proved.

Question 53: Prove that 2(sin⁶ A + cos⁶ A) − 3(sin⁴ A + cos⁴ A) + 1 = 0.

We have, 2(sin⁶ A + cos⁶ A) − 3(sin⁴ A + cos⁴ A) + 1
Using sin² A + cos² A = 1,
sin⁶ A + cos⁶ A = (sin² A + cos² A)³ − 3 sin² A cos² A(sin² A + cos² A)
= 1 − 3 sin² A cos² A
Also, sin⁴ A + cos⁴ A = (sin² A + cos² A)² − 2 sin² A cos² A
= 1 − 2 sin² A cos² A
Therefore,
2(sin⁶ A + cos⁶ A) − 3(sin⁴ A + cos⁴ A) + 1
= 2(1 − 3 sin² A cos² A) − 3(1 − 2 sin² A cos² A) + 1
= 2 − 6 sin² A cos² A − 3 + 6 sin² A cos² A + 1
= 0
∴ Hence proved.

Question 54: If cot A + cos A = j and cot A − cos A = k, prove that (j² − k²)² = 16jk.

We have, j = cot A + cos A and
k = cot A − cos A
Therefore, j² − k² = (j + k)(j − k)
= [(cot A + cos A) + (cot A − cos A)] [(cot A + cos A) − (cot A − cos A)]
= (2 cot A)(2 cos A)
= 4 cot A cos A
= 4 cos² A/sin A
Therefore, (j² − k²)² = 16 cos⁴ A/sin² A
Now, jk = (cot A + cos A)(cot A − cos A)
= cot² A − cos² A
= cos² A/sin² A − cos² A
= [cos² A − sin² A cos² A]/sin² A
= cos² A(1 − sin² A)/sin² A
Using the identity 1 − sin² A = cos² A,
= cos⁴ A/sin² A
Therefore, (j² − k²)² = 16 cos⁴ A/sin² A
= 16jk
∴ (j² − k²)² = 16jk
Hence proved.

Question 55: When 0° < A < 90°, solve tan² A = 3(sec A − 1).[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row admin_label="SE_Answer" _builder_version="4.27.6" _module_preset="default" custom_margin="-50px||||false|false" global_colors_info="{}"][et_pb_column type="4_4" _builder_version="4.27.6" _module_preset="default" global_colors_info="{}"][et_pb_text _builder_version="4.27.6" _module_preset="default" text_font="|300|||||||" text_font_size="18px" custom_padding="0.5%||0.5%|1%|false|false" hover_enabled="0" text_font_size_tablet="17px" text_font_size_phone="16px" text_font_size_last_edited="on|phone" border_radii="on|3px|3px|3px|3px" border_width_all="1px" border_color_all="#d97706" border_width_all_tablet="1px" border_width_all_phone="2px" border_width_all_last_edited="on|phone" global_colors_info="{}" sticky_enabled="0"]We have, tan² A = 3(sec A − 1) Using the identity tan² A = sec² A − 1, sec² A − 1 = 3(sec A − 1) = sec² A − 1 = 3sec A − 3 Rearranging, sec² A − 3sec A + 2 = 0 Factoring, (sec A − 1)(sec A − 2) = 0 Therefore, sec A = 1 or sec A = 2 Since 0° < A < 90°, we have sec A > 1.
Hence, sec A = 2
⇒ cos A = 1/2
⇒ A = 60°
∴ The value of A is 60°.

What are Trigonometric Identities?

Trigonometric Identities are equations involving trigonometric ratios that are true for all angles for which the expressions are defined. Unlike ordinary equations, which are true only for specific values of the variable, trigonometric identities remain valid regardless of the value of the angle, provided the ratios involved are defined. These identities establish relationships between the six trigonometric ratios—sin θ, cos θ, tan θ, cosec θ, sec θ, and cot θ

Reciprocal Identities

The reciprocal identities express one trigonometric ratio in terms of the reciprocal of another. These identities follow directly from the definitions of the six trigonometric ratios.

cosec θ = 1/sin θ
sin θ = 1/cosec θ
sec θ = 1/cos θ
cos θ = 1/sec θ
cot θ = 1/tan θ
tan θ = 1/cot θ

These identities are useful for converting one trigonometric ratio into another and for simplifying expressions.

Quotient Identities

The quotient identities relate tan θ and cot θ to sin θ and cos θ.
Formulae:

tan θ = sin θ/cos θ
cot θ = cos θ/sin θ

Pythagorean Identities

The Pythagorean identities are the most important trigonometric identities. They are derived from the Pythagorean theorem and establish relationships among sin θ, cos θ, tan θ, sec θ, cot θ and cosec θ.

sin² θ + cos² θ = 1
1 + tan² θ = sec² θ
1 + cot² θ = cosec² θ

Difference Between an Identity and an Equation

A trigonometric identity is true for every value of θ for which the expressions are defined.
Example:
sin² θ + cos² θ = 1
This identity is always true.
A trigonometric equation is true only for specific values of θ.
Example:
sin θ = ½
This equation is true only for particular angles such as 30° and 150° (within 0° to 360°).

Summary of Trigonometric Identities

Identity	Formula
Reciprocal Identity (sin θ)	cosec θ = 1/sin θ, sin θ = 1/cosec θ
Reciprocal Identity (cos θ)	sec θ = 1/cos θ, cos θ = 1/sec θ
Reciprocal Identity (tan θ)	cot θ = 1/tan θ, tan θ = 1/cot θ
Quotient Identity	tan θ = sin θ/cos θ, cot θ = cos θ/sin θ
Pythagorean Identity	sin² θ + cos² θ = 1
Pythagorean Identity	1 + tan² θ = sec² θ
Pythagorean Identity	1 + cot² θ = cosec² θ
Rearranged Form	sin² θ = 1 − cos² θ, cos² θ = 1 − sin² θ
Rearranged Form	tan² θ = sec² θ − 1, sec² θ = 1 + tan² θ
Rearranged Form	cot² θ = cosec² θ − 1, cosec² θ = 1 + cot² θ

Common Mistakes

Confusing Identities with Equations:
Students often treat a trigonometric identity as if it is true only for specific values of θ. Remember that an identity is true for every value of θ for which the expressions are defined, whereas an equation is true only for particular values of θ.
Using the Wrong Identity:
Students sometimes apply an incorrect identity while simplifying an expression. Always identify whether the problem requires a reciprocal, quotient or Pythagorean identity before substituting.
Formulae:
sin² θ + cos² θ = 1
1 + tan² θ = sec² θ
1 + cot² θ = cosec² θ
Forgetting the Rearranged Forms:
Students often remember the basic Pythagorean identities but forget their rearranged forms. These are frequently needed to find an unknown trigonometric ratio.
Formulae:
sin² θ = 1 − cos² θ
cos² θ = 1 − sin² θ
tan² θ = sec² θ − 1
cot² θ = cosec² θ − 1
Confusing Reciprocal and Quotient Identities:
Students sometimes assume that tan θ = cos θ/sin θ or cot θ = sin θ/cos θ. Remember that tan θ = sin θ/cos θ and cot θ = cos θ/sin θ, while sec θ, cosec θ and cot θ are reciprocal ratios.
Not Simplifying Step by Step:
Students often substitute several identities at once, making algebraic mistakes. Replace only one identity at a time, simplify the expression after each substitution, and then apply the next identity if required. This reduces errors and makes the solution easier to verify.

Practice Questions

Question 1: If sin θ = 5/13 and θ is an acute angle, find the value of (1 + tan² θ)(1 − sin² θ)

Question 2: Simplify (sec² θ − tan² θ) + (sin² θ + cos² θ) + (cosec² θ − cot² θ)

Question 3: If tan θ = 12/5 and θ is an acute angle, find the value of (sec² θ − 1)/(1 − cos² θ)

Question 4: Simplify [(1 + tan² θ)(1 − sin² θ)] + [(1 + cot² θ)(1 − cos² θ)]

Question 5: If sin θ = 15/17 and θ is an acute angle, find the value of [(1 − cos² θ) + (sec² θ − 1)] ÷ tan² θ