Trigonometry JEE Foundation Important Solved Problems and Examples

Master JEE Foundation-level Trigonometry with expertly solved problems. Build strong basics, improve accuracy, and boost your confidence for JEE, NDA competitive exams and CBSE Class 10 with our detailed step-by-step solutions.

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Prove the Following Problems

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 sin⁴ A – cos⁴ A = 2 sin² A – 1

Solution:

Taking L.H.S,

sin⁴ A – cos⁴ A

= (sin² A)² – (cos² A)²

= (sin² A + cos² A) (sin² A – cos² A)

= sin²A – cos²A

= sin²A – (1 – sin²A) [Since, cos² A = 1 – sin² A]

= 2sin² A – 1

– Hence Proved

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(1 – tan A)² + (1 + tan A)² = 2 sec² A

Solution:

Taking L.H.S,

(1 – tan A)² + (1 + tan A)²

= (1 + tan² A + 2 tan A) + (1 + tan² A – 2 tan A)

= 2 (1 + tan² A)

= 2 sec² A [Since, 1 + tan² A = sec² A]

– Hence Proved

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cosec⁴ A – cosec² A = cot⁴ A + cot² A

Solution:

cosec⁴ A – cosec² A

= cosec² A(cosec² A – 1)

= (1 + cot² A) (1 + cot² A – 1)

= (1 + cot² A) cot² A

= cot⁴ A + cot² A = R.H.S

– Hence Proved

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(cosec A + sin A) (cosec A – sin A) = cot² A + cos² A

Solution:

Taking L.H.S,

(cosec A + sin A) (cosec A – sin A)

= cosec² A – sin² A

= (1 + cot² A) – (1 – cos² A)

= cot² A + cos² A = R.H.S

– Hence Proved

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(sec A – cos A)(sec A + cos A) = sin² A + tan² A

Solution:

Taking L.H.S,

(sec A – cos A)(sec A + cos A)

= (sec² A – cos² A)

= (1 + tan² A) – (1 – sin² A)

= sin² A + tan² A = RHS

– Hence Proved

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(cos A + sin A)² + (cosA – sin A)² = 2

Solution:

Taking L.H.S,

(cos A + sin A)² + (cosA – sin A)²

= cos2 A + sin2 A + 2cos A sin A + cos2 A – 2cosA.sinA

= 2 (cos² A + sin² A) = 2 = R.H.S

– Hence Proved

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(sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

Solution:

Taking LHS,

(sin A + cosec A)² + (cos A + sec A)²

= sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2cos A sec A

= (sin² A + cos² A ) + cosec² A + sec² A + 2 + 2

= 1 + cosec² A + sec² A + 4

= 5 + (1 + cot² A) + (1 + tan² A)

= 7 + tan² A + cot² A = RHS

– Hence Proved

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sec² A. cosec² A = tan² A + cot² A + 2

Solution:

Taking,

RHS = tan² A + cot² A + 2 = tan² A + cot² A + 2 tan A. cot A

= (tan A + cot A)² = (sin A/cos A + cos A/ sin A)²

= (sin2 A + cos2 A/ sin A.cos A)² = 1/ cos² A. sin² A

= sec² A. cosec² A = LHS

– Hence Proved

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 cot² A – 1/sin² A + 1 = 0

Solution:

cot² A – 1/sin² A + 1 = 0

LHS = cot² A – 1/sin² A + 1

We know that

1/sin A = cosec A

= cot² A – cosec² A + 1

= (1 + cot² A) – cosec² A

We know that 1 + cot² A = cosec² A

= cosec² A – cosec² A

= 0

= RHS

Therefore, LHS = RHS.

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1/(1 + tan² A) + 1/(1 + cot² A) = 1

Solution:

LHS = 1/(1 + tan² A) + 1/(1 + cot² A)

We know that

sec² A – tan² A = 1

sec² A = 1 + tan² A

cosec² A – cot² A = 1

cosec² A = 1 + cot² A

So we get

= 1/sec² A + 1/cosec² A

Here 1/sec A = cos A and 1/cosec A = sin A

= cos² A + sin² A

= 1

= RHS

Therefore, LHS = RHS.

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If sin A + cosec A = 2; Find the value of sin² A + cosec² A.

Solution:

Given, sin A + cosec A = 2

On squaring on both sides, we have

(sin A + cosec A)² = 2²

sin² A + cosec² A + 2sinA. cosec A = 4

sin² A + cosec² A + 2 = 4 [As sin A. cosec A = sin A x 1/sin A = 1]

sin² A + cosec² A = 4 – 2 = 2

Hence, the value of (sin² A + cosec² A) is 2

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If x cos A + y sin A = m and x sin A – y cos A = n, then prove that: x² + y² = m² + n² 

Solution:

Taking RHS,

m² + n²

= (x cos A + y sin A)² + (x sin A – y cos A)²

= x² cos² A + y² sin² A + 2xy cos A sin A + x² sin² A + y² cos² A – 2xy sin A cos A

= x² (cos² A + sin² A) + y² (sin² A + cos² A)

= x² + y² [Since, cos² A + sin² A = 1]

= RHS

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If m = a sec A + b tan A and n = a tan A + b sec A, prove that m² – n² = a² – b²

Solution:

Taking LHS,

m² – n²

= (a sec A + b tan A)² – (a tan A + b sec A)²

= a² sec² A + b² tan² A + 2 ab sec A tan A – a² tan² A – b² sec² A – 2ab tan A sec A

= a² (sec² A – tan² A) + b² (tan² A – sec² A)

= a² (1) + b² (-1) [Since, sec² A – tan² A = 1]

= a² – b²

= RHS

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

Solution:

It is given that

sin θ + cosec θ = 2

sin θ + 1/sin θ = 2

By further calculation

sin² θ + 1 = 2 sin θ

sin² θ – 2 sin θ + 1 = 0

So we get

(sin θ – 1)² = 0

sin θ – 1 = 0

sin θ = 1

Here

sin² θ + cosec² θ = sin² θ + 1/sin² θ

Substituting the values

= 1² + 1/1²

= 1 + 1/1

= 1 + 1

= 2

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If x = a cos θ + b sin θ and y = a sin θ – b cos θ, prove that x² + y² = a² + b².

Solution:

It is given that

x = a cos θ + b sin θ …. (1)

y = a sin θ – b cos θ …. (2)

By squaring and adding both the equations

x² + y² = (a cos θ + b sin θ)² + (a sin θ – b cos θ)²

Using the formula

(a + b)² = a² + b² + 2ab and (a – b)² = a² + b² – 2ab

= [(a cos θ)² + (b sin θ)² + 2 (a cos θ) (b sin θ)] + [(a sin θ)² + (b cos θ)² – 2 (a sin θ) (b cos θ)]

By further calculation

= a² cos² θ + b² sin² θ + 2 ab sin θ cos θ + a² sin² θ + b² cos² θ – 2 ab sin θ cos θ

= a² cos² θ + b² sin² θ + a² sin² θ + b² cos² θ

So we get

= a² (cos² θ + sin² θ) + b² (sin² θ + cos² θ)

Here sin² θ + cos² θ = 1

= a² (1) + b² (1)

= a² + b²

Therefore, x² + y² = a² + b².

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 If 2 cos θ = √3, prove that 3 sin θ – 4 sin³ θ = 1.

Solution:

It is given that

2 cos θ = √3

cos θ = √3/2

We know that

sin² θ = 1 – cos² θ

Substituting the values

sin² θ = 1 – (√3/2)²

sin² θ = 1 – 3/4

sin² θ = ¼

sin θ = √ 1/4 = 1/2

Consider

LHS = 3 sin θ – 4 sin³ θ

It can be written as

= sin θ (3 – 4 sin² θ)

Substituting the values

= 1/2 (3 – 4 × ¼)

= 1/2 (3 – 1)

= 1/2 × 2

= 1

= RHS

Therefore, proved.

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 If sin θ + cosec θ = 10/3, find the value of sin² θ + cosec² θ.

Solution:

It is given that

sin θ + cosec θ = 10/3

By squaring on both sides

(sin θ + cosec θ)² = (10/3)²

Expanding using formula (a + b)² = a² + b² + 2ab

sin² θ + cosec² θ + 2 sin θ cosec θ = 100/9

We know that sin θ = 1/cosec θ

sin² θ + cosec² θ + 2 sin θ × 1/ sin θ = 100/9

By further calculation

sin² θ + cosec² θ + 2 = 100/9

sin² θ + cosec² θ = 100/9 – 2

Taking LCM

sin² θ + cosec² θ = (100 – 18)/9 = 82/9

So we get

sin² θ + cosec² θ = 82/9

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Prove that : tan A + cot A = sec A cosec A 

Solution:

L.H.S. = tan A + cot A

= sin A/cos A + cos A/sin A

= (sin² A + cos² A)/ (sin A cos A)

= 1/ (sin A cos A)

= sec A cosec A

= R.H.S

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Prove that : (1 + tan A)² + (1 – tan A)² = 2 sec² A

Solution:

L.H.S. = (1 + tan A)² + (1 – tan A)²

= 1 + 2 tan A + tan² A + 1 – 2 tan A + tan² A

= 2 + 2 tan² A

= 2(1 + tan² A) [As 1 + tan² A = sec² A]

= 2 sec² A

= R.H.S.

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Prove that : sec² A + cosec² A = sec² A. cosec² A

Solution:

L.H.S = sec² A + cosec² A

= 1/cos² A + 1/sin² A

= (sin² A + cos² A)/ (sin² A cos² A)

= 1/ (sin² A cos² A)

= sec² A cosec² A = R.H.S

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Prove that : cosec⁴ θ – cosec² θ = cot⁴ θ + cot² θ 

Solution:

L.H.S. = cosec⁴ θ – cosec² θ

= cosec² θ (cosec² θ – 1)

= cosec² θ cot² θ [cosec² θ – 1 = cot² θ]

= (cot² θ + 1) cot² θ

= cot⁴ θ + cot² θ

= R.H.S.

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Prove that : 2 sec² θ – sec⁴ θ – 2 cosec² θ + cosec⁴ θ = cot⁴ θ – tan⁴ θ.

Solution:

L.H.S. = 2 sec² θ – sec⁴ θ – 2 cosec² θ + cosec⁴ θ

= 2 (tan² θ + 1) – (tan² θ + 1)² – 2 (1 + cot² θ) + (1 + cot² θ)²

= 2 tan² θ + 2 – (tan⁴ θ + 2 tan² θ + 1) – 2 – 2 cot² θ + (1 + 2 cot² θ + cot⁴ θ)

= 2 tan² θ + 2 – tan⁴ θ – 2 tan² θ – 1 – 2 – 2 cot² θ + 1 + 2 cot² θ + cot⁴ θ

= cot⁴ θ – tan⁴ θ = R.H.S.

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Prove that : sec⁴ A – tan⁴ A = 1 + 2 tan² A

Solution:

sec⁴ A – tan⁴ A = 1 + 2 tan² A

L.H.S. = sec⁴ A – tan⁴ A

= (sec² A – tan² A) (sec² A + tan² A)

= (1 + tan⁴ A – tan⁴ A) (1 + tan⁴ A + tan⁴ A) [As sec² A = tan⁴ A + 1]

= 1 (1 + 2 tan² A)

= 1 + 2 tan² A = R.H.S.

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 If 7 sin² θ + 3 cos² θ = 4, 0° ≤ θ ≤ 90°, then find the value of θ.

Solution:

Given,

7 sin² θ + 3 cos² θ = 4, 0° ≤ θ ≤ 90°

3 sin² θ + 3 cos² θ + 4 sin² θ = 4

3 (sin² θ + 3 cos² θ) + 4 sin² θ = 4

3 (1) + 4 sin² θ = 4

4 sin² θ = 4 – 3

sin² θ = ¼

Taking square-root on both sides, we get

sin θ = ½

Thus, θ = 30^o

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If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.

Solution:

Given,

sec θ + tan θ = m

sec θ – tan θ = n

Now,

mn = (sec θ + tan θ) (sec θ – tan θ)

= sec² θ – tan² θ = 1

Thus, mn = 1

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 If x = h + a cos θ and y = k + a sin θ, prove that (x – h)² + (y – k)² = a².

Solution:

Given,

x = h + a cos θ

y = k + a sin θ

Now,

x – h = a cos θ

y – k = a sin θ

On squaring and adding, we get

(x – h)² + (y – k)² = a² cos² θ + a² sin² θ

= a² (sin² θ + cos² θ)

= a² (1) [Since, sin² θ + cos² θ = 1]

– Hence proved

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Prove that : sin² θ + cos⁴ θ = cos² θ + sin⁴ θ 

Solution :

sin² θ + cos⁴ θ = cos² θ + sin⁴ θ

L.H.S. = sin² θ + cos⁴ θ

= (1 – cos² θ) + cos⁴ θ

= cos⁴ θ – cos² θ + 1

= cos² θ (cos² θ – 1) + 1

= cos² θ (- sin² θ) + 1

= 1 – sin² θ cos² θ

Now,

R.H.S. = cos² θ + sin⁴ θ

= (1 – sin² θ) + sin⁴ θ

= sin⁴ θ – sin² θ + 1

= sin² θ (sin² θ – 1) + 1

= sin² θ (-cos² θ) + 1

= 1 – sin² θ cos² θ

Hence, L.H.S. = R.H.S.

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Prove that : 2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = 0

Solution:

Given,

2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = 0

L.H.S. = 2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1

= 2 [(sin² θ)³ + (cos² θ)³] – 3 (sin⁴ θ + cos⁴ θ) + 1

= 2 [(sin² θ + cos² θ) (sin⁴ θ + cos⁴ θ – sin² θ cos² θ)] – 3 (sin⁴ θ + cos⁴ θ) + 1

= 2 (sin⁴ θ + cos⁴ θ – sin² θ cos² θ) – 3 (sin⁴ θ + cos⁴ θ) + 1

= 2 sin⁴ θ + 2 cos⁴ θ – 2 sin² θ cos² θ – 3 sin⁴ θ – 3 cos⁴ θ + 1

= 1 – sin⁴ θ – cos⁴ θ – 2 sin² θ cos² θ

= 1 – [sin⁴ θ + cos⁴ θ + 2 sin² θ cos² θ]

= 1 – 1

= 0 = R.H.S.

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When 0° < θ < 90°, solve the following equations: 

(i) 2 cos² θ + sin θ – 2 = 0 

(ii) 3 cos θ = 2 sin² θ 

(iii) sec² θ – 2 tan θ = 0 

(iv) tan² θ = 3 (sec θ – 1).

Solution:

Given, 0° < θ < 90°

(i) 2 cos² θ + sin θ – 2 = 0

2 (1 – sin² θ) + sin θ – 2 = 0

2 – 2 sin² θ + sin θ – 2 = 0

– 2 sin² θ + sin θ = 0

sin θ (1 – 2 sin θ) = 0

So, either sin θ = 0 or 1 – 2 sin θ = 0

If sin θ = 0

⇒ θ = 0^o

And, if 1 – 2 sin θ = 0

sin θ = ½

⇒ θ = 30^o

Thus, θ = 0^o or 30^o

(ii) 3 cos θ = 2 sin² θ

3 cos θ = 2 (1 – cos² θ)

3 cos θ = 2 – 2 cos² θ

2 cos² θ + 3 cos θ – 2 = 0

2 cos² θ + 4 cos θ – cos θ – 2 = 0

2 cos θ (cos θ + 2) – 1(cos θ + 2)

(2 cos θ – 1) (cos θ + 2) = 0

So, either 2 cos θ – 1 = 0 or cos θ + 2 = 0

If 2 cos θ – 1 = 0

cos θ = ½

⇒ θ = 60^o

And, for cos θ + 2 = 0

⇒ cos θ = -2 which is not possible being out of range.

Thus, θ = 60^o

(iii) sec² θ – 2 tan θ = 0

(1 + tan² θ) – 2 tan θ = 0

tan² θ – 2 tan θ + 1 = 0

(tan θ – 1)² = 0

tan θ – 1 = 0

⇒ tan θ = 1

Thus, θ = 45^o

(iv) tan² θ = 3 (sec θ – 1)

(sec² θ – 1) = 3 sec θ – 3

sec² θ – 1 – 3 sec θ + 3 = 0

sec² θ – 3 sec θ + 2 = 0

sec² θ – 2 sec θ – sec θ + 2 = 0

sec θ (sec θ – 2) – 1 (sec θ = 2) = 0

(sec θ – 1) (sec θ – 2) = 0

So, either sec θ – 1 = 0 or sec θ – 2 = 0

If sec θ – 1 = 0

sec θ = 1

⇒ θ = 0^o

And, if sec θ – 2 = 0

sec θ = 2

⇒ θ = 60^o

Thus, θ = 0^o or 60^o

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💡 Do You Know?

• The values of trigonometric ratios are exact for 0°, 30°, 45°, 60°, and 90°.
• These ratios are frequently used in geometry, physics, astronomy, and engineering.
• The identities like sin²θ + cos²θ = 1 help derive all other formulas in trigonometry.

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📘 Quick Summary Table

Angle	sin	cos	tan	sec	cosec	cot
0°	0	1	0	1	–	–
30°	1/2	√3/2	1/√3	2/√3	2	√3
45°	1/√2	1/√2	1	√2	√2	1
60°	√3/2	1/2	√3	2	2/√3	1/√3
90°	1	0	–	–	1	0

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Trigonmetric Identities Table Summary

Identities Name	Identities
Pythagorean Identities	sin2θ + cos2θ = 1
	1 + tan2θ = sec2θ
	1 + cot2θ = cosec2θ
Reciprocal Identities	cosec θ = 1/sin θ
	sec θ = 1/cos θ
	cot θ = 1/tan θ
Quotient Identities	tan θ = sin θ/cos θ
	cot θ = cos θ/sin θ

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