We have provided you with Extra and Important Questions from Class 10 Maths Chapter 8 Introduction to Trigonometry. This Extra and Important Questions will help you to score 100% in your Board Exams. These extra questions will be helpful to revise the important topics and concepts.

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**Introduction to Trigonometry Class 10 Important Questions with Answers Maths Chapter 8**

**Extra Questions for Class 10 Maths Chapter 8 Very Short Answer Type**

**Question:**If sin θ = cos θ, then find the value of 2 tan θ + cos^{2} θ.

**Solution:**

sin θ = cos θ (Given)

It means value of θ = 45°

Now, 2 tan θ + cos^{2} θ = 2 tan 45° + cos^{2} 45°

**Question:**If sin (x – 20)° = cos (3x – 10)°, then find the value of x.

**Solution:**

sin (x – 20)° = cos (3x – 10)°

⇒ cos [90° – (x – 20)°] = cos (3x – 10)°

By comparing the coefficient

90° – x° + 20° = 3x° – 10° = 110° + 10° = 3x° + x°

120° = 4x°

⇒ 120∘/4 = 30°

**Question:**If x = a cos θ, y = b sin θ, then find the value of b^{2}x^{2} + a^{2}y^{2} – a^{2}b^{2}.

**Solution:**Given x = acos θ, y = b sin θ

b^{2}x^{2} + a^{2}y^{2} – a^{2}b^{2} = b^{2}(acos θ)^{2} + a^{2}(b sin θ)^{2} – a^{2}b^{2}

= a^{2}b^{2} cos^{2}θ + a^{2}b^{2} sin^{2} θ – a^{2}b^{2} = a^{2}b^{2} (sin^{2} θ + cos^{2} θ) – a^{2}b^{2}

= a^{2}b^{2} – a^{2}b^{2} = θ (∵ sin^{2} θ + cos^{2} θ = 1)

**Question:**If tan A = cot B, prove that A + B = 90°.

**Solution:**We have

tan A = cot B

⇒ tan A = tan (90° – B)

A = 90° – B

[∵ Both A and B are acute angles]

⇒ A + B = 90°

**Question:**In a ∆ABC, if ∠C = 90°, prove that sin^{2} A + sin^{2} B = 1.

**Solution:**Since ∠C = 90°

∴ ∠A + ∠B = 180° – ∠C = 90°

Now, sin^{2} A + sin^{2} B = sin^{2} A + sin^{2} (90° – A) = sin^{2} A + cos^{2} A = 1

**Question:**If sec 4A = cosec (A – 20°) where 4 A is an acute angle, find the value of A.

**Solution:**

We have

sec 4 A = cosec (A – 20°)

⇒ cosec (90° – 4 A) = cosec (A – 20°)

∴ 90° – 4 A = A – 20°

⇒ 90° + 20° = A + 4 A

⇒ 110° = 5 A

∴ A = 110/5 = 22°

**Question:**If sin A = 3/4, calculate cos A and tan A.

**Solution:**

Let us first draw a right ∆ABC in which ∠C = 90°.

Now, we know that

**Question:**Given 15 cot A = 8, find sin A and sec A.

**Solution:**

Let us first draw a right ∆ABC in which ∠B = 90°.

Now, we have, 15 cot A = 8

**Question:**If sin θ + cos θ = √3 , then prove that tan θ + cot θ = 1.

**Solution:**

sin θ + cos θ = √3

⇒ (sin θ + cos θ)^{2} = 3

⇒ sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 3

⇒ 2 sin cos θ = 2 (∵ sin^{2} θ + cos^{2} θ = 1)

⇒ sin θ. cos θ = 1 = sin^{2} θ + cos^{2} θ

⇒ 1 = tan θ + cot θ = 1

Therefore tan θ + cot θ = 1

**Question:**

**Solution:**

**Extra Questions for Class 10 Maths Chapter 8 Short Answer Type**

**Question:**

**Solution:**

**Question:**

**Solution:**

**Question:**Evaluate: sin 25° cos 65° + cos 25° sin 65°.

**Solution:**sin 25°. cos 65° + cos 25° . sin 65°

= sin (90° – 65°). cos 65° + cos (90° – 65°). sin 65°

= cos 65° . cos 65° + sin 65°. sin 65°

= cos^{2} 65° + sin^{2} 65° = 1.

**Question:**If sin 30 = cos (θ – 6°) where 30 and (θ – 6°) are both acute angles, find the value of θ.

**Solution:**

According to question:

sin 3θ = cos (θ – 6°)

cos (90° – 30) = cos (θ – 6°) [∵ cos (90° – θ ) = sin θ]

90° – 3θ = θ – 6° [comparing the angles)

= 4θ = 90° + 6° = 96°

θ = 96/4 = 24°

Hence, θ = 24°

**Question:**

Find an acute angle θ, when

**Solution:**

On comparing we get

⇒ tan θ = √3

⇒ tan θ = tan 60°

= θ = 60°

**Question:**

Prove that

**Solution:**

n:

**Question:**

Prove that

**Solution:**

We have,

**Question:**

Prove that:

**Solution:**

**Question:**

Prove that:

**Solution:**

**Question:**

Evaluate the following:

**Solution:**

n:

**Extra Questions for Class 10 Maths Chapter 8 Long Answer Type**

**Question:**

Prove that:

**Solution:**

n:

**Question:**Prove that: (sin A + cosec A)^{2} + (cos A + sec A)^{2} = 7 + tan^{2} A + cot^{2}A.

**Solution:**LHS = (sin A + cosec A)^{2} + (cos A + sec A)^{2}

= sin^{2} A + coses^{2} A + 2sin A . cosec A + cos^{2} A + sec^{2} A + 2 cos A . sec A

= (sin^{2} A + coses^{2} A + 2) + (cos^{2} A + sec^{2} A + 2) [sin A. cosec A = 1]

= (sin^{2} A + cos^{2} A) + (coses^{2} A + sec^{2} A) + 4 [cos A. sec A = 1]

= 1 + 1 + cot^{2}A + 1 + tan^{2} A + 4

= 7 + tan^{2} A + cot^{2} A = RHS [∵ 1 + cot^{2}A = coses^{2} A and 1 + tan^{2} A = sec^{2} A]

**Question:**

**Solution:**

**Question:**

**Solution:**

**Question: **Prove that

**Solution:**

**Question: **Evaluate the following:

**Solution:**