Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8

Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8

Extra Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

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Introduction to Trigonometry Class 10 Extra Questions Very Short Answer Type

Question 1.
From the given figure, find the value of x:

Answer:
In the given fig., only one side is known which is hypotenuse and side to be evaluated is BC which is perpendicular with reference to given angle ZA = 30°.
∴ sin 30° = \(\frac{x}{15}\) ⇒ x = 15 sin 30°
x = 15 × \(\frac{1}{2}\) = 7\(\frac{1}{2}\) cm

Question 2.
If tan A = cot B, prove that A + B = 90°.
Answer:
∵ tan A = cot B
∴ tan A = tan (90°-B)
⇒ A = 90°-B
⇒ A + B = 90°. [Hence proved]

Question 3.
Evaluate cos 60° sin 30° + sin 60° cos 30°. (AI CBSE 2010)
Answer:
We have: cos 60° . sin 30° + sin 60°. cos 30°

Question 4.
If cos A = \(\frac{2}{5}\), find the value of 4 + 4 tan²A [CBSE Sample Paper-2017]
Answer:
sec A = \(\frac{1}{\cos \mathrm{A}}=\frac{5}{2}\)
4 + 4 tan² A = 4 (1 + tan² A)
= 4 (sec² A) = 4 \(\left(\frac{5}{2}\right)^2\) = 25

Question 5.
If 3x = sec θ and 9\(\left(x^2-\frac{1}{x^2}\right)\) = tan θ, then find \(\).
Answer:

Question 6.
What is the value of (cos² 67° – sin² 23°)?
Answer:
∵ 67° + 23° = 90°
∴ cos² 67° – sin² 23° = cos² (90° – 23°) – sin²23°
= sin² 23° – sin² 23°
= 0

Question 7.
Find the value of sin 38° – cos 52°? [CBSE 2016]
Answer:
sin 38° – cos 52°
= sin38° – cos (90° – 38°) [∵ cos (90° – θ = sin θ]
= sin 38° – sin 38°
= 0

Question 8.
If 7 sin² θ + 3 cos² θ = 4, then find the value of tan θ.
Answer:
7 sin² θ + 3 cos² θ = 4
Dividing both sides by cos² θ
\(\frac{7 \sin ^2 \theta}{\cos ^2 \theta}\) + 3 = \(\frac{4}{\cos ^2 \theta}\)
7 tan² θ + 3 = 4 sec² θ
= 4(1 + tan² θ)
⇒ 3 tan² θ = 1
tan θ = \(\frac{1}{\sqrt{3}}\)

Question 9.
Write the value of \(\frac{5}{\cot ^2 \theta}-\frac{5}{\cos ^2 \theta}\)
Answer:
\(\frac{5}{\cot ^2 \theta}-\frac{5}{\cos ^2 \theta}\) = 5 tan² θ – 5 sec² θ
= 5(tan² θ – sec² θ)
= – 5(sec² θ – tan² θ)
= – 5(1)
= – 5

Question 10.
Given that tan θ = \(\frac{1}{\sqrt{3}}\) then find the value of \(\frac{{cosec}^2 \theta-\cot ^2 \theta}{{cosec}^2 \theta+\sec ^2 \theta}\).
Answer:

Question 11.
If tan A = cot (A + 10), find A. [CBSE Delhi 2016]
Answer:
tan A = cot (A + 10)
⇒ cot (90° – A) = cot (A + 10) [ ∵ cot (90 – θ) = tan θ]
⇒ 90° – A = A + 10
⇒ 90° – 10 = A + A
⇒ 80° = 2A
\(\frac{80^{\circ}}{2}\) = A
∴ A = 40°

Question 12.
Express sin 67° + cos 75° in terms of t-ratios of angles between 0° and 45°.
Answer:
∵ 67 = 90 – 23 and 75 = 90 – 15
∴ sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°.

Question 13.
Find the value of sin (60° + θ) – cos (30° – θ).
Answer:
Observe (60° + θ) = 90° – (30° – θ)
∴ sin (60°+ θ) – cos (30° – θ)
= sin {90° – (30° – θ)} – cos (30° – θ)
= cos (30° – θ) – cos (30° – θ)
= 0

Introduction to Trigonometry Class 10 Extra Questions Short Answer Type-1

Question 1.
Express cot 85° + cos 75° in terms of trigonometric ratio of angles between 0° and 45°.
Answer:
We have
cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
[∵ cot (90° – 0) = tan θ
and cos (90° – θ) = sin θ]
which is the required result.

Question 2.
Write the simplest value of \(\frac{\cos ^2 \theta}{\sin \theta}\) + sin θ
Answer:
\(\frac{\cos ^2 \theta+\sin ^2 \theta}{\sin \theta}\) = \(\frac{1}{\sin \theta}\) = cosec θ

Question 3.
If sin 3θ = cos (θ – 6)° and 30 and (θ – 6)° are acute angles, find the value of θ.
Answer:
We have:
sin 30 = cos (θ – 6)°
= sin [90°- (θ – 6)°]
[∵ sin (90° – θ) = cos θ]
⇒ 3θ = 90° – (θ – 6)°
⇒ 3θ = 90 – θ + 6
⇒ 3θ + θ = 96
⇒ 4θ = 96
⇒ θ = \(\frac{96}{4}\) = 24
Thus θ = 24°.

Question 4.
Prove the following identity:
\(\left(1+\frac{1}{\tan ^2 A}\right)\left(1+\frac{1}{\cot ^2 A}\right)\) = \(\frac{1}{\cos ^2 A-\cos ^4 A}\) [CBSE 2016]
Answer:
As, 1 + cot² A = cosec² A
1 + tan² A = sec² A
sin² A + cos² A = 1

L.H.S = R.H.S.
Hence Proved.

Question 5.
If tan θ = cot (30° + θ), find the value of θ.
Answer:
We have:
tan θ = cot (30° + θ)
= tan [90° – (30° + θ)]
= tan [90° – 30°- θ]
= tan (60° – θ)
⇒ 0 = 60° – θ
⇒ θ + θ = 60°
⇒ 20 = 60°
⇒ θ = \(\frac{60^{\circ}}{2}\) = 30°
∴ θ = 30°.

Question 6.
If 3 cot A = 4, find the value of \(\frac{{cosec}^2 A+1}{{cosec}^2 A-1}\)
Answer:

Question 7.
Without using trigonometric tables, prove that \(\frac{\cos ^2 49^{\circ}+\cos ^2\left(90^{\circ}-49^{\circ}\right)}{\sin ^2 31^{\circ}+\sin ^2\left(90^{\circ}-31^{\circ}\right)}\) + 2 tan 35° tan 55° = 3.
Answer:

Question 8.
Prove that: \(\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}\) = 2 cosec θ
Answer:

Question 9.
Show that: tan 10° tan 15° tan 75° tan 80° = 1
Answer:
We have:
LHS = tan 10° tan 15° tan 75° tan 80°
= tan (90° – 80°) tan 15° tan (90° -15°) tan 80°
= cot 80° tan 15° cot 15° tan 80°
= (cot 80° × tan 80°) × (tan 15° × cot 15°)
= 1 × 1 = 1 = RHS

Question 10.
Evaluate cosec (65° + θ) – sec (25° – θ) – tan (55° – θ) + cot (35° + θ). [C.B.S.E. 2000]
Answer:
Here observe that:
(65° + θ) + (25° – θ) = 90°
⇒ 65° + θ = 90° – (25° – θ)
(55° – θ) + (35° + θ) = 90°
⇒ 35° + θ = 90° – (55° – θ)
∴ Given expression
cosec [9θ° – (25 – θ)] – sec (25° – θ) – tan (55° – θ) + cot [9θ° – (55° – θ)]
= sec (25° – θ) – sec (25° – θ) – tan (55° – θ) + tan (55° – θ)
[∵ cosec (9θ° – θ) = sec θ cot (9θ° – θ) = tan θ
= 0 – 0
= θ

Question 11.
Prove the following identities:
(i) (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan² θ + cot² θ
(ii) (sin θ + sec θ)² + (cos θ + cosec θ)² = (1 + sec θ cosec θ)² [CBSE 2001C]
Answer:
(i) LHS = (sin θ + cosec θ)² + (cos θ + sec θ)²
= (sin² θ + cosec² θ + 2 sin θ cosec θ) + (cos² θ + sec² θ + 2 cos θ sec θ)
= (sin² θ + cos² θ) + (cosec² θ + sec² θ) + 2(sin θ cosec θ + cos θ sec θ)
= 1 + (1 + cot² θ + 1 + tan² θ) + 2(1 + 1)
= 3 + cot² θ + tan² θ + 4
[∵ sin θ cosec θ = 1 = cos θ sec θ]
= (3 + 4) + tan² θ + cot² θ
= 7 + tan² θ + cot² θ
= RHS
Hence Proved.

(ii) LHS = (sin θ + sec θ)² + (cos θ + cosec θ)²
= (sin² θ + sec² θ + 2 sin θ sec θ) + (cos² θ + cosec² θ + 2 cos θ cosec θ)
= (sin² θ + cos² θ) + (sec² θ + cosec² θ) + 2 sin θ sec θ + 2 cos θ cosec θ
= 1 + (1 + tan² θ + 1 + cot² θ) + \(\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)\)
= 1 + (tan² θ + cot² θ + 2 tan θ cot θ) + 2\(\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}\right)\)
= 1 + (tan θ + cot θ)² + 2 cosec θ sec θ
= 1 + \(\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)^2\) + 2 cosec θ sec θ
= 1 + sec² θ cosec² θ + 2 cosec θ sec θ
= (1 + sec θ cosec θ)²
= RHS
Hence proved

Question. 12.
If cos θ + sin θ = √2 cos θ, show that cos θ – sin θ = √2 sin θ.
Answer:
Given, cos θ + sin θ = √2 cos θ ……. (i)
Squaring both sides
cos² θ + sin² θ + 2 sin θ cos θ = 2 cos² θ
⇒ cos² θ – sin² θ = 2 sin θ cos θ
⇒ (cos θ – sin θ) (cos θ + sin θ) = 2 sin θ cos θ
⇒ (cos θ – sin θ) (√2 cos θ) = 2 sin θ cos θ (Using (i))
⇒ cos θ – sin θ = \(\frac{2 \sin \theta \cos \theta}{\sqrt{2} \cos \theta}\)
⇒ cos θ – sin θ = √2 sin θ

Introduction to Trigonometry Class 10 Extra Questions Short Answer Type-2

Question 1
If 7 sin² A + 3 cos² A = 4, show that tan A = \(\frac{1}{\sqrt{3}}\). [CBSE Delhi 2016]
Answer:
7 sin² A + 3 cos² A = 4
Dividing both sides by cos² A

⇒ 7 tan² A + 3 = 4 sec² A
⇒ 7 tan² A + 3 = 4(1 + tan²A) [∵sec²θ = 1 + tan² θ]
⇒ 7 tan² A + 3 – 4(1 + tan² A) = 0
⇒ 7 tan² A + 3 – 4 – 4 tan² A = 0
⇒ 3 tan² A – 1 = 0
⇒ 3 tan² A = 1
⇒ tan² A = \(\frac{1}{3}\)
∴ tan A = \(\frac{1}{\sqrt{3}}\)
∴ Hence proved.

Question 2.
Prove (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A. [CBSE Delhi 2016]
Answer:
We take,
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A cosec A) + (cos² A + sec² A + 2 cos A sec A)
= sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A
= (sin² A + cos² A) + (cosec² A + sec² A) + 2 (sin A cosec A + cos A sec A)
= 1 + (1 + cot² A + 1 + tan² A) + 2(1 + 1)
= 3 + cot² A + tan² A + 4
[∵ sin θ cosec θ = 1
cos θ sec θ = 1]
= (3 + 4) + (cot² A + tan² A)
= 7 + tan² A + cot² A
∴ = R.H.S.
∴ Hence proved.

Question 3.
If sin θ + cos θ = √2, then evaluate tan θ + cot θ [CBSE Sample Paper-2017]
Answer:
sin θ + sos θ = √2
⇒ (sin θ + cos θ)² = (√2)²
⇒ sin² θ + cos² θ + 2 sin θ cos θ = 2
⇒ 1 + 2 sin θ cos θ = 2
⇒ sin θ cos θ = \(\frac{1}{2}\) ………… (i)
we know, sin² θ + cos² θ = 1 ……………… (ii)
Dividing (ii) by (i) wet get

⇒ tan θ + cot θ = 2

Question 4.
If sec 3A = cosec (A – 10°) where 3A is an acute angle, find the value of A.
Answer:
sec 3A = cosec (A -10°)
sec 3A = sec {90° – (A – 10)}
[∵sec (90° – θ) = cosec θ]
⇒ sec 3A = sec (100° – A)
⇒ 3A = 100° – A
⇒ 3A + A = 100°
4A = 100°
A = \(\frac{100^{\circ}}{4}\) = 25°

Question 5.
Prove that:
sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A [C.B.S.E. 2000 (F)]
Answer:
LHS = sin A (1 + tan A) + cos A (1 + cot A)

Question 6.
Evaluate
\(\frac{{cosec}^2 63^{\circ}+\tan ^2 24^{\circ}}{\cot ^2 66^{\circ}+\sec ^2 27^{\circ}}\) + \(\frac{\sin ^2 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}+\sin 27^{\circ} \sec 63^{\circ}}{2\left({cosec}^2 65^{\circ}-\tan ^2 25^{\circ}\right)}\) [CBSE Sample Paper-2017]
Answer:

Question .7.
If cos² θ – sin² θ = tan² Φ, prove that
cos Φ = \(\frac{1}{\sqrt{2} \cos \theta}\)
Answer:
Consider cos² θ – sin² θ = tan² Φ
⇒ cos² θ – (1 – cos² θ) = sec² Φ – 1
[∵ sin² θ + cos² θ = 1, sec² Φ = 1 + tan² Φ]
⇒ 2cos² θ – 1 = sec²Φ – 1
⇒ 2 cos² θ = sec² Φ
Taking square root, we get
√2 cos θ = sec Φ
⇒ √2 cos θ = \(\frac{1}{\cos \phi}\)
⇒ cos Φ = \(\frac{1}{\sqrt{2} \cos \theta}\)

Question 8.
If tan 2A = cot (A -18°), where 2A is an acute angle, find the value of A. [CBSE 2018]
Answer:
We know that tan θ = cot (90 – θ)
∴ tan 2A = cot (A – 18°)
⇒ cot (90 – 2A) = cot (A – 18°)
⇒ 90° – 2A = A – 18°
⇒ 3A = 108°
A = 36°

Question 9.
Without using trigonometric tables, evaluate the following:
\(\frac{\sec 39^{\circ}}{{cosec} 51^{\circ}}+\frac{2}{\sqrt{3}}\) = tan 17° tan 38° tan 60° tan 52° tan 73° – 3 (sin² 31° + sin² 59°) [CBSE 2006(C)]
Answer:

= 1 + 2 (tan 17° cot 17°) (tan 38° cot 38°) – 3 × 1
= 1 + 2 × 1 × 1 – 3 = 3 – 3 = 0 [∵ tan θ cot θ = 1]

Question 10.
Prove that: a² + b² = x² + y² when a cos θ – b sin θ = x and a sin θ + b cos θ = y.
Answer:
RHS = x² + y²
= [a cos θ – b sin θ]² + [a sin θ + b cos θ]²
= a² cos² θ + b² sin² θ – 2ab sin θ cos θ + a² sin² θ + b² cos² θ + 2ab sin θ cos θ
= a² cos² θ + b² sin² θ + a² sin² θ + b² cos² θ
= a² [cos² θ + sin² θ] + b² [sin² θ + cos² θ]
= a²[1] + b²[1] [ ∵ sin² θ + cos² θ = 1]
= a² + b²
= LHS
Hence proved

Introduction to Trigonometry Class 10 Extra Questions Long Answer Type 1

Question 1.
If 3 tan A = 4, check whether = \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\) = cos²A – sin²A or not? [CBSE Delhi 2016]
Answer:

Now, take R.H.S. = cos² A – sin² A
Firstly, we find the value of cos A and sin A.
By using Pythagoras Theorem.

(AC)² = (AB)² + (BC)²
⇒ (AC)² = (3)² + (4)² = 9 + 16 = 25
AC = √25
∴ AC = 5

∴ cos² A – sin² A = \(\frac{-7}{25}\)
∴ Hence, L.H.S. = R.H.S.
Hence proved.

Question 2.
If x = a sin θ and y = b tan θ, then prove that \(\frac{a^2}{x^2}-\frac{b^2}{y^2}\) = 1.
Answer:

Question 3.
Prove that \(\frac{2}{\cos ^2 \theta}-\frac{1}{\cos ^4 \theta}-\frac{2}{\sin ^2 \theta}+\frac{1}{\sin ^4 \theta}\) = cot⁴ θ – tan⁴ θ
Answer:

= 2 sec² θ – sec⁴ θ – 2 cosec² θ + cosec⁴ θ
= 2(1 + tan² θ) – (1 + tan² θ)² – 2(1 + cot² θ) + (1 + cot² θ)²
= (1 + tan² θ) [2 – (1 + tan² θ)] – (1 + cot² θ) [2 – (1 + cot² θ)]
= (1 + tan² θ) (1 – tan² θ) – (1 + cot² θ) (1 – cot² θ)
= 1 – tan⁴ θ – (1 – cot⁴ θ) = cot⁴ θ – tan⁴ θ
= RHS
Hence proved

Question 4.
Evaluate sec 41° sin 49° + cos 49° cosec 41° + \(\frac{2}{\sqrt{3}}\) = tan 20° tan 60° tan 70° – 3 (cos² 45° – sin² 90°).
Answer:
The given expression is
sec 41° sin 49° + cos 49° cosec 41° + \(\frac{2}{\sqrt{3}}\) tan 20° tan 60° tan 70° – 3 (cos² 45° – sin² 90°)
= sec 41° × sin (90° – 41°) + cos (90° – 41°) × cosec 41° + \(\frac{2}{\sqrt{3}}\) tan 20° × tan 70° × tan 60° – 3 (cos²45° – (1)²)
= sec 41° × cos 41° + sin 41° × cosec 41°

Question 5.
Determine the value of x such that
2 cosec² 30° + x sin² 60° – \(\frac{3}{4}\) tan² 30° = 10.
Answer:
We have,
2 cosec2² 30° + x sin² 60° – \(\frac{3}{4}\) tan² 30° = 10

Introduction to Trigonometry Class 10 Extra Questions HOTS

Question 1.
Show that cosec² θ + sec² θ can never be less than 2.
Answer:
Let, if possible, cosec² θ + sec² θ < 2
⇒ (1 + cot² θ) + (1 + tan² θ) < 2
⇒ cot² θ + tan² θ + 2 < 2
⇒ cot² θ + tan² θ < θ
But sum of squares of two real numbers is always non negative.
∴ cot² θ + tan² θ < 0 is not possible.
Therefore, our supposition is wrong.
Hence cosec² θ + sec² θ can never be less than 2.

Question 2.
If sin θ + sin²θ + sin³θ = 1, then find the value of cos⁶ θ – 4 cos4 θ + 8 cos² θ.
Answer:
We have, sin θ + sin³ θ = 1- sin² θ
⇒ sin θ (1 + sin² θ) = cos² θ
⇒ sin θ (2 – cos² θ) = cos² θ
Squaring both sides, we get
sin² θ (2 – cos² θ)² = cos⁴ θ
⇒ (1 – cos² θ) (4 + cos⁴ θ – 4 cos² θ) = cos⁴ θ
⇒ 4 + cos⁴ θ – 4 cos² θ – 4 cos² θ – cos⁶ θ + 4 cos⁴ θ = cos⁴ θ
⇒ cos⁶ θ – 4 cos⁴ θ + 8 cos² θ = 4.

Question 3.
If 7 sin² θ + 3 cos² 0 = 4, then, show that tan θ = \(\frac{1}{\sqrt{3}}\), where θ is an acute angle.
Answer:
7 sin² θ + 3 cos² θ = 4
⇒ 7 sin² θ + 3 cos² θ = 4 × 1 = 4 × (sin² θ + cos² θ)
⇒ 3 sin² θ = cos² θ

Question 4.
Prove that: 3 (sin θ – cos θ)⁴ + 6 (sin θ + cos θ)² + 4 (sin⁶ θ + cos⁶ θ) = 13.
Answer:
(sin θ – cos θ)⁴
= [(sin θ – cos θ)²]²
= [sin²θ + cos²θ – 2 sin θ cos θ]²
= [1 – 2 sin θ cos θ]²
= 1 + 4 sin²θ cos²θ – 4 sin θ cos θ Also (sin θ + cos θ)²
= sin² θ + cos² θ + 2 sin θ cos θ and (sin⁶ θ + cos⁶ θ)
= (sin²θ)³ + (cos²θ)³
= (sin²θ + cos²θ) [(sin²θ)² – sin² θ cos² θ + (cos² θ)²]
= (1) [sin⁴ θ – sin² θ cos² θ + cos⁴ θ]
= [(sin²θ + cos²θ)²– 2 sin² θ cos² θ – sin² θ cos² θ]
= 1 – 3 sin² θ cos² θ
∴ LHS
= 3 (1 + 4 sin² θ cos² θ – 4 sin θ cos θ) + 6 (1 + 2 sin θ cos θ) + 4 (1 – 3 sin² θ cos² θ)
= 3 + 6 + 4 = 13
= RHS Hence proved

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