Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8 with Solutions

Extra Questions for Class 10 Maths Introduction to Trigonometry with Answers

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.

You can also download Class 10 Maths to help you to revise complete syllabus and score more marks in your examinations.

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

Multiple Choice Questions

Choose the correct option out of four given in each of the following:

Question 1.
The value of 0, for sin2 0 = 1, where 0° < θ < 90° is
(a) 30°
(b) 60°
(c) 45°
(d) 135°
Answer:
(c) 45°

Question 2.
If cos A = \(\frac{4}{5}\), then value of tan A is
(a) \(\frac{3}{5}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{4}{3}\)
(d) \(\frac{5}{3}\)
Answer:
(b) \(\frac{3}{4}\)

Question 3.
In ∆ABC, ∠B = 90°. If AB = 14 cm and AC = 50 cm then tan A equals:
(a) \(\frac{24}{25}\)
(b) \(\frac{24}{7}\)
(c) \(\frac{7}{24}\)
(d) \(\frac{25}{24}\)
Answer:
(c) \(\frac{7}{24}\)

Question 4.
If 4 cot θ = 3 then \(\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}\) is equal to :
(a) \(\frac{5}{6}\)
(b) \(\frac{1}{7}\)
(c) \(\frac{6}{7}\)
(d) \(\frac{3}{4}\)
Answer:
(b) \(\frac{1}{7}\)

Question 5.
Value of cosec² 27° – tan² 63° is equal to :
(a) 1
(b) 0
(c) – 1
(d) 90
Answer:
(c) – 1

Question 6.
The value of the expression [cosec (85° + θ) – sec (5° – θ) – tan (42° + θ) + cot (48° – θ)] is
(a) – 1
(b) 0
(c) 1
(d) \(\frac{3}{2}\)
Answer:
(b) 0

Question 7.
If A + B = 90° and cot B = \(\frac{3}{4}\) then tan A is equal to
(a) \(\frac{3}{4}\)
(b) \(\frac{4}{3}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{1}{3}\)
Answer:
(a) \(\frac{3}{4}\)

Question 8.
If ∆ABC is right angled at C, then value of cos(A + B) is
(a) 0
(b) 1
(c) \(\frac{1}{2}\)
(d) \(\frac{\sqrt{3}}{2}\)
Answer:
(a) 0

Question 9.
Maximum of \(\frac{1}{{cosec} \theta}\), 0° < θ < 90° is
(a) – 1
(b) 2
(c) 1
(d) \(\frac{1}{2}\)
Answer:
(c) 1

Question 10.
The value of tan 1° tan 2° tan 3° ………….. tan 89° is
(a) 0
(b) 1
(c) 2
(d) 0
Answer:
(b) 1

Question 11.
Given that sin α = \(\frac{\sqrt{3}}{2}\) and cos β = \(\frac{1}{2}\) then the value of (α + β) is
(a) 90°
(b) 30°
(c) 60°
(d) 120°
Answer:
(d) 120°

Question 12.
(1 – cos θ) (cot θ + cosec θ) is equal to :
(a) sin θ
(b) sec θ
(c) tan θ
(d) cot θ.
Answer:
(a) sin θ

Question 13.
If 3 cos θ = 1, then the value of \(\frac{6 \sin ^2 \theta+\tan ^2 \theta}{4 \cos \theta}\)
(a) 5
(b) 10
(c) -10
(d) 0
Answer:
(b) 10

Question 14.
If sec² θ (1 + sin θ) (1 – sin θ) = k then value of k is
(a) 2
(b) 3
(c) θ
(d) 1
Answer:
(d) 1

Question 15.
If sin θ = \(\frac{1}{3}\), then value of (2 cot² θ + 2) is
(a) 9
(b) 18
(c) 27
(d) 20
Answer:
(b) 18

Fill in the Blanks:

Question 1.
5 sin θ° + cos 9θ° = ___________ .
Answer:
0

Question 2.
cosec³ (9θ° – θ) cos³ (9θ° – θ) = ___________ .
Answer:
tan³θ

Question 3.
Value of tan 90° is ___________.
Answer:
undefined/∞

Question 4.
If sin θ = 1, then θ = ___________
Answer:
90°

Question 5.
1 + tan² θ = sec² θ is not valid for θ = ___________ .
Answer:
90°

Question 6.
2 tan² 45° + 2 cos² 45° – 3 sin 9θ° = ___________ .
Answer:
0

Question 7.
sin θ ___________ when θ increases from θ° to 9θ°.
Answer:
increases

Question 8.
cos θ ___________ when θ decreases from θ° to 9θ°.
Answer:
decreases

Question 9.
(1 + tan² θ) (1 + sin θ) (1 – sin θ) = ___________ .
Answer:
1

Question 10.
\(\frac{\cos 58^{\circ}}{\sin 32^{\circ}}-\frac{\tan 31^{\circ}}{\cot 59^{\circ}}\) ___________ .
Answer:
0

Extra Questions for Class 10 Maths

NCERT Solutions for Class 10 Maths