• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • RBSE Model Papers
    • RBSE Class 12th Board Model Papers 2022
    • RBSE Class 10th Board Model Papers 2022
    • RBSE Class 8th Board Model Papers 2022
    • RBSE Class 5th Board Model Papers 2022
  • RBSE Books
  • RBSE Solutions for Class 10
    • RBSE Solutions for Class 10 Maths
    • RBSE Solutions for Class 10 Science
    • RBSE Solutions for Class 10 Social Science
    • RBSE Solutions for Class 10 English First Flight & Footprints without Feet
    • RBSE Solutions for Class 10 Hindi
    • RBSE Solutions for Class 10 Sanskrit
    • RBSE Solutions for Class 10 Rajasthan Adhyayan
    • RBSE Solutions for Class 10 Physical Education
  • RBSE Solutions for Class 9
    • RBSE Solutions for Class 9 Maths
    • RBSE Solutions for Class 9 Science
    • RBSE Solutions for Class 9 Social Science
    • RBSE Solutions for Class 9 English
    • RBSE Solutions for Class 9 Hindi
    • RBSE Solutions for Class 9 Sanskrit
    • RBSE Solutions for Class 9 Rajasthan Adhyayan
    • RBSE Solutions for Class 9 Physical Education
    • RBSE Solutions for Class 9 Information Technology
  • RBSE Solutions for Class 8
    • RBSE Solutions for Class 8 Maths
    • RBSE Solutions for Class 8 Science
    • RBSE Solutions for Class 8 Social Science
    • RBSE Solutions for Class 8 English
    • RBSE Solutions for Class 8 Hindi
    • RBSE Solutions for Class 8 Sanskrit
    • RBSE Solutions

RBSE Solutions

Rajasthan Board Textbook Solutions for Class 5, 6, 7, 8, 9, 10, 11 and 12

  • RBSE Solutions for Class 7
    • RBSE Solutions for Class 7 Maths
    • RBSE Solutions for Class 7 Science
    • RBSE Solutions for Class 7 Social Science
    • RBSE Solutions for Class 7 English
    • RBSE Solutions for Class 7 Hindi
    • RBSE Solutions for Class 7 Sanskrit
  • RBSE Solutions for Class 6
    • RBSE Solutions for Class 6 Maths
    • RBSE Solutions for Class 6 Science
    • RBSE Solutions for Class 6 Social Science
    • RBSE Solutions for Class 6 English
    • RBSE Solutions for Class 6 Hindi
    • RBSE Solutions for Class 6 Sanskrit
  • RBSE Solutions for Class 5
    • RBSE Solutions for Class 5 Maths
    • RBSE Solutions for Class 5 Environmental Studies
    • RBSE Solutions for Class 5 English
    • RBSE Solutions for Class 5 Hindi
  • RBSE Solutions Class 12
    • RBSE Solutions for Class 12 Maths
    • RBSE Solutions for Class 12 Physics
    • RBSE Solutions for Class 12 Chemistry
    • RBSE Solutions for Class 12 Biology
    • RBSE Solutions for Class 12 English
    • RBSE Solutions for Class 12 Hindi
    • RBSE Solutions for Class 12 Sanskrit
  • RBSE Class 11

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

April 21, 2022 by Safia Leave a Comment

Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry Important Questions and Answers.

RBSE Class 10 Maths Chapter 8 Important Questions Introduction to Trigonometry

Objective Type Questions—

Question 1.
\(\frac{\tan \theta}{\sqrt{1+\tan ^{2} \theta}}\) is equal to—
(A) cos θ
(B) sin θ
(C) sec θ
(D) cot θ
Answer:
(B) sin θ

Question 2.
\(\frac{\sqrt{{cosec}^{2} \theta-1}}{{cosec} \theta}\) is equal to—
(A) cos θ
(B) sec θ
(C) sin θ
(D) cosec θ
Answer:
(A) cos θ

Question 3.
sin θ cosec θ + cos θ sec θ is equal to—
(A) 2
(B) 1
(C) \(\frac{1}{2}\)
(D) -1
Answer:
(A) 2

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 4.
2 cosec 30° sec 30° is equal to—
(A) \(\frac{2}{\sqrt{3}}\)
(B) \(\frac{\sqrt{3}}{2}\)
(C) 8
(D) \(\frac{8}{\sqrt{3}}\)
Answer:
(D) \(\frac{8}{\sqrt{3}}\)

Question 5.
If sin θ = \(\frac{\sqrt{3}}{2}\). then the value of θ is—
(A) 30°
(B) 45°
(C) 60°
(D) 90°
Answer:
(C) 60°

Question 6.
The value of \(\frac{3 \sec 51^{\circ}}{{cosec} 39^{\circ}}\) is—
(A) 1
(B) 2
(C) 3
(D) 0
Answer:
(C) 3

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 7.
If cos (90° – θ) = \(\frac{1}{2}\) then the value of θ is—
(A) 90°
(B) 60°
(C) 45°
(D) 30°
Answer:
(D) 30°

Question 8.
sin2 50° + cos2 50° + 1 is equal to—
(A) 2
(B) 1
(C) \(\frac{1}{2}\)
(D) 0
Answer:
(A) 2

Very Short Answer Type Questions—

Question 1.
If sin A = cos (A – 26°), where 3A is an acute angle then find the value of A.
Solution:
Given
sin 3A = cos (A – 26°)
∵ sin 3A = cos (90° – 3A)
∴ cos (90° – 3A) = cos (A – 26°)
Since 90° – 3A and A – 26° both are acute angles
∴ 90° – 3A = A – 26°
or 4A = 90° + 26° – 116°
A = \(\frac{116^{\circ}}{4}\) = 29°

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 2.
Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
cot 85° + cos 75° = cot
(90°- 5°) + cos (90° – 15°) = tan 5° + sin 15°

Question 3.
Find the value of sin 25° . cos 65° + cos 25° . sin 65° + sin2 25° + sin2 65°.
Solution:
sin 25° cos 65° + cos 25°.sin 65° + sin2 25° + sin2 65°
= sin (90° – 65°) . cos 65° + cos (90° – 65°) . sin 65° + sin2 (90° – 65°) + sin2 65°
= cos 65° . cos 65° + sin 65° sin 65° + cos2 65° + sin2 65°
= cos2 65° + sin2 65° + cos2 65° + sin2 65°
= 1 + 1 = 2

Question 4.
if ∠A < 90° and ∠B < 90° and sin A = cos B, then prove that A + B = 90°.
Solution:
∵ sin A = cos B
⇒ sin A = sin (90° – B)
⇒ A = 90° – B
⇒ A + B = 90° Hence Proved

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 5.
If sin θ = cos θ then find the value of θ.
Solution:
∵ sin θ = cos θ
⇒ sin θ = sin (90° – θ)
⇒ θ = 90° – θ0
⇒ 2θ = 90°
⇒ θ = \(\frac{90^{\circ}}{2}\) = 45°

Question 6.
Write the value of 4 sin 18° sec 72°.
Solution:
4 sin 18° sec 72° = 4 sin 18° sec (90° – 18°)
= 4 sin 18° . cosec 18°
= 4 sin 18° × \(\frac{1}{\sin 18^{\circ}}\) = 4 Ans.

Question 7.
Write the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) + 1.
Solution:
sin θ sin θ + cos θ cos θ + 1
= sin2 θ + cos2 θ + 1
= 1 + 1 = 2

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 8.
Write the value of sin2 60° + cos2 30°.
Solution:
\(\left(\frac{\sqrt{3}}{2}\right)^{2}\) + \(\left(\frac{\sqrt{3}}{2}\right)^{2}\) = \(\frac{3}{4}\) + \(\frac{3}{4}\) = \(\frac{6}{4}\) = \(\frac{3}{2}\)

Question 9.
Find the value of cos2 50° + cos2 40°.
Solution:
cos2 50° + cos2(90° – 50°)
= cos2 50° + sin2 50°
= 1

Question 10.
Write the simplest value of \(\frac{\sqrt{1-\sin ^{2} 40^{\circ}}}{\cos 40^{\circ}}\)
Solution:
\(\frac{\sqrt{\cos ^{2} 40^{\circ}}}{\cos 40^{\circ}}\) = \(\frac{\cos 40^{\circ}}{\cos 40^{\circ}}\) = 1

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 11.
Find the value of 3sin 60° – 4sin3 60°.
Solution:
3 sin 60° – 4 sin3 60°
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 1

Question 12.
Find the value of sin θ . cosec θ – cos θ sec θ.
Solution:
sin θ . cosec θ – cos θ . sec θ
= sin θ. \(\frac{1}{\sin \theta}\) – cos θ. \(\frac{1}{\cos \theta}\) = 1 – 1 = 0

Question 13.
Write the value of (1 – sin2 θ) sec2 θ.
Solution:
(1 – sin2 θ) sec2 θ
= cos2 θ . sec2 θ
= cos2 θ. \(\frac{1}{\cos ^{2} \theta}\) = 1

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 14.
Write the value of \(\frac{1}{\sqrt{cosec}^{2} \theta-1}\)
Solution:
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 2

Question 15.
Write the value of \(\frac{\cos \theta}{\sin \theta}\) × cot θ × tan2 θ.
Solution:
\(\frac{\cos \theta}{\sin \theta}\) × \(\frac{\cos \theta}{\sin \theta}\) × \(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\)
= \(\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\) × \(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\) = 1

Question 16.
Write the value of 2 sin2 A (1 + cot2 A).
Solution:
2 sin2 A × cosec2 A
= 2 sin2 A × \(\frac{1}{\sin ^{2} A}\) = 2

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Short Answer Type Questions—

Question 1.
In a right triangle ABC, right-angled at B, if tan A = 1, then verify that
2 sin A cos A = 1
Solution:
In △ABC
tan A = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = 1
i.e., BC = AB
Let AB = BC = K
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 3
Where K is a positive number.
By Pythagoras theorem
(AC)2 = (AB)2 + (BC)2
= (K)2 + K2
(AC)2 = 2K2
∴ AC = K\(\sqrt {2}\)
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 4
1, which is the required value. LHS = RHS

Question 2.
In △OPQ, right-angled at P, OP = 7 cm. and OQ – PQ = 1 cm. (see figure). Determine the values of sin θ and cos θ.
Solution:
From the given figure OQ2 = OP2 + PQ2
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 5
But given that
OQ – PQ = 1
∴ OQ = 1 + PQ
i.e., (1 + PQ)2 = OP2 + PQ2
i.e., 1 + PQ2 + 2PQ = OP2 + PQ2
i.e., 1 + 2PQ = (7)2
1 + 2PQ = 49 ∴ PQ = 24 cm.
and OQ = 1 + PQ
= 1 + 24 = 25 cm.
sin θ = \(\frac{\mathrm{OP}}{\mathrm{OQ}}\) = \(\frac{7}{25}\)
and cos θ = \(\frac{\mathrm{PQ}}{\mathrm{OQ}}\) = \(\frac{24}{25}\)

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 3.
If tan A = \(\frac{3}{4}\), then find the value of sec A (1 – sin A) (sec A + tan A)
Solution:
Let ABC be a right triangle in which angle B is a right angle.
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 6
For ∠A
base = AB,
perpendicular = BC
and hypotenuse = AC
Given that,tan A = \(\frac{3}{4}\)
But tan A = \(\frac{perpendicular}{base}\) = \(\frac{BC}{AB}\)
\(\frac{BC}{AB}\) = \(\frac{3}{4}\) = k (say)
∴ AB = 4k, BC = 3k
In right △ABC, by Pythagoras Theorem
(AC)2 = (AB)2 + (BC)2
= (4k)2 + (3k)2
(AC)2 =16k2 + 9k2 = 25k2
∴ AC = ±\(\sqrt{25 k^{2}}\)
= ± 5k
AC = 5k
[∵ AC ≠ – 5k, since side cannot be negative]
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 7

Question 4.
In △PQR, where angle Q is a right angle. In the given figure, PQ = 3cm and PR = 6 cm. Find ∠QPR and ∠PRQ.
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 8
Solution:
Given
PQ = 3 cm and PR = 6cm
Therefore, \(\frac{PQ}{PR}\) = sin R
sin R = \(\frac{3}{6}\) = \(\frac{1}{2}\) = sin 30°
So ∠PRQ = 30°
and therefore∠QPR = 180° – (90° + 30°)
= 60°

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 5.
Find the value of 6 tan 20° tan 70° – 3 sec2 45°.
Solution:
6 tan 20° tan 70° – 3 sec2 45°
= 6 tan 20° . tan (90° – 20°) – 3 sec2 45°
= 6 tan 20° . cot 20° – 3 (\(\sqrt {2}\) )2
∵ tan 0 x cot 0 = 1
= 6 – 6 = 0

Question 6.
Find the value of sin 30° . cos2 30° + tan 45° . cos2 60°.
Solution:
sin 30° . cos2 30° + tan 45° . cos2 60°
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 9

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 7.
If sin (A – B) = \(\frac{1}{2}\), cos (A + B) = \(\frac{1}{2}\), 0° < A + B ≤ 90°, A > B, then find the values of A and B.
Solution:
Since
sin (A – B) = \(\frac{1}{2}\)
∴ sin (A – B) = sin 30°
⇒ A – B = 30° …. (i)
Similarly cos (A + B) = \(\frac{1}{2}\)
∴ cos (A + B) = cos 60°
⇒ A + B = 60° …. (ii)
Adding equation (i) and (ii)
A – B + A + B = 30° + 60° = 90°
2A = 90°
A = 45°
From equation (ii) B = 15°

Question 8.
Find the value of (sec2 30° + cosec2 45°) (2 cos 60° + sin 90° + tan 45°).
Solution:
(sec2 30° + cosec2 45°) (2 cos 60° + sin 90° + tan 45°)
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 10

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 9.
Find the value of sec2 65° – cot2 25° – 2 sin 30° cos 60°
Solution:
sec2 65° – cot2 25° – 2 sin 30° cos 60°
Here taking 25° = 90° – 65°
cot (25°) = cot (90° – 65°)
cot 25° = tan 65°
[∵ cot (90° – θ) = tan θ]
Now the expression will become as follows—
(sec2 65° – tan2 65°) – 2 sin 30° cos 60°
= 1 – 2 × \(\frac{1}{2}\) × \(\frac{1}{2}\)
[∵ sec2 θ – tan2 θ = 1]
= 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\)

Question 10.
Find the value of 5\(\frac{\sin 17^{\circ}}{\cos 73^{\circ}}\) + 2\(\frac{\cos 67^{\circ}}{\sin 23^{\circ}}\) – 6\(\frac{\sin 15^{\circ}}{\cos 75^{\circ}}\).
Solution:
sin 17° = sin (90° – 73°)
or sin 17° = cos 73° …. (i)
cos 67° = cos (90° – 23°)
or cos 67° = sin 23° …. (ii)
sin 15° = sin (90° – 75°)
or sin 15° = cos 75° …. (iii)
From (i), (ii) and (iii) putting the values of sin 17°, cos 67° and sin 15° in the original expression. Then we get
= 5\(\frac{\cos 73^{\circ}}{\cos 73^{\circ}}\) + 2\(\frac{\sin 23^{\circ}}{\sin 23^{\circ}}\) – 6\(\frac{\cos 75^{\circ}}{\cos 75^{\circ}}\)
= 5 (1) + 2 (1) – 6 (1) = 5 + 2 – 6
= 7 – 6 = 1

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 11.
Prove that sec A (1 – sin A) (sec A + tan A) = 1
Solution:
L.H.S. = sec A (1 – sin A) (sec A + tan A)
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 11

Question 12.
If sin θ = \(\frac{1}{2}\) then write the value of (tan θ + cot θ)2.
Solution:
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 12

Question 13.
If sin A = \(\frac{3}{5}\), then find cos A and cosec A.
Solution:
Let ABC be any right angled triangle in which there is a right angle at B.
∵ sin A = \(\frac{3}{5}\)
But sin A = \(\frac{perpendicular}{hypotenuse}\) = \(\frac{BC}{AC}\)
∴ \(\frac{BC}{AC}\) = \(\frac{3}{4}\)
Let BC = 3k
AC = 5k
By Pythagoras Theorem
AC2 = AB2 + BC2
or (5k2) = (AB)2 + (3k)2
or 25k2 = AB2 + 9k2
or 25k2 – 9k2 = AB2
or 16k2 = AB2
or AB = 4k
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 13

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 14.
Find the value of \(\frac{\tan 65^{\circ}}{\cot 25^{\circ}}\)
Solution:
\(\frac{\tan 65^{\circ}}{\cot 25^{\circ}}\)
We know that cot A = tan (90° – A)
So cot 25° = tan (90° – 25°)
= tan 65°
i.e., \(\frac{\tan 65^{\circ}}{\cot 25^{\circ}}\) = \(\frac{\tan 65^{\circ}}{\tan 65^{\circ}}\)
= 1

Question 15.
Find the value of sin 35° cos 55° + cot 35° sin 55°.
Solution:
sin 35° cos 55° + cot 35° sin 55°
= sin 35° × cos (90° – 35°) + cos 35° × sin(90° – 35°)
[∵ cos (90° – θ) = sinθ
sin (90° – θ) = cosθ]
= sin 35° × sin 35° + cos 35° × cos 35°
= sin2 35° + cos2 35°
= 1

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 16.
If sinθ = \(\frac{1}{2}\), then find the value of \(\frac{1-2 \sin ^{2} \theta}{\sin \theta}\).
Solution:
Given equation = \(\frac{1-2 \sin ^{2} \theta}{\sin \theta}\)
Here putting sin θ = \(\frac{1}{2}\)
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 14
= 1

Question 17.
Find the value of cos2 12° + cos2 78°.
Solution:
cos2 12° + cos2 78°
= cos2 12° + {cos (90° – 12°)}2
= cos2 12° + sin2 12°
[∵ cos (90° – θ) = sin θ]
= 1

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 18.
Show that tan 36° tan 17° tan 54° tan 73° = 1
Solution:
tan 36° tan 17° tan 54° tan 73°
= tan 36° tan 17° tan (90° – 36°) . tan (90° – 17°)
= tan 36° . tan 17° cot 36° cot 17°
= tan 36° . cot 36° . tan 17° . cot 17°
∵ tanθ.cotθ = 1
= 1.1 = 1

Question 19.
Show that sin 28° cos 62° + cos 28° sin 62° = 1.
Solution:
sin 28° cos 62° + cos 28° sin 62°
= sin 28° × cos (90° – 28°) + cos 28° × sin (90° – 28°)
[∵ cos (90° – θ) = sinθ
and sin (90° – θ) = cos θ]
= sin 28° × sin 28° + cos 28° × cos 28°
= sin2 28° + cos2 28°
= 1

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 20.
Find the value of \(\frac{\tan 67^{\circ}}{\cot 23^{\circ}}\).
Solution:
\(\frac{\tan 67^{\circ}}{\cot 23^{\circ}}\)
Wc know that cotA = tan (90° – A)
∴ cot 23° = tan (90° – 23°) = tan 67°
Therefore \(\frac{\tan 67^{\circ}}{\cot 67^{\circ}}\) = 1

Question 21.
If 3cot A = 4, then evaluate \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\).
Solution:
∵ 3 cot A = 4 ∴ cot A = \(\frac{4}{3}\)

Long Answer Type Questions—

Question 1.
In △ABC, right angled at B, AB = 5 cm and ∠ACB = 30° (see Figure). Determine the lengths of the sides BC and AC.
Solution:
To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB. Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore
\(\frac{AB}{BC}\) = tan C
i.e., \(\frac{5}{BC}\) = tan 30° = \(\frac{1}{\sqrt{3}}\)
which gives BC = 5\(\sqrt{3}\)cm
To find the length of the side AC,
∵ sin 30 = \(\frac{AB}{BC}\)
i.e., \(\frac{1}{2}\) = \(\frac{5}{AC}\)
i.e., AC = 10 cm

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 2.
If ∠B and ∠Q are acute angles such that sin B = sin Q, then prove that ∠B = ∠Q.
Solution:
Let us consider two right triangles ABC and PQR where sin B = sin Q (see figure)
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 15
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 16
From (i) and (ii)
\(\frac{AC}{PR}\) = \(\frac{AB}{PQ}\) = \(\frac{BC}{QR}\)
Then using SSS similarity theorem
△ACB ~ △PRQ
Consequently,∠B = ∠Q Hence Proved

Question 3.
If θ = 30°, then find the value of the following—
\(\frac{3 \cot \left(90^{\circ}-\theta\right)-\tan ^{3} \theta}{1-3 \cot ^{2}\left(90^{\circ}-\theta\right)}\)
Solution:
Putting θ = 30° the expression will become
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 17

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 4.
Prove that—
tan 6° . tan 26° tan 64° . tan84° = 1
Solutions:
L.H.S = tan6° . tan26° tan64° tan84°
∵ 6° = 90° – 84°
∴ tan 6° = tan(90° – 84°)
tan 6° = cot 84°
∵ tan (90° – θ) = cot θ …. (i)
and 26° = 90° – 64°
tan 26° = tan (90° – 64°)
tan 26° = cot 64° …. (ii)
From (i) and (ii) putting the values in the given expression
= tan 6° . tan 26° tan 64° . tan 84°
= cot 84° . tan 84° . cot 64° . tan 64°
[∵ tanθ. cotθ = 1]
= 1 . 1 = 1= R.H.S. Hence Proved

Question 5.
From the following equation find the value of x—
cosec (90° – θ) + x cos θ cot (90° – θ) = sin (90° – θ)
Solution:
cosec (90° – θ) + x cos θ cot (90° – θ) = sin (90° – θ)
⇒ sec θ + x cos θ tan θ = cos θ
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 18

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 6.
Find the value of the following—
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
Solution:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 19
= 2

Question 7.
Prove that—
tan2 A – tan2 B = \(\frac{\cos ^{2} B-\sin ^{2} A}{\cos ^{2} B \cos ^{2} A}\) = \(\frac{\sin ^{2} A-\sin ^{2} B}{\cos ^{2} A \cos ^{2} B}\)
Solution:
L.H.S. = tan2 A – tan2 B
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 20
= R.H.S Hence Proved

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 8.
If tan θ = \(\frac{1}{\sqrt{7}}\), then find the value of \(\frac{{cosec}^{2} \theta-\sec ^{2} \theta}{{cosec}^{2} \theta+\sec ^{2} \theta}\)
Solution:
Here tanθ = \(\frac{1}{\sqrt{7}}\)
We know that
sec2 θ = 1 + tan2 θ
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 21

Question 9.
Prove the following identity
\(\frac{\sin A+\cos A}{\sin A-\cos A}\) + \(\frac{\sin A-\cos A}{\sin A+\cos A}\) = \(\frac{2}{\sin ^{2} A-\cos ^{2} A}\)
Solution:
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 22
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 23
∴ L.H.S:= R.H.S. Hence Proved

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 10.
Prove that \(\frac{\cot A-\cos A}{\cot A+\cos A}\) = \(\frac{cosec A- 1}{cosec A + 1}\)
Solution:
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 24

Question 11.
Using the identity sec2 θ = 1 + tan2 θ, prove that
\(\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}\) = \(\frac{1}{\sec \theta-\tan \theta}\)
Solution:
Since we will apply the identity involving sec θ and tan θ, let us first convert the LHS of the identity in terms of sec θ and tan θ by dividing numerator and denominator by cos θ.
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 25
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 26

RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

Question 12.
Prove that \(\sqrt{\frac{1+\cos \mathrm{A}}{1-\cos \mathrm{A}}}\) = cosec A + cot A.
Solution:
L.H.S = \(\sqrt{\frac{1+\cos \mathrm{A}}{1-\cos \mathrm{A}}}\)
multiplying the numerator and the denominator by 1 + cos A within the square root
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 27
RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry 28
= cosec A + cot A
= R.H.S.
∴ L.H.S. =R.H.S. Hence Proved.

Share this:

  • Click to share on WhatsApp (Opens in new window)
  • Click to share on Twitter (Opens in new window)
  • Click to share on Facebook (Opens in new window)

Related

Filed Under: Class 10

Reader Interactions

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

Rajasthan Board Questions and Answers

Recent Posts

  • RBSE Class 9 Science Important Questions in Hindi & English Medium
  • RBSE Class 10 Maths Important Questions in Hindi & English Medium
  • RBSE Class 10 Science Important Questions in Hindi & English Medium
  • RBSE Class 9 Social Science Important Questions in Hindi & English Medium
  • RBSE Class 9 Social Science Notes in Hindi Medium & English Medium Pdf Download 2021-2022
  • RBSE Class 10 Social Science Important Questions in Hindi & English Medium
  • RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2
  • RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1
  • RBSE Class 10 Maths Important Questions Chapter 10 Circles
  • RBSE Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2
  • RBSE Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

Footer

RBSE Solutions for Class 12
RBSE Solutions for Class 11
RBSE Solutions for Class 10
RBSE Solutions for Class 9
RBSE Solutions for Class 8
RBSE Solutions for Class 7
RBSE Solutions for Class 6
RBSE Solutions for Class 5
RBSE Solutions for Class 12 Maths
RBSE Solutions for Class 11 Maths
RBSE Solutions for Class 10 Maths
RBSE Solutions for Class 9 Maths
RBSE Solutions for Class 8 Maths
RBSE Solutions for Class 7 Maths
RBSE Solutions for Class 6 Maths
RBSE Solutions for Class 5 Maths
Target Batch
RBSE Class 11 Political Science Notes
RBSE Class 11 Geography Notes
RBSE Class 11 History Notes

Copyright © 2022 RBSE Solutions

 

Loading Comments...