RBSE Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry

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

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

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°

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

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

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

Question 11.
Find the value of 3sin 60° – 4sin³ 60°.
Solution:
3 sin 60° – 4 sin³ 60°

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

Question 14.
Write the value of \(\frac{1}{\sqrt{cosec}^{2} \theta-1}\)
Solution:

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

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

Where K is a positive number.
By Pythagoras theorem
(AC)² = (AB)² + (BC)²
= (K)² + K²
(AC)² = 2K²
∴ AC = K\(\sqrt {2}\)

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

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 = \(\frac{\mathrm{PQ}}{\mathrm{OQ}}\) = \(\frac{24}{25}\)

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.

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]

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.

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°

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°

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°)

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

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)

Question 12.
If sin θ = \(\frac{1}{2}\) then write the value of (tan θ + cot θ)².
Solution:

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

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

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 = \(\frac{1}{2}\)

= 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

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

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

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)


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

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

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)

= 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

= R.H.S Hence Proved

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

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:


∴ L.H.S:= R.H.S. Hence Proved

Question 10.
Prove that \(\frac{\cot A-\cos A}{\cot A+\cos A}\) = \(\frac{cosec A- 1}{cosec A + 1}\)
Solution:

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.

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


= cosec A + cot A
= R.H.S.
∴ L.H.S. =R.H.S. Hence Proved.