## RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Other Exercises

Mark the correct alternative in each of the following :

Question 1.
If 6 θ is an acute angle such that cos θ =

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.
If 6 is an acute angle such that tan2 6 = 8 $$\frac { 8 }{ 7 }$$, then the value of

Solution:

Question 9.
If 3 cos θ = 5 sin 6, then the value of

Solution:

Question 10.
If tan2 45° – cos2 30° = x sin 45° cos 45°, then x =

Solution:

Question 11.
The value of cos217° – sin2 73° is

Solution:
cos2 17° – sin2 73° = cos2 (90° – 73°) – sin2 73°
= sin2 73° – sin2 73° = 0 (c)

Question 12.

Solution:

Question 13.

Solution:

Question 14.
If A and B are complementary angles then
(a) A – sin B
(b) cos A = cos B
(c) A = tan B
(d) sec A = cosec B
Solution:
∵ A and B are complementary angles
∴ A + B = 90°
⇒ A – 90° – B
sec A = sec (90° – B) = cosec B (d)

Question 15.
If x sin (90° – θ) cot (90° – 6) – cos (90° – θ), then x =
(a) 0
(b) 1

(c) -1
(d) 2,

Solution:

Question 16.
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to

Solution:

Question 17.
If angles A, B, C of a AABC form an increasing AP, then sin B =

Solution:

Question 18.
If 6 is an acute angle such that sec2 θ =

Solution:

Question 19.
The value of tan 1° tan 2° tan 3s tan 89° is
(a) 1
(b) -1

(c) 0
(d) None of these

Solution:
tan 1° tan 2° tan 3° tan 44° tan 45° tan 46° tan 89°
= tan 1° tan 2° tan 3° tan 44° tan 45° tan
(90° – 44°) tan (90° – 43°) tan (90° – 1°)’
= tan 1° tan 2° tan 3° tan 44° tan 45° cot
44° cot 43°….. cot 1°

Question 20.
The value of cos 1″ cos 2° cos 3° cos 180° is
(a) 1
(b) 0
(c) -1
(d) None of these
Solution:
None of these, because we deal here only an angle O<θ ≤ 90° (d)

Question 21.
The value of tan 10° tan 15° tan 75° tan 80° is
(a) -1
(b) 0
(c) 1
(d) None of these
Solution:
tan 10° tan 15° tan 75° tan 80°
= tan 10° tan 15° tan (90° – 15°) tan (90° – 10°) .
= tan 10° tan 15° cot 15° cot 10°
= tan 10° cot 10° tan 15° cot 15°
{∵ tan θ cot θ = 1}
= 1×1 = 1 (C)

Question 22.
The value of

Solution:

Question 23.
If 6 and 20 – 45° arc acute angles such that sin 0 = cos (20 – 45°), then tan 0 is equal to

Solution:

Question 24.
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2sin 3θ-√3 tan 3θ is equal to
(a) 1
(b) 0
(c) -1
(d) 1 + √3
Solution:

Question 25.
If A + B = 90°, then
Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.
sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°

(c) 45°
(d) 60°

Solution:

Question 29.

Solution:

Question 30.
If A, B and C are interior angles of a

Solution:

Question 31.
If cos θ = $$\frac { 2 }{ 3 }$$, then 2 sec2 θ + 2 tan2 θ-7 is equal to
(a) 1
(b) 0
(c) 3
(d) 4
Solution:

Question 32.
tan 5° x tan 30° x 4 tan 85° is equal to

Solution:

Question 33.

(a)-2
(b) 2
(c) 1
(d) 0
Solution:

Question 34.
In the figure, the value of cos Φ is

Solution:

Question 35.
In the figure, AD = 4 cm, BD = 3 cm and CB = 12 cm, find cot θ.

Solution:

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS are helpful to complete your math homework.

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