## Rajasthan Board RBSE Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise

Question 1.

A right angle is:

(A) equal to a radian

(B) equal to 90 degree

(C) equal to 18°

(D) equal to 90 radian

Solution :

(B) equal to 90 degree

Question 2.

Which trigonometric function is positive in third quadrant:

(A) sin θ

(B) tan θ

(C) cos θ

(D) sec θ

Solution:

(B) tan θ

Question 3.

cosec (- θ) is equal to:

(A) sin θ

(B) tan θ

(C) cos θ

(D) -cosec θ

Solution:

(D) -cosec θ

Question 4.

tan (90° – θ) is equal to:

(A) -tan θ

(B) cot θ

(C) tan θ

(D) -cot θ

Solution :

(B) cot θ

Question 5.

If cos θ = then value of θ will be:

(A)

(B)

(C) –

(D)

Solution:

(A)

Question 6.

If n is an even integer, then the value of sin (2nπ ± θ) will be:

(A) ± cos θ

(B) ± tan θ

(C) ± sin θ

(D) ± cot θ

Solution:

(C) ± sin θ

Question 7.

The value of cot 15° will be:

(A) 2 + √3

(B) – 2 + √3

(C) 2 – √3

(D) – 2 – √3

Solution:

(A) 2 + √3

Question 8.

The value of cos 15° will be:

(A)

(B)

(C)

(D)

Solution:

(A)

Question 9.

The value of 2 sin cos will be:

(A) 1

(B)

(C) -1

(D) +1

Solution:

(D) +1

2 sin cos

= 2 sin 75° cos 15°

= 2 sin (90° – 15°) cos 15°

= 2 cos 15° cos 15°

= 2 cos^{2} 15°

= cos^{2} × 15° + 1

= cos 30° + 1

= +1

Question 10.

The value of cos sin will be:

(A)

(B) 0

(C)

(D)

Solution:

(D)

cos – sin = cos 15°- sin 15°

= cos (45° – 30°) – sin (45° – 30°)

= (cos 45° cos 30° + sin 45° sin 30°)

= (sin 45° cos 30° – cos 45° sin 30°)

Question 11.

If sin A = than value sin 2A will be:

(A)

(B)

(C)

(D)

solution:

Hence,Option (C) is correct.

Question 12.

If sin A = than value of sin 3A will be:

(A)

(B) –

(C)

(D)

Solution:

sin 3A

3 sin A – 4 sin^{3} A

Hence,Option (A) is correct.

Question 13.

If tan A = then value of tan 3A will be:

(A)

(B)

(C)

(D)

Solution:

Hence,Option (B) is correct.

Question 14.

If A + B = then value (1 + tan A) (1 + tan B) will be:

(A) 3

(B) 2

(C) 4

(D) 1

Solution:

(1 + tan A) (1 + tan B)

= 1 + tan A + tan B + tan A tan B ….. (i)

From equation (1) and (2)

(1 + tan A) (1 + tan B) = 2.

Question 15.

General value of θ in equation sec^{2} θ = 2 will be:

Solution:

Hence,Option (A) is correct.

Question 16.

Prove that:

(i) cos θ + sin (27θ° + θ) – sin (27θ° – θ)+ cos (18θ° + θ) = θ

Solution:

(i) sin (27θ° + θ) → IV quardant, -ive = -cosθ … (i)

sin (27θ° – θ) → III quardant, -ive = – cos θ …. (ii)

cos (18θ° + θ) → III quardrant, -ive = – cos θ … (iii)

L.H.S. = cos θ + sin (27θ° + θ) – sin (27θ° – θ) + cos (18θ° + 9)

= cos θ – cos θ + cos θ – cos θ [from (i), (ii) and (iii)]

= θ

= R.H.S.

Hence Proved.

= sec (27θ° – θ) sec (θ – 45θ°)+ tan (45θ° + θ) tan (θ – 27θ°)

= – cosec θ sec (θ – 45θ°)+ tan (36θ° + 9θ° + θ) tan (θ – 27θ°)

= – cosec θ sec (45θ° – θ) – tan (9θ° + θ) tan (27θ° – θ)

= – cosec θ sec (36θ° + 9θ° – θ) – (- cot θ) cot θ

= – cosec θ sec (9θ° – θ) + cot^{2} θ

= – cosec θ cosec θ + cot^{2} θ

= – cosec^{2} θ + cot2 θ

= – 1 (∵ 1 + cot^{2} θ = cosec^{2} θ cot2 θ – cosec^{2} θ = – 1)

R.H.S.

Hence Proved.

Question 17.

Find the value of sin , where n is an integer.

Solution:

Question 18.

If sin A + sin 5 = a and cos A + cos 5 = b, then prove that:

(i) sin (A + B) =

(ii) cos (A + B) =

Solution:

Question 19.

If A + B + C – 180°, then prove that:

(i) cos 2A + cos 2B – cos 2C = 1 – 4 sin A sin 5 cos C

(ii) sin A – sin B + sin C = 4 sin cos sin

Solution:

(i) L.H.S.

= cos 2A + cos 2B – cos 2C

= (cos 2A + cos 2B) – cos 2C

= 2 cos (A + B) cos (A – B) – cos 2C

= 2 cos (180° – C) cos (A – B) – (2 cos^{2} C – 1)

= -2 cos C cos (A – B) -2 cos^{2} C + 1

= 1 – 2 cos C cos (A – B) -2 cos^{2} C

= 1 – 2 cos C [cos (A – B) + cos C]

Question 20.

If A + B + C = 2π, then prove that cos^{2} B + cos^{2} C – sin^{2} A = 2 cos A cos B cos C.

Solution:

We know that:

= [cos^{2}B + cos (2π + B) cos (A – C)]

= cos^{2}B + cos B cos (A – C)

= cos B (cos B + cos (A – C)

= cos B (cos (2π – (A – C) + cos (A – C) – cos B [cos (A – B) + cos (A + C)]

= cos B [(cos (A + B) + cos (A – C)]

= cos B [2 cos A cos C]

= 2 cos A cos B cos C

Hence Proved

Question 21.

Find the following equation 2 tan θ – cot θ +1 = 0.

Solution:

When tan θ + 1 = 0

then tan θ = – 1 = tan 135° or tan

⇒ Hence, principal value of θ =

And general value of θ = nπ + ….. (i)

when 2 tan θ – 1 =0

⇒ tan θ =

Hence,principal value of θ = tan^{-1}

and general value of θ = nπ + tan^{-1} ….. (ii)

From (i) and (ii), solution of given equation is

nπ + tan^{-1} and nπ + , where n ∈ Z

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