RD Sharma Solutions for Class 11 Chapter 5 - Trigonometric Functions Exercise 5.3

In Exercise 5.3, we shall discuss problems based on signs of trigonometric functions and variations in values of trigonometric functions in different quadrants. Students can refer to RD Sharma Class 11 Solutions which is designed by our expert tutors for the purpose of strengthening their concepts and problem-solving abilities. These solutions are developed with utmost care to help students analyze the problems competently and hence, score good marks in their exams. Students can download the PDFs of RD Sharma Class 11 from the links provided below and start practising offline.

RD Sharma Solutions for Class 11 Maths Exercise 5.3 Chapter 5 – Trigonometric Functions

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Exercise 5.2

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1. Find the values of the following trigonometric ratios:

(i) sin 5π/3

(ii) sin 17π

(iii) tan 11π/6

(iv) cos (-25π/4)

(v) tan 7π/4

(vi) sin 17π/6

(vii) cos 19π/6

(viii) sin (-11π/6)

(ix) cosec (-20π/3)

(x) tan (-13π/4)

(xi) cos 19π/4

(xii) sin 41π/4

(xiii) cos 39π/4

(xiv) sin 151π/6

Solution:

(i) sin 5π/3

5π/3 = (5π/3 × 180)^o

= 300^o

= (90×3 + 30)^o

Since 300^o lies in the IV quadrant in which the sine function is negative.

sin 5π/3 = sin (300)^o

= sin (90×3 + 30)^o

= – cos 30^o

= – √3/2

(ii) sin 17π

Sin 17π = sin 3060^o

= sin (90×34 + 0)^o

Since 3060^o lies in the negative direction of the x-axis, i.e., on the boundary line of the II and III quadrants.

Sin 17π = sin (90×34 + 0)^o

= – sin 0^o

= 0

(iii) tan 11π/6

tan 11π/6 = (11/6 × 180)^o

= 330^o

Since 330^o lies in the IV quadrant in which the tangent function is negative.

tan 11π/6 = tan (300)^o

= tan (90×3 + 60)^o

= – cot 60^o

= – 1/√3

(iv) cos (-25π/4)

cos (-25π/4) = cos (-1125)^o

= cos (1125)^o

Since 1125^o lies in the I quadrant in which the cosine function is positive.

cos (1125)^o = cos (90×12 + 45)^o

= cos 45^o

= 1/√2

(v) tan 7π/4

tan 7π/4 = tan 315^o

= tan (90×3 + 45)^o

Since 315^o lies in the IV quadrant in which the tangent function is negative.

tan 315^o = tan (90×3 + 45)^o

= – cot 45^o

= -1

(vi) sin 17π/6

sin 17π/6 = sin 510^o

= sin (90×5 + 60)^o

Since 510^o lies in the II quadrant in which the sine function is positive.

sin 510^o = sin (90×5 + 60)^o

= cos 60^o

= 1/2

(vii) cos 19π/6

cos 19π/6 = cos 570^o

= cos (90×6 + 30)^o

Since 570^o lies in the III quadrant in which the cosine function is negative.

cos 570^o = cos (90×6 + 30)^o

= – cos 30^o

= – √3/2

(viii) sin (-11π/6)

sin (-11π/6) = sin (-330^o)

= – sin (90×3 + 60)^o

Since 330^o lies in the IV quadrant in which the sine function is negative.

sin (-330^o) = – sin (90×3 + 60)^o

= – (-cos 60^o)

= – (-1/2)

= 1/2

(ix) cosec (-20π/3)

cosec (-20π/3) = cosec (-1200)^o

= – cosec (1200)^o

= – cosec (90×13 + 30)^o

Since 1200^o lies in the II quadrant in which the cosec function is positive.

cosec (-1200)^o = – cosec (90×13 + 30)^o

= – sec 30^o

= -2/√3

(x) tan (-13π/4)

tan (-13π/4) = tan (-585)^o

= – tan (90×6 + 45)^o

Since 585^o lies in the III quadrant in which the tangent function is positive.

tan (-585)^o = – tan (90×6 + 45)^o

= – tan 45^o

= -1

(xi) cos 19π/4

cos 19π/4 = cos 855^o

= cos (90×9 + 45)^o

Since 855^o lies in the II quadrant in which the cosine function is negative.

cos 855^o = cos (90×9 + 45)^o

= – sin 45^o

= – 1/√2

(xii) sin 41π/4

sin 41π/4 = sin 1845^o

= sin (90×20 + 45)^o

Since 1845^o lies in the I quadrant in which the sine function is positive.

sin 1845^o = sin (90×20 + 45)^o

= sin 45^o

= 1/√2

(xiii) cos 39π/4

cos 39π/4 = cos 1755^o

= cos (90×19 + 45)^o

Since 1755^o lies in the IV quadrant in which the cosine function is positive.

cos 1755^o = cos (90×19 + 45)^o

= sin 45^o

= 1/√2

(xiv) sin 151π/6

sin 151π/6 = sin 4530^o

= sin (90×50 + 30)^o

Since 4530^o lies in the III quadrant in which the sine function is negative.

sin 4530^o = sin (90×50 + 30)^o

= – sin 30^o

= -1/2

2. prove that:
(i) tan 225^o cot 405^o + tan 765^o cot 675^o = 0

(ii) sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2

(iii) cos 24^o + cos 55^o + cos 125^o + cos 204^o + cos 300^o = 1/2

(iv) tan (-125^o) cot (-405^o) – tan (-765^o) cot (675^o) = 0

(v) cos 570^o sin 510^o + sin (-330^o) cos (-390^o) = 0

(vi) tan 11π/3 – 2 sin 4π/6 – 3/4 cosec² π/4 + 4 cos² 17π/6 = (3 – 4√3)/2

(vii) 3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1

Solution:

(i) tan 225^o cot 405^o + tan 765^o cot 675^o = 0

Let us consider LHS:

tan 225° cot 405° + tan 765° cot 675°

tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8 + 45°) cot (90° × 7 + 45°)

We know that when n is odd, cot → tan.

tan 45° cot 45° + tan 45° [-tan 45°]

tan 45° cot 45° – tan 45° tan 45°

1 × 1 – 1 × 1

1 – 1

0 = RHS

∴ LHS = RHS

Hence proved.

(ii) sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2

Let us consider LHS:

sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6

sin 480° cos 690° + cos 780° sin 1050°

sin (90° × 5 + 30°) cos (90° × 7 + 60°) + cos (90° × 8 + 60°) sin (90° × 11 + 60°)

We know that when n is odd, sin → cos and cos → sin.

cos 30° sin 60° + cos 60° [-cos 60°]

√3/2 × √3/2 – 1/2 × 1/2

3/4 – 1/4

2/4

1/2

= RHS

∴ LHS = RHS

Hence proved.

(iii) cos 24^o + cos 55^o + cos 125^o + cos 204^o + cos 300^o = 1/2

Let us consider LHS:

cos 24^o + cos 55^o + cos 125^o + cos 204^o + cos 300^o

cos 24° + cos (90° × 1 – 35°) + cos (90° × 1 + 35°) + cos (90° × 2 + 24°) + cos (90° × 3 + 30°)

We know that when n is odd, cos → sin.

cos 24° + sin 35° – sin 35° – cos 24° + sin 30°

0 + 0 + 1/2

1/2

= RHS

∴ LHS = RHS

Hence proved.

(iv) tan (-125^o) cot (-405^o) – tan (-765^o) cot (675^o) = 0

Let us consider LHS:

tan (-125^o) cot (-405^o) – tan (-765^o) cot (675^o)

We know that tan (-x) = -tan (x) and cot (-x) = -cot (x).

tan (225°) cot (405°) + tan (765°) cot (675°)

tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8 + 45°) cot (90° × 7 + 45°)

tan 45° cot 45° + tan 45° [-tan 45°]

1 × 1 + 1 × (-1)

1 – 1

0

= RHS

∴ LHS = RHS

Hence proved.

(v) cos 570^o sin 510^o + sin (-330^o) cos (-390^o) = 0

Let us consider LHS:

cos 570^o sin 510^o + sin (-330^o) cos (-390^o)

We know that sin (-x) = -sin (x) and cos (-x) = +cos (x).

cos 570^o sin 510^o + [-sin (330^o)] cos (390^o)

cos 570^o sin 510^o – sin (330^o) cos (390^o)

cos (90° × 6 + 30°) sin (90° × 5 + 60°) – sin (90° × 3 + 60°) cos (90° × 4 + 30°)

We know that cos is negative at 90° + θ i.e. in Q₂ and when n is odd, sin → cos and cos → sin.

-cos 30° cos 60° – [-cos 60°] cos 30°

-cos 30° cos 60° + cos 60° cos 30°

0

= RHS

∴ LHS = RHS

Hence proved.

(vi) tan 11π/3 – 2 sin 4π/6 – 3/4 cosec² π/4 + 4 cos² 17π/6 = (3 – 4√3)/2

Let us consider LHS:

tan 11π/3 – 2 sin 4π/6 – 3/4 cosec² π/4 + 4 cos² 17π/6

tan (11 × 180^o)/3 – 2 sin (4 × 180^o)/6 – 3/4 cosec² 180^o/4 + 4 cos² (17 × 180^o)/6

tan 660^o – 2 sin 120^o – 3/4 (cosec 45^o)² + 4 (cos 510^o)²

tan (90° × 7 + 30°) – 2 sin (90° × 1 + 30°) – 3/4 [cosec 45°]² + 4 [cos (90° × 5 + 60°)]²

We know that tan and cos is negative at 90° + θ i.e. in Q₂ and when n is odd, tan → cot, sin → cos and cos → sin.

– cot 30° – 2 cos 30° – 3/4 [cosec 45°]² + [sin 60°]²

-√3 – 2√3/2 – 3/4 (√2)² + 4 (√3/2)²

-√3 – √3 – 6/4 + 12/4

(3 – 4√3)/2

= RHS

∴ LHS = RHS

Hence proved.

(vii) 3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1

Let us consider LHS:

3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4

3 sin 180^o/6 sec 180^o/3 – 4 sin 5(180^o)/6 cot 180^o/4

3 sin 30° sec 60° – 4 sin 150° cot 45°

3 sin 30° sec 60° – 4 sin (90° × 1 + 60°) cot 45°

We know that when n is odd, sin → cos.

3 sin 30° sec 60° – 4 cos 60° cot 45°

3 (1/2) (2) – 4 (1/2) (1)

3 – 2

1

= RHS

∴ LHS = RHS

Hence proved.

3. Prove that:

(i)

(ii)

(iii)

(iv)

(v)

Solution:

(i)

1 = RHS

∴ LHS = RHS

Hence proved.

(ii)

1 + 1

2 = RHS

∴ LHS = RHS

Hence proved.

(iii)

1 = RHS

∴ LHS = RHS

Hence proved.

(iv)

{1 + cot x – (-cosec x)} {1 + cot x + (-cosec x)}

{1 + cot x + cosec x} {1 + cot x – cosec x}

{(1 + cot x) + (cosec x)} {(1 + cot x) – (cosec x)}

By using the formula, (a + b) (a – b) = a² – b²

(1 + cot x)² – (cosec x)²

1 + cot² x + 2 cot x – cosec² x

We know that 1 + cot² x = cosec² x

cosec² x + 2 cot x – cosec² x

2 cot x = RHS

∴ LHS = RHS

Hence proved.

(v)

1 = RHS

∴ LHS = RHS

Hence proved.

4. Prove that: sin² π/18 + sin² π/9 + sin² 7π/18 + sin² 4π/9 = 2

Solution:

Let us consider LHS:

sin² π/18 + sin² π/9 + sin² 7π/18 + sin² 4π/9

sin² π/18 + sin² 2π/18 + sin² 7π/18 + sin² 8π/18

sin² π/18 + sin² 2π/18 + sin² (π/2 – 2π/18) + sin² (π/2 – π/18)

We know that when n is odd, sin → cos.

sin² π/18 + sin² 2π/18 + cos² 2π/18 + cos² 2π/18

when rearranged,

sin² π/18 + cos² 2π/18 + sin² π/18 + cos² 2π/18

We know that sin² + cos²x = 1.

So,

1 + 1

2 = RHS

∴ LHS = RHS

Hence proved.