RD Sharma Solutions for Class 11 Maths Chapter 5 Trigonometric Functions

RD Sharma Solutions Class 11 Maths Chapter 5 – Free PDF Download

RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions are provided here for students to study and score good marks in their board exams. In earlier classes, students studied trigonometric ratios for acute angles as the ratio of the sides of a right-angled triangle. In this chapter, we will extend the definitions of trigonometric ratios to any angle in terms of radian measure and study them as trigonometric functions. Experts at BYJU’S have designed RD Sharma Class 11 Solutions in a very lucid and clear manner that helps students solve problems in the most efficient possible ways. Students can download solutions in PDF format from the links provided below.

RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions contain three exercises. For effective learning of concepts, students are advised to practise RD Sharma Solutions as many times as possible and secure good marks in exams. Now, let us have a look at the concepts discussed in this chapter.

• Trigonometric functions of a real number
• Values of trigonometric functions
• Trigonometric identities
  • Fundamental trigonometric identities
• Signs of trigonometric functions
• Variations in values of trigonometric functions in different quadrants
• Values of trigonometric functions at allied angles
• Periodic functions
• Even and odd functions

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

Download PDF

carouselExampleControls111

Previous



Next

Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions

Exercise 5.1

Exercise 5.2

Exercise 5.3

Access answers to RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions

EXERCISE 5.1 PAGE NO: 5.18

Prove the following identities:

1. sec⁴ x – sec² x = tan⁴ x + tan² x

Solution:

Let us consider LHS: sec⁴ x – sec² x

(sec² x)² – sec² x

By using the formula, sec² θ = 1 + tan² θ.

(1 + tan² x) ² – (1 + tan² x)

1 + 2tan² x + tan⁴ x – 1 – tan² x

tan⁴ x + tan² x

= RHS

∴ LHS = RHS

Hence proved.

2. sin⁶ x + cos⁶ x = 1 – 3 sin² x cos² x

Solution:

Let us consider LHS: sin⁶ x + cos⁶ x

(sin² x) ³ + (cos² x) ³

By using the formula, a³ + b³ = (a + b) (a² + b² – ab)

(sin² x + cos² x) [(sin² x) ² + (cos² x) ² – sin² x cos² x]

By using the formula, sin² x + cos² x = 1 and a² + b² = (a + b) ² – 2ab

1 × [(sin² x + cos² x) ² – 2sin² x cos² x – sin² x cos² x

1² – 3sin² x cos² x

1 – 3sin² x cos² x

= RHS

∴ LHS = RHS

Hence proved.

3. (cosec x – sin x) (sec x – cos x) (tan x + cot x) = 1

Solution:

Let us consider LHS: (cosec x – sin x) (sec x – cos x) (tan x + cot x)

By using the formulas

cosec θ = 1/sin θ;

sec θ = 1/cos θ;

tan θ = sin θ / cos θ;

cot θ = cos θ / sin θ

Now,

1 = RHS

∴ LHS = RHS

Hence proved.

4. cosec x (sec x – 1) – cot x (1 – cos x) = tan x – sin x

Solution:

Let us consider LHS: cosec x (sec x – 1) – cot x (1 – cos x)

By using the formulas

cosec θ = 1/sin θ;

sec θ = 1/cos θ;

tan θ = sin θ / cos θ;

cot θ = cos θ / sin θ

Now,

By using the formula, 1 – cos²x = sin²x;

= RHS

∴ LHS = RHS

Hence Proved.

5.

Solution:

Let us consider the LHS.

By using the formula,

cosec θ = 1/sin θ;

sec θ = 1/cos θ;

Now,

sin x

= RHS

∴ LHS = RHS

Hence Proved.

6.

Solution:

Let us consider the LHS.

By using the formula,

tan θ = sin θ / cos θ;

cot θ = cos θ / sin θ

Now,

By using the formula, a³ – b³ = (a – b) (a² + b² + ab)

By using the formula,

cosec θ = 1/sin θ,

sec θ = 1/cos θ;

cosec x × sec x + 1

sec x cosec x + 1

=RHS

∴ LHS = RHS

Hence Proved.

7.

Solution:

Let us consider LHS.

By using the formula a³ ± b³ = (a ± b) (a² + b²∓ ab)

We know sin²x + cos²x = 1.

1 – sinx cosx + 1 + sinx cosx

2

= RHS

∴ LHS = RHS

Hence proved.

8. (sec x sec y + tan x tan y)² – (sec x tan y + tan x sec y)² = 1

Solution:

Let us consider LHS

(sec x sec y + tan x tan y)² – (sec x tan y + tan x sec y)²

Expanding the above equation, we get,

sec² x sec² y – sec² x tan² y + tan² x tan² y – tan² x sec² y

sec² x (sec² y – tan² y) + tan² x (tan² y – sec² y)

sec² x (sec² y – tan² y) – tan² x (sec² y – tan² y)

We know, sec² x – tan² x = 1.

sec² x × 1 – tan² x × 1

sec² x – tan² x

1 = RHS

∴ LHS = RHS

Hence proved.

9.

Solution:

Let us consider RHS.

= LHS

∴ LHS = RHS

Hence proved.

10.

Solution:

Let us consider LHS.

By using the formulas,

1 + tan²x = sec²x and 1 + cot²x = cosec²x

= RHS

∴ LHS = RHS

Hence proved.

11.

Solution:

Let us consider LHS.

By using the formula,

tan θ = sin θ / cos θ

cot θ = cos θ / sin θ

Now,

By using the formula, a³ + b³ = (a + b) (a² + b²– ab)

We know, sin² x + cos² x = 1.

1 – 1 + sin x cos x

Sin x cos x

= RHS

∴ LHS = RHS

Hence proved.

12.

Solution:

Let us consider LHS.

By using the formula,

cosec θ = 1/sin θ

sec θ = 1/cos θ

= RHS

∴ LHS = RHS

Hence proved.

13. (1 + tan α tan β) ² + (tan α – tan β) ² = sec² α sec² β

Solution:

Let us consider LHS: (1 + tan α tan β) ² + (tan α – tan β) ²

1+ tan² α tan² β + 2 tan α tan β + tan² α + tan² β – 2 tan α tan β

1 + tan² α tan² β + tan² α + tan² β

tan² α (tan² β + 1) + 1 (1 + tan² β)

(1 + tan² β) (1 + tan² α)

We know, 1 + tan² θ = sec² θ

So,

sec² α sec² β

= RHS

∴ LHS = RHS

Hence proved.

EXERCISE 5.2 PAGE NO: 5.25

1. Find the values of the other five trigonometric functions in each of the following:

(i) cot x = 12/5, x in quadrant III

(ii) cos x = -1/2, x in quadrant II

(iii) tan x = 3/4, x in quadrant III

(iv) sin x = 3/5, x in quadrant I

Solution:

(i) cot x = 12/5, x in quadrant III

In the third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative.

By using the formulas,

tan x = 1/cot x

= 1/(12/5)

= 5/12

cosec x = –√(1 + cot² x)

= –√(1 + (12/5)²)

= –√(25+144)/25

= –√(169/25)

= -13/5

sin x = 1/cosec x

= 1/(-13/5)

= -5/13

cos x = – √(1 – sin² x)

= – √(1 – (-5/13)²)

= – √(169-25)/169

= – √(144/169)

= -12/13

sec x = 1/cos x

= 1/(-12/13)

= -13/12

∴ sin x = -5/13, cos x = -12/13, tan x = 5/12, cosec x = -13/5, sec x = -13/12

(ii) cos x = -1/2, x in quadrant II

In the second quadrant, sin x and cosec x are positive. tan x, cot x, cos x, sec x are negative.

By using the formulas,

sin x = √(1 – cos² x)

= √(1 – (-1/2)²)

= √(4-1)/4

= √(3/4)

= √3/2

tan x = sin x/cos x

= (√3/2)/(-1/2)

= -√3

cot x = 1/tan x

= 1/-√3

= -1/√3

cosec x = 1/sin x

= 1/(√3/2)

= 2/√3

sec x = 1/cos x

= 1/(-1/2)

= -2

∴ sin x = √3/2, tan x = -√3, cosec x = 2/√3, cot x = -1/√3 sec x = -2

(iii) tan x = 3/4, x in quadrant III

In the third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative.

By using the formulas,

sin x = √(1 – cos² x)

= – √(1-(-4/5)²)

= – √(25-16)/25

= – √(9/25)

= – 3/5

cos x = 1/sec x

= 1/(-5/4)

= -4/5

cot x = 1/tan x

= 1/(3/4)

= 4/3

cosec x = 1/sin x

= 1/(-3/5)

= -5/3

sec x = -√(1 + tan² x)

= – √(1+(3/4)²)

= – √(16+9)/16

= – √ (25/16)

= -5/4

∴ sin x = -3/5, cos x = -4/5, cosec x = -5/3, sec x = -5/4, cot x = 4/3

(iv) sin x = 3/5, x in quadrant I

In the first quadrant, all trigonometric ratios are positive.

So, by using the formulas,

tan x = sin x/cos x

= (3/5)/(4/5)

= 3/4

cosec x = 1/sin x

= 1/(3/5)

= 5/3

cos x = √(1-sin² x)

= √(1 – (-3/5)²)

= √(25-9)/25

= √(16/25)

= 4/5

sec x = 1/cos x

= 1/(4/5)

= 5/4

cot x = 1/tan x

= 1/(3/4)

= 4/3

∴ cos x = 4/5, tan x = 3/4, cosec x = 5/3, sec x = 5/4, cot x = 4/3

2. If sin x = 12/13 and lies in the second quadrant, find the value of sec x + tan x.

Solution:

Given:

Sin x = 12/13, and x lies in the second quadrant.

We know, in the second quadrant, sin x and cosec x are positive, and all other ratios are negative.

By using the formulas,

Cos x = √(1-sin² x)

= – √(1-(12/13)²)

= – √(1- (144/169))

= – √(169-144)/169

= -√(25/169)

= – 5/13

We know,

tan x = sin x/cos x

sec x = 1/cos x

Now,

tan x = (12/13)/(-5/13)

= -12/5

sec x = 1/(-5/13)

= -13/5

Sec x + tan x = -13/5 + (-12/5)

= (-13-12)/5

= -25/5

= -5

∴ Sec x + tan x = -5

3. If sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2 find the value of 8 tan x -√5 sec y.

Solution:

Given:

sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2

We know that x is in the second quadrant and y is in the third quadrant.

In the second quadrant, cos x and tan x are negative.

In the third quadrant, sec y is negative.

By using the formula,

cos x = – √(1-sin² x)

tan x = sin x/cos x

Now,

cos x = – √(1-sin² x)

= – √(1 – (3/5)²)

= – √(1 – 9/25)

= – √((25-9)/25)

= – √(16/25)

= – 4/5

tan x = sin x/cos x

= (3/5)/(-4/5)

= 3/5 × -5/4

= -3/4

We know that sec y = – √(1+tan² y)

= – √(1 + (1/2)²)

= – √(1 + 1/4)

= – √((4+1)/4)

= – √(5/4)

= – √5/2

Now, 8 tan x – √5 sec y = 8(-3/4) – √5(-√5/2)

= -6 + 5/2

= (-12+5)/2

= -7/2

∴ 8 tan x – √5 sec y = -7/2

4. If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x.

Solution:

Given:

Sin x + cos x = 0, and x lies in the fourth quadrant.

Sin x = -cos x

Sin x/cos x = -1

So, tan x = -1 (since, tan x = sin x/cos x)

We know that, in the fourth quadrant, cos x and sec x are positive, and all other ratios are negative.

By using the formulas,

Sec x = √(1 + tan² x)

Cos x = 1/sec x

Sin x = – √(1- cos² x)

Now,

Sec x = √(1 + tan² x)

= √(1 + (-1)²)

= √2

Cos x = 1/sec x

= 1/√2

Sin x = – √(1 – cos² x)

= – √(1 – (1/√2)²)

= – √(1 – (1/2))

= – √((2-1)/2)

= – √(1/2)

= -1/√2

∴ sin x = -1/√2 and cos x = 1/√2

5. If cos x = -3/5 and π<x<3π/2 find the values of the other five trigonometric functions and hence, evaluate

Solution:

Given:

cos x= -3/5 and π <x < 3π/2

We know that in the third quadrant, tan x and cot x are positive, and all other rations are negative.

By using the formulas,

Sin x = – √(1-cos² x)

Tan x = sin x/cos x

Cot x = 1/tan x

Sec x = 1/cos x

Cosec x = 1/sin x

Now,

Sin x = – √(1-cos² x)

= – √(1-(-3/5)²)

= – √(1-9/25)

= – √((25-9)/25)

= – √(16/25)

= – 4/5

Tan x = sin x/cos x

= (-4/5)/(-3/5)

= -4/5 × -5/3

= 4/3

Cot x = 1/tan x

= 1/(4/3)

= 3/4

Sec x = 1/cos x

= 1/(-3/5)

= -5/3

Cosec x = 1/sin x

= 1/(-4/5)

= -5/4

∴
= [(-5/4) + (3/4)] / [(-5/3) – (4/3)]

= [(-5+3)/4] / [(-5-4)/3]

= [-2/4] / [-9/3]

= [-1/2] / [-3]

= 1/6

EXERCISE 5.3 PAGE NO: 5.39

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.