RD Sharma Solutions Class 11 Maths Chapter 5 – Trigonometric Functions (Updated For 2024)

RD Sharma Solutions Class 11 Maths Chapter 5

RD Sharma Solutions Class 11 Maths Chapter 5-Trigonometric Functions: RD Sharma Solutions Class 11 Maths Chapter 5 is here for you to prepare the 5th chapter (Trigonometric Functions) from your class 11 Mathematics syllabus. Chapter 5 from class 11 talks about Trigonometric Functions which is one of the Important chapters which needs clarity. And with a significant guide by side, getting thorough with the chapter might make you confident with the subject.

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RD Sharma Solutions Class 11 Maths Chapter 5

 


Exercise wise: RD Sharma Solutions Class 11 Maths Chapter 5

Class 11 RD Sharma Chapter 5 Exercise 5.1
Class 11 RD Sharma Chapter 5 Exercise 5.2
Class 11 RD Sharma Chapter 5 Exercise 5.3

Access RD Sharma Solutions Class 11 Maths Chapter 5

EXERCISE 5.1 PAGE NO: 5.18

Prove the following identities:

1. secx – secx = tanx + tanx

Solution:

Let us consider LHS: secx – secx

(secx)2 – secx

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

(1 + tanx) 2 – (1 + tanx)

1 + 2tanx + tanx – 1 – tanx

tanx + tanx

= RHS

∴ LHS = RHS

Hence proved.

2. sinx + cosx = 1 – 3 sinx cosx

Solution:

Let us consider LHS: sinx + cosx

(sinx) 3 + (cosx) 3

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

(sinx + cosx) [(sinx) 2 + (cosx) 2 – sinx cosx]

By using the formula, sinx + cosx = 1 and a2 + b2 = (a + b) 2 – 2ab

1 × [(sinx + cosx) 2 – 2sinx cosx – sinx cosx

12 – 3sinx cosx

1 – 3sinx cosx

= 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,

rd sharma solutions for class 11 maths chapter 5

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,

rd 51

By using the formula, 1 – cos2x = sin2x;

rd 52

= RHS

∴ LHS = RHS

Hence Proved.

5. 

rd 53

Solution:

Let us consider the LHS:

rd 54

By using the formula,

cosec θ = 1/sin θ;

sec θ = 1/cos θ;

Now,

rd 55

rd 56

sin x

= RHS

∴ LHS = RHS

Hence Proved.

6.

rd 5

Solution:

Let us consider the LHS:

rd 57

By using the formula,

tan θ = sin θ / cos θ;

cot θ = cos θ / sin θ

Now,

rd 58

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

rd 59

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. 

rd 510

Solution:

Let us consider LHS:

rd 511

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

rd 512

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,

[(sec x sec y)2 + (tan x tan y)2 + 2 (sec x sec y) (tan x tan y)] – [(sec x tan y)2 + (tan x sec y)2 + 2 (sec x tan y) (tan x sec y)] [secx sec2 y + tanx tan2 y + 2 (sec x sec y) (tan x tan y)] – [secx tan2 y + tanx sec2 y + 2 (secx tan2 y) (tan x sec y)]

secx sec2 y – secx tan2 y + tanx tan2 y – tanx sec2 y

secx (sec2 y – tan2 y) + tanx (tan2 y – sec2 y)

secx (sec2 y – tan2 y) – tanx (sec2 y – tan2 y)

We know, secx – tanx = 1.

secx × 1 – tanx × 1

secx – tanx

1 = RHS

∴ LHS = RHS

Hence proved.

9.

rd 513

Solution:

Let us Consider RHS:

rd 514

rd 515 1

rds 516

rd 517

= LHS

∴ LHS = RHS

Hence Proved.

10.

rd 518

Solution:

Let us consider LHS:

rd 518 1

By using the formulas,

1 + tan2x = sec2x and 1 + cot2x = cosec2x

rd 520

rd 521

= RHS

∴ LHS = RHS

Hence Proved.

11

rd 5 1

Solution:

Let us consider LHS:

rd 522

By using the formula,

tan θ = sin θ / cos θ;

cot θ = cos θ / sin θ

Now,

rd 523

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

rd 524

We know, sinx + cosx = 1.

1 – 1 + sin x cos x

Sin x cos x

= RHS

∴ LHS = RHS

Hence proved.

12. 

rd 5 2

Solution:

Let us consider LHS:

rd 525

By using the formula,

cosec θ = 1/sin θ,

sec θ = 1/cos θ;

rd 526

rd 527

= 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 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 other five trigonometric functions and hence evaluate RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions image - 32

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

∴ RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions image - 33
= [(-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

= 300o

= (90×3 + 30)o

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

sin 5π/3 = sin (300)o

= sin (90×3 + 30)o

= – cos 30o

= – √3/2

(ii) sin 17π

Sin 17π = sin 3060o

= sin (90×34 + 0)o

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

Sin 17π = sin (90×34 + 0)o

= – sin 0o

= 0

(iii) tan 11π/6

tan 11π/6 = (11/6 × 180)o

= 330o

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

tan 11π/6 = tan (300)o

= tan (90×3 + 60)o

= – cot 60o

= – 1/√3

(iv) cos (-25π/4)

cos (-25π/4) = cos (-1125)o

= cos (1125)o

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

cos (1125)o = cos (90×12 + 45)o

= cos 45o

= 1/√2

(v) tan 7π/4

tan 7π/4 = tan 315o

= tan (90×3 + 45)o

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

tan 315o = tan (90×3 + 45)o

= – cot 45o

= -1

(vi) sin 17π/6

sin 17π/6 = sin 510o

= sin (90×5 + 60)o

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

sin 510o = sin (90×5 + 60)o

= cos 60o

= 1/2

(vii) cos 19π/6

cos 19π/6 = cos 570o

= cos (90×6 + 30)o

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

cos 570o = cos (90×6 + 30)o

= – cos 30o

= – √3/2

(viii) sin (-11π/6)

sin (-11π/6) = sin (-330o)

= – sin (90×3 + 60)o

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

sin (-330o) = – sin (90×3 + 60)o

= – (-cos 60o)

= – (-1/2)

= 1/2

(ix) cosec (-20π/3)

cosec (-20π/3) = cosec (-1200)o

= – cosec (1200)o

= – cosec (90×13 + 30)o

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

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

= – sec 30o

= -2/√3

(x) tan (-13π/4)

tan (-13π/4) = tan (-585)o

= – tan (90×6 + 45)o

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

tan (-585)o = – tan (90×6 + 45)o

= – tan 45o

= -1

(xi) cos 19π/4

cos 19π/4 = cos 855o

= cos (90×9 + 45)o

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

cos 855o = cos (90×9 + 45)o

= – sin 45o

= – 1/√2

(xii) sin 41π/4

sin 41π/4 = sin 1845o

= sin (90×20 + 45)o

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

sin 1845o = sin (90×20 + 45)o

= sin 45o

= 1/√2

(xiii) cos 39π/4

cos 39π/4 = cos 1755o

= cos (90×19 + 45)o

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

cos 1755o = cos (90×19 + 45)o

= sin 45o

= 1/√2

(xiv) sin 151π/6

sin 151π/6 = sin 4530o

= sin (90×50 + 30)o

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

sin 4530o = sin (90×50 + 30)o

= – sin 30o

= -1/2

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

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

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

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

(v) cos 570o sin 510o + sin (-330o) cos (-390o) = 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 225o cot 405o + tan 765o cot 675o = 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 24o + cos 55o + cos 125o + cos 204o + cos 300o = 1/2

Let us consider LHS:

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

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

Let us consider LHS:

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

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

[-tan (225°)] [-cot (405°)] – [-tan (765°)] cot (675°)

 

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

Let us consider LHS:

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

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

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

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

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

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

 

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

rd 5 5

(ii)

 

rd 5 6

(iii)

rd 5 7

(iv)

rd 5 8

(v)

rd 5 9

Solution:

(i) 

rd 528

1 = RHS

∴ LHS = RHS

Hence proved.

(ii)

rd 529

1 + 1

2 = RHS

∴ LHS = RHS

Hence proved.

(iii)

rd 531

1 = RHS

∴ LHS = RHS

Hence proved.

(iv)

rd 532

{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 + cotx + 2 cot x – cosecx

We know that 1 + cotx = cosecx

cosecx + 2 cot x – cosecx

2 cot x = RHS

∴ LHS = RHS

Hence proved.

(v)

rd 533

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.

Important Topics Class 11 Maths Chapter 23

RD Sharma Solutions Class 11 Maths Chapter 5, undoubtedly is one of the chapters that students need to take seriously from the very beginning of the course curriculum. So, to make it better, have a quick look at the important topics that are there, that need to be discussed several times before appearing for the exam. Besides helping you with scoring well, these RD Sharma Solutions will let you enhance your exam performance.

  • 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.

Access Other Important Chapters of RD Sharma Solutions Class 11 Maths 

Hence paying focus in this chapter is what you need to do right now. Have your copy of RD Sharma Solutions CBSE Class 11 Maths Chapter 5 PDF and begin your exam preparation.

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