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

• Last Updated : 05 Mar, 2021

### (i) sin 5π/3

Solution:

We have, sin 5π/3

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

=sin (2π-π/3)                                             [∵sin(2π-θ)=-sinθ]

=-sin(π/3)

= – √3/2

### (ii) sin 17π

Solution:

We have, sin 17π

⇒sin 17π=sin (34×π/2)

Since, 17π lies in the negative x-axis i.e. between 2nd and 3rd quadrant

=sin 17π                                                [∵ sin nπ=0]

= 0

### (iii) tan11π/6

Solution:

Clearly, tan(11π/6) = tan ((12π-π)/6)

=tan (4π/2-π/6)

clearly, the angle lies in IV quadrant in which tangent  function is negative and the multiple of π/2 is even.

=tan (4π/2-π/6)= -cot (π/6)

=-1/√3

### (iv) cos (-25π/4)

Solution:

The cosine function is an even function, Therefore,

cos (-25π/4)=cos (25π/4)

Now, 25π/4=(12×π/2+π/4)

25π/4 lies in the I quadrant and even multiple of π/2

cos (25π/4)=cos (12×π/2+π/4)=cos π/4=1/√2

### (v) tan (7π/4)

Solution:

We have, 7π/4=(8π-π)/4 = 2π-π/4

=tan (2π-π/4)                                        [∵ tan(2π-θ)=-tanθ]

=-tan π/4

=-1

### (vi) sin 17π/6

Solution:

sin 17π/6= sin (3π-π/6)

=sin (2π+(π-π/6))

=sin (π-π/6)                                           [∵ sin(2π+θ)=sinθ]

=sin π/6                                                [∵ sin(π-θ)=sinθ]

=1/2

### (vii) cos 19π/6

Solution:

cos 19π/6 = cos (3π+(π+π/6))

= cos (2π+(π+π/6))

=cos (π+π/6)                       [∵ cos(2π+θ)= cosθ)]

=-cos π/6                            [∵ cos(π+θ)=-cosθ]

=-√3/2

### (viii) sin (-11π/6)

Solution:

sin (-11π/6) = sin (-(2π-π/6))

=sin (2π-π/6)                                         [∵ sin(-θ)= -sinθ]

=-(-sin π/6)                                           [∵ sin(2π-θ)= -sinθ]

=sin π/6

=1/2

### (ix) cosec (-20π/3)

Solution:

cosec (-20π/3)= cosec (-(7π-π/3))

= cosec (7π-π/3)                                     [∵ cosec(-θ) = -cosecθ]

= – cosec (2×3π+ (π-π/3))

= – cosec (π-π/3)

= – cosec π/3                                           [∵ cosec(π-θ)= cosecθ]

= – 2/√3

### (x) tan (-13π/4)

Solution:

tan (-13π/4) = -tan (13π/4)                     [∵ tan(-θ)=-tanθ]

=-tan (3π+π/4)

=- tan (2π+(π+π/4)                                 [∵ tan(2π+θ)=tanθ]

=-tan π/4                                                [∵ tan(π+θ)=tanθ]

= -1

### (xi) cos 19π/4

Solution:

cos 19π/4 = cos (5π-π/4))

= cos (2×2π+(π-π/4))                             [∵ cos(2nπ+θ)= cosθ , n ∈ N]

=cos (π-π/4)                                           [∵ cos(π-θ)= -cosθ]

=-cos π/4

=-1/√2

### (xii) sin (41π/4)

Solution:

sin (41π/4) = sin (10π+π/4)

=sin (2×5π+π/4)                                     [∵ sin(-θ)= -sinθ]

=sin π/4                                                  [∵ sin(2π-θ)= -sinθ]

=1/√2

### (xiii) cos 39π/4

Solution:

cos 39π/4 = cos (10π-π/4))

= cos (2×5π-π/4)

=cos π/4                                                 [∵ cos(2nπ-θ)= cosθ , n ∈ N]

=1/√2

### (xiv) sin (151π/6)

Solution:

sin (151π/6) = sin (25π+π/6)

=sin (2×12π+ (π +π/6))                          [∵ sin(2nπ+θ)= sinθ , n ∈ N]

=sin (π +π/6)                                          [∵ sin(π+θ)= -sinθ]

=-sin π/6

=-1/2

### (i) tan 225° cot 405°+tan 765° cot 675°=0

Solution:

Taking LHS

tan 225° cot 405°+tan 765° cot 675°

=tan (π+π/4) cot (2π+π/4)+tan (4π+π/4) cot (4π-π/4)

=tan(π/4)×cot(π/4)+tan(π/4)×{-cot(π/4)}                 [∵ cot(4π-π/4)=-cot(π/4)]

=1×1+1×(-1)

=0 = RHS     (Hence Proved)

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

Solution:

Taking LHS

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

=sin (3π-π/3) cos (4π-π/6)+cos (4π+π/3) sin (6π-π/6)

=sin (π/3) cos (π/6)+cos (π/3) {-sin (π/6)}        [∵ sin(6π-θ)= -sinθ]

=√3/2×√3/2+1/2×(-1/2)

=3/4-1/4

=2/4

=1/2= RHS        (Hence Proved)

### (iii) cos 24° + cos55° + cos125° + cos204° + cos300°=1/2

Solution:

Taking LHS

cos 24° + cos55° + cos125° + cos204° + cos300°

=cos 24° – cos ( π+24°) + cos 55° +cos (π-55°) + cos ( 2π-π/3)

=cos 24° – cos 24° + cos 55° – cos 55° + cos π/3

= cos π/3

= 1/2 = RHS          (Hence Proved)

### (iv) tan (-225°) cot (-405°)-tan (-765°) cot (675°) = 0

Solution:

Taking LHS

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

=-tan 225° {-cot 405°}+tan 765° cot 675°

=tan (π+π/4) cot (2π+π/4)+tan (4π+π/4) cot (4π-π/4)

=tan(π/4) cot(π/4)+tan(π/4)×{-cot(π/4)}                      [∵ cot(4π-π/4)=-cot(π/4)]

=1×1+1×(-1)

=1-1

=0 = RHS     (Hence Proved)

### (v) cos 570° sin 510° + sin (-330°) cos (-390°)=0

Solution:

Taking LHS

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

=cos (3π+π/6) sin (3π-π/6) – sin 330° cos 390°            [∵ sin(-θ)= -sinθ and cos(-θ)= cosθ]

=-cos π/6 sin π/6 + sin π/6 cos π/6                              [∵ sin(2π-θ)= -sinθ]

=0=RHS (Hence Proved)

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

Solution:

Taking LHS

tan (4π-π/3)- 2sin (2π/3)-3/4×(√2)2+4cos2 (3π-π/6)

=-tan π/3- 2sin (π-π/3)-3/4×2+4cos2 π/6                    [∵ tan(nπ-θ)=-tanθ   ∵cos(2nπ-θ)= -cosθ , n ∈ N]

=-√3 – 2sin π/3 -3/2+4×(√3/2)2

=-√3 – 2×(√3/2) -3/2+4×(3/4)

=-√3 – √3 -3/2+3

=-2√3+(-3+6)/2

=-2√3+3/2

=(3-4√3)/2=RHS (Hence Proved)

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

Solution:

Taking LHS

3sin (π/6) sec (π/3)- 4sin (5π/6) cot (π/4)

=3×(1/2)×2- 4sin (π-π/6)×1

=3 – 4sin π/6

=3-4×1/2

=3-2=1=RHS (Hence Proved)

### Question 3. Prove that:

(i) Solution: [∵tan(π/2+θ)= -cotθ]

= [∵sec(π/2+θ)= -cosecθ]

=1

=RHS (Hence Proved)

(ii) =2

Solution:

Taking LHS = [∵cot(π/2+θ)= -tanθ    ∵cot(2π+θ)= cotθ]]

=(sec x+cot(π/2+x))/(secx-tanx)+1

=(sec x-tanx)/(secx-tanx)+1

=1+1

=2=RHS (Hence Proved)

{iii} =1

Solution:

Taking LHS [∵ tan(π/2-θ)= cotθ     ∵sin(π/2+θ)= -cosθ]

= [∵ cotθ= cosθ/sinθ ∵ cosecθ= 1/sinθ]

= = 1 = RHS (Hence Proved)

### (iv) {1+cot x -sec(π/2+x)}{1+cot x + sec(π/2+x)}=2cot x

Solution:

Taking LHS

{1+cot x -sec(π/2+x)}{1+cot x + sec(π/2+x)}                             [∵ sec(π/2+θ)= -cosecθ]

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

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

=(1+cot x)2 -cosec2 x

=1+cot2 x+2cot x -cosec2 x                                                           [∵ 1+cot2 θ=cosec2θ]

=cosec2 x+2cot x -cosec2 x

=2cot x=RHS (Hence Proved)

(v) =1

Solution:

Taking LHS = =1=RHS (Hence Proved)

### Question 4. Prove that: sin2π/18+sin2π/9+sin27π/18+sin24π/9=2

Solution:

Taking LHS

sin2π/18+sin2π/9+sin27π/18+sin24π/9

=sin2(π/2-4π/9)+sin24π/9+sin2π/9+sin2(π/2-π/9)

=cos24π/9+sin24π/9+sin2π/9+cos2π/9

=1+1=2=RHS  (Hence Proved)

### Question 5. Prove that: sec(3π/2-x)sec(x-5π/2)+tan(5π/2+x)tan(x-3π/2)=-1

Solution:

Taking LHS:

sec(3π/2-x)sec(x-5π/2)+tan(5π/2+x)tan(x-3π/2)

=sec(3π/2-x)sec(-(5π/2-x))+tan(5π/2+x)tan(-(3π/2-x))         [∵ sec(-θ)= secθ]

=-cosec x sec(5π/2-x)-cot x (-tan(3π/2-x))

=-cosec x cosec x-cot x (-cot x)

=-cosec2 x + cot2 x

=-cosec2 x + cosec2 x – 1                                                      [∵1+cot2 θ=cosec2θ]

=-1=RHS (Hence Proved)

### (i) cos (A+B) + cos C = 0

Solution:

A+B+C=π

A+B=π-C ———-(1)

Taking LHS

cos (A+B) + cos C

Putting the value of A+B

cos (π-C) + cos C                                                 [∵ cos(π-θ)= -cosθ]

=-cos C + cos C

=0 = RHS (Hence Proved)

### (ii) cos (A+B)/2=sin C/2

Solution:

Taking LHS

cos (A+B)/2

Putting the value of A+B from (1)

=cos (π-C)/2

=cos (π/2-C/2)                                                       [∵ cos(π/2+θ)= sinθ]

=sin C/2 = RHS (Hence Proved)

### (iii) tan (A+B)/2=cot C/2

Solution:

Taking LHS

tan (A+B)/2

Putting the value of A+B from (1)

=tan(π-C)/2

=tan (π/2-C/2)                                                     [∵ tan(π/2-θ)= cotθ]

=cot C/2 = RHS (Hence Proved)

### cos(180°-A)+cos(180°+B)+cos(180°+C)-sin(90°+D)=0

Solution:

Since, A, B, C, D are the angles of a cyclic quadrilateral

Therefore, A+B+C+D=2π

or A+B=π or C+D=π

A=π-B also C=π-D

Taking LHS

cos(180°-A)+cos(180°+B)+cos(180°+C)-sin(90°+D)

=cos(π-(π-B))+cos(π+B)+cos(π+(π-D))-sin(π/2+D)                [∵ cos(π+θ)= -cosθ]

=cos B +(-cos B) +cos D -cos D

=cos B – cos B +0

=0 =RHS (Hence Proved)

### (i) cosec (π/2+θ) + x cos θ cot(π/2+θ)=sin(π/2+θ)

Solution:

We have,

cosec (π/2+θ) + x cos θ cot(π/2+θ)=sin(π/2+θ)

⇒ sec θ + x cos θ (-tanθ)=cos θ

⇒1/cosθ – x cos θ (sinθ/cosθ)=cos θ

⇒1/cosθ – x sinθ=cos θ

⇒1-x sinθcosθ/cosθ =cos θ

⇒1-x sinθcosθ =cos2 θ

⇒1-cos2θ =x sinθcosθ

⇒sin2θ =x sinθcosθ

⇒x=sinθ/cosθ

⇒x=tanθ

### (ii) x cot (π/2+θ) + tan (π/2+θ)sin θ+ cosec(π/2+θ)=0

Solution:

We have,

x cot (π/2+θ) + tan (π/2+θ)sin θ+ cosec(π/2+θ)=0

⇒-x tan θ – cot θ sin θ+ sec θ=0

⇒-x sin θ/cos θ – (cos θ/sin θ) sin θ+ 1/cos θ=0

⇒-x sin θ/cos θ – cos θ + 1/cos θ=0

⇒(-x sin θ – cos2θ + 1)/cos θ=0

⇒-x sin θ +1- cos2θ =0

⇒-x sin θ + sin2θ =0

⇒x sin θ = sin2θ =0

⇒x = sin θ

### (i) tan 4π – cos (3π/2)-sin (5π/6)cos (2π/3)=1/4

Solution:

Taking LHS

tan 4π – cos (3π/2)-sin (5π/6)cos (2π/3)                    [∵ tan nπ= 0, ∀ n∈ Z ]

=0- cos (π+π/2)-sin (π-π/6)cos(π/2-π/6)

=0- (cos π/2)- (sin π/6)(-sin π/6)

=0-0+sin2 π/6

=(1/2)2

=1/4=RHS (Hence Proved)

### (ii) sin (13π/3) sin (8π/3) + cos (2π/3)sin (5π/6)=1/2

Solution:

Taking LHS

sin (13π/3) sin (8π/3) + cos (2π/3)sin (5π/6)

=sin (4π+π/3) sin (3π-π/3) + cos (π/2+π/6)sin (π-π/6)         [∵ sin (4π+θ)= sinθ  & sin (3π-θ)= sinθ]

=sin π/3 sin π/3 + (-sin π/6) sin π/6

=(√3/2)×(√3/2)-(1/2)×(1/2)

=3/4-1/4

=2/4=1/2=RHS (Hence Proved)

### (iii) sin (13π/3) sin (2π/3) + cos (4π/3)sin (13π/6)=1/2

Solution:

Taking LHS

sin (13π/3) sin (2π/3) + cos (4π/3)sin (13π/6)

=sin (4π+π/3) sin (π/2-π/6) + cos (π+π/6)sin (2π+π/6)

=sin π/3 cos π/6 – cos π/3 sin π/6

=(√3/2)×(√3/2)-(1/2)×(1/2)

=3/4-1/4

=2/4=1/2=RHS (Hence Proved)

### (iv) sin (10π/3) cos (13π/6) + cos (8π/3)sin (5π/6)=-1

Solution:

Taking LHS

sin (10π/3) cos (13π/6) + cos (8π/3)sin (5π/6)

=sin (3π+π/3) cos (2π+π/6) + cos (3π-π/3)sin (π-π/6)

=-sin (π/3) cos (π/6) + cos π/3 (- sin π/6)                           [∵ sin (3π+θ)= -sinθ  & cos (3π-θ)= -cosθ]

=(-√3/2)×(-√3/2)-(1/2)×(1/2)

=-3/4-1/4

=-4/4=-1=RHS (Hence Proved)

### (V) tan (5π/4) cot (9π/4) + tan (17π/4) cot (15π/4)=0

Solution:

Taking LHS

tan (5π/4) cot (9π/4) + tan (17π/4) cot (15π/4)

=tan (π+π/4) cot (2π+π/4) + tan (4π+π/4) cot (4π-π/4)

=(tan π/4) (cot π/4) + (tan π/4) (-cot π/4)

=1.1+1.(-1)

=1-1=0 RHS (Hence Proved)

My Personal Notes arrow_drop_up