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

**Question 1. Find the values of the following trigonometric ratios:**

**(i) sin 5π/3**

**Solution:**

We have, sin 5π/3

=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

**Question 2. Prove that:**

**(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)-3/4cosec**^{2} (π/4)+4cos^{2} (17π/6)=(3-4√3)/2

^{2}(π/4)+4cos

^{2}(17π/6)=(3-4√3)/2

**Solution:**

Taking LHS

tan (4π-π/3)- 2sin (2π/3)-3/4×(√2)

^{2}+4cos^{2}(3π-π/6)=-tan π/3- 2sin (π-π/3)-3/4×2+4cos

^{2}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}-cosec^{2}x=1+cot

^{2}x+2cot x -cosec^{2}x [∵ 1+cot^{2}θ=cosec^{2}θ]=cosec

^{2}x+2cot x -cosec^{2}x=2cot x=RHS (Hence Proved)

**(v) **=1

**Solution:**

Taking LHS

=

=1=RHS (Hence Proved)

**Question 4. Prove that: sin**^{2}π/18+sin^{2}π/9+sin^{2}7π/18+sin^{2}4π/9=2

^{2}π/18+sin

^{2}π/9+sin

^{2}7π/18+sin

^{2}4π/9=2

**Solution:**

Taking LHS

sin

^{2}π/18+sin^{2}π/9+sin^{2}7π/18+sin^{2}4π/9=sin

^{2}(π/2-4π/9)+sin^{2}4π/9+sin^{2}π/9+sin^{2}(π/2-π/9)=cos

^{2}4π/9+sin^{2}4π/9+sin^{2}π/9+cos^{2}π/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)

=-cosec

^{2}x + cot^{2}x=-cosec

^{2}x + cosec^{2}x – 1 [∵1+cot^{2}θ=cosec^{2}θ]=-1=RHS (Hence Proved)

**Question 6. In a △ABC, prove that:**

**(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)

**Question 7. If A,B,C,D be the angles of a cyclic quadrilateral, taken in order, prove that**

**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)

**Question 8. Find x from the following equations**

**(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 =cos

^{2}θ⇒1-cos

^{2}θ =x sinθcosθ⇒sin

^{2}θ =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 – cos

^{2}θ + 1)/cos θ=0⇒-x sin θ +1- cos

^{2}θ =0⇒-x sin θ + sin

^{2}θ =0⇒x sin θ = sin

^{2}θ =0⇒x = sin θ

**Question 9. Prove that:**

**(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+sin

^{2}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)

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