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

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

(i) sin 5Ï€/3

Solution:

We have, sin 5Ï€/3

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

=-sin(Ï€/3)

= – √3/2

(ii) sin 17Ï€

Solution:

We have, sin 17Ï€

⇒sin 17π=sin (34×π/2)

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

=sin 17Ï€                                                [∵ sin nÏ€=0] 

= 0

(iii) tan11Ï€/6

Solution:

Clearly, tan(11Ï€/6) = tan ((12Ï€-Ï€)/6)

=tan (4Ï€/2-Ï€/6)

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

=tan (4Ï€/2-Ï€/6)= -cot (Ï€/6)

=-1/√3

(iv) cos (-25Ï€/4)

Solution:

The cosine function is an even function, Therefore, 

cos (-25Ï€/4)=cos (25Ï€/4)

Now, 25Ï€/4=(12×π/2+Ï€/4)  

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

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

(v) tan (7Ï€/4)

Solution:

We have, 7Ï€/4=(8Ï€-Ï€)/4 = 2Ï€-Ï€/4

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

=-tan π/4

=-1

(vi) sin 17Ï€/6

Solution:

sin 17Ï€/6= sin (3Ï€-Ï€/6)

=sin (2Ï€+(Ï€-Ï€/6))

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

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

=1/2

(vii) cos 19Ï€/6 

Solution:

cos 19Ï€/6 = cos (3Ï€+(Ï€+Ï€/6))

= cos (2Ï€+(Ï€+Ï€/6))

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

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

=-√3/2

(viii) sin (-11Ï€/6)

Solution:

sin (-11Ï€/6) = sin (-(2Ï€-Ï€/6))

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

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

=sin π/6

=1/2

(ix) cosec (-20Ï€/3)

Solution:

cosec (-20Ï€/3)= cosec (-(7Ï€-Ï€/3))

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

= – cosec (2×3Ï€+ (Ï€-Ï€/3))

= – cosec (Ï€-Ï€/3)           

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

= – 2/√3

(x) tan (-13Ï€/4)

Solution:

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

=-tan (3Ï€+Ï€/4)

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

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

= -1

(xi) cos 19Ï€/4  

Solution:

cos 19Ï€/4 = cos (5Ï€-Ï€/4))

= cos (2×2Ï€+(Ï€-Ï€/4))                             [∵ cos(2nÏ€+θ)= cosθ , n ∈ N]  

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

=-cos π/4

=-1/√2

(xii) sin (41Ï€/4)

Solution:

sin (41Ï€/4) = sin (10Ï€+Ï€/4)

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

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

=1/√2

(xiii) cos 39Ï€/4  

Solution:

cos 39Ï€/4 = cos (10Ï€-Ï€/4))

= cos (2×5π-π/4)

=cos Ï€/4                                                 [∵ cos(2nÏ€-θ)= cosθ , n ∈ N]

=1/√2

(xiv) sin (151Ï€/6)

Solution:

sin (151Ï€/6) = sin (25Ï€+Ï€/6)

=sin (2×12Ï€+ (Ï€ +Ï€/6))                          [∵ sin(2nÏ€+θ)= sinθ , n ∈ N]  

=sin (Ï€ +Ï€/6)                                          [∵ sin(Ï€+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 (Ï€+Ï€/4) cot (2Ï€+Ï€/4)+tan (4Ï€+Ï€/4) cot (4Ï€-Ï€/4)

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

=1×1+1×(-1)

=0 = RHS     (Hence Proved)

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

Solution:

Taking LHS

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

=sin (3Ï€-Ï€/3) cos (4Ï€-Ï€/6)+cos (4Ï€+Ï€/3) sin (6Ï€-Ï€/6)

=sin (Ï€/3) cos (Ï€/6)+cos (Ï€/3) {-sin (Ï€/6)}        [∵ sin(6Ï€-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 ( Ï€+24°) + cos 55° +cos (Ï€-55°) + cos ( 2Ï€-Ï€/3)

=cos 24° – cos 24° + cos 55° – cos 55° + cos Ï€/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 (Ï€+Ï€/4) cot (2Ï€+Ï€/4)+tan (4Ï€+Ï€/4) cot (4Ï€-Ï€/4)

=tan(Ï€/4) cot(Ï€/4)+tan(Ï€/4)×{-cot(Ï€/4)}                      [∵ cot(4Ï€-Ï€/4)=-cot(Ï€/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 (3Ï€+Ï€/6) sin (3Ï€-Ï€/6) – sin 330° cos 390°            [∵ sin(-θ)= -sinθ and cos(-θ)= cosθ]

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

=-√3 – 2sin Ï€/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 (Ï€/6) sec (Ï€/3)- 4sin (5Ï€/6) cot (Ï€/4)=1

Solution:

Taking LHS

3sin (Ï€/6) sec (Ï€/3)- 4sin (5Ï€/6) cot (Ï€/4)

=3×(1/2)×2- 4sin (π-π/6)×1

=3 – 4sin Ï€/6

=3-4×1/2

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

Question 3. Prove that:

(i) \frac{cos(2π+x)cosec(2π+x)tan(\frac{π}{2}+x)}{sec(\frac{π}{2}+x)cos x cot(π+x)}

Solution:

\frac{cos(2π+x)cosec(2π+x)tan(\frac{π}{2}+x)}{sec(\frac{π}{2}+x)cos x cot(π+x)}                                              [∵tan(Ï€/2+θ)= -cotθ]

=\frac{(cos x) (cosec x)(-cot x)}{(-cosec x)(cos x)(cot(π+x))}                                                     [∵sec(Ï€/2+θ)= -cosecθ]

=1

=RHS (Hence Proved)

(ii) \frac{cosec(90°+x)+cot(450°+x)}{cosec(90°-x)+tan(180°-x)}+\frac{tan(180°+x)+sec(180°-x)}{tan(360°+x)-sec(-x)}     =2

Solution: 

Taking LHS

 \frac{cosec(90°+x)+cot(450°+x)}{cosec(90°-x)+tan(180°-x)}+\frac{tan(180°+x)+sec(180°-x)}{tan(360°+x)-sec(-x)}

= \frac{sec x+cot(2π+π/2+x)}{secx-tanx}+\frac{tanx-secx}{tanx-secx}                                 [∵cot(Ï€/2+θ)= -tanθ    âˆµcot(2Ï€+θ)= cotθ]]

=(sec x+cot(Ï€/2+x))/(secx-tanx)+1

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

=1+1

=2=RHS (Hence Proved)

{iii} \frac{sin(π+x)cos(\frac{π}{2}+x)tan(\frac{3π}{2}-x)cot(2π-x)}{sin(2π-x)cos(2π+x)cosec(-x)sin(\frac{3π}{2}-x)}    =1

Solution:

Taking LHS

 \frac{sin(π+x)cos(\frac{π}{2}+x)tan(\frac{3π}{2}-x)cot(2π-x)}{sin(2π-x)cos(2π+x)cosec(-x)sin(\frac{3π}{2}-x)}             [∵ tan(Ï€/2-θ)= cotθ     ∵sin(Ï€/2+θ)= -cosθ]

= \frac{(sinx)(-sinx)(cotx)(-cotx)}{(-sinx)(cosx)(-cosecx)(-cosx)}                                  [∵ cotθ= cosθ/sinθ ∵ cosecθ= 1/sinθ]

= \frac{(sinx)(-sinx)(cosx)(-cosx)(-sinx)}{(-sinx)(cosx)(-cosx)(sinx)(-sinx)}

= 1 = RHS (Hence Proved)

(iv) {1+cot x -sec(Ï€/2+x)}{1+cot x + sec(Ï€/2+x)}=2cot x 

Solution:

Taking LHS

{1+cot x -sec(Ï€/2+x)}{1+cot x + sec(Ï€/2+x)}                             [∵ sec(Ï€/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) \frac{tan(\frac{π}{2}-x)sec(π-x)sin(-x)}{sin(π+x)cot(2π-x)cosec(\frac{π}{2}-x)}    =1

Solution:

Taking LHS

\frac{tan(\frac{π}{2}-x)sec(π-x)sin(-x)}{sin(π+x)cot(2π-x)cosec(\frac{π}{2}-x)}

=\frac{(cotx)(-secx)(-sinx)}{(-sinx)(-cotx)(secx)}

=1=RHS (Hence Proved)

Question 4. Prove that: sin2Ï€/18+sin2Ï€/9+sin27Ï€/18+sin24Ï€/9=2

Solution:

Taking LHS

sin2Ï€/18+sin2Ï€/9+sin27Ï€/18+sin24Ï€/9

=sin2(Ï€/2-4Ï€/9)+sin24Ï€/9+sin2Ï€/9+sin2(Ï€/2-Ï€/9)

=cos24Ï€/9+sin24Ï€/9+sin2Ï€/9+cos2Ï€/9

=1+1=2=RHS  (Hence Proved)

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

Solution:

Taking LHS:

sec(3Ï€/2-x)sec(x-5Ï€/2)+tan(5Ï€/2+x)tan(x-3Ï€/2)

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

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

Question 6. In a â–³ABC, prove that:

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

Solution:

A+B+C=Ï€

A+B=Ï€-C ———-(1)

Taking LHS

cos (A+B) + cos C

Putting the value of A+B 

cos (Ï€-C) + cos C                                                 [∵ cos(Ï€-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 (Ï€-C)/2

=cos (Ï€/2-C/2)                                                       [∵ cos(Ï€/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(Ï€-C)/2

=tan (Ï€/2-C/2)                                                     [∵ tan(Ï€/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=2Ï€

or A+B=Ï€ or C+D=Ï€

A=Ï€-B also C=Ï€-D

Taking LHS

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

=cos(Ï€-(Ï€-B))+cos(Ï€+B)+cos(Ï€+(Ï€-D))-sin(Ï€/2+D)                [∵ cos(Ï€+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 =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 

Question 9. Prove that:

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

Solution:

Taking LHS

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

=0- cos (Ï€+Ï€/2)-sin (Ï€-Ï€/6)cos(Ï€/2-Ï€/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 (13Ï€/3) sin (8Ï€/3) + cos (2Ï€/3)sin (5Ï€/6)=1/2

Solution:

Taking LHS

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

=sin (4Ï€+Ï€/3) sin (3Ï€-Ï€/3) + cos (Ï€/2+Ï€/6)sin (Ï€-Ï€/6)         [∵ sin (4Ï€+θ)= sinθ  & sin (3Ï€-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 (13Ï€/3) sin (2Ï€/3) + cos (4Ï€/3)sin (13Ï€/6)=1/2

Solution:

Taking LHS

sin (13Ï€/3) sin (2Ï€/3) + cos (4Ï€/3)sin (13Ï€/6)

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

=sin Ï€/3 cos Ï€/6 – cos Ï€/3 sin Ï€/6

=(√3/2)×(√3/2)-(1/2)×(1/2)

=3/4-1/4

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

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

Solution:

Taking LHS

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

=sin (3Ï€+Ï€/3) cos (2Ï€+Ï€/6) + cos (3Ï€-Ï€/3)sin (Ï€-Ï€/6)

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

=(-√3/2)×(-√3/2)-(1/2)×(1/2)

=-3/4-1/4

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

(V) tan (5Ï€/4) cot (9Ï€/4) + tan (17Ï€/4) cot (15Ï€/4)=0

Solution:

Taking LHS

tan (5Ï€/4) cot (9Ï€/4) + tan (17Ï€/4) cot (15Ï€/4)

=tan (Ï€+Ï€/4) cot (2Ï€+Ï€/4) + tan (4Ï€+Ï€/4) cot (4Ï€-Ï€/4)

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

=1.1+1.(-1)

=1-1=0 RHS (Hence Proved)



Last Updated : 05 Mar, 2021
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