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

  • Last Updated : 05 Mar, 2021

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

(i) sin 5π/3

Solution:

We have, sin 5π/3

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=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)-3/4cosec2 (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) \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(π/2+θ)= -cotθ]

=\frac{(cos x) (cosec x)(-cot x)}{(-cosec x)(cos x)(cot(π+x))}                                                     [∵sec(π/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(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} \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(π/2-θ)= cotθ     ∵sin(π/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(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) \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: 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)

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 =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 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)




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