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

### (i) sin 5Ï€/3

Solution:

We have, sin 5Ï€/3

=sin (2Ï€-Ï€/3)                                             [âˆµsin(2Ï€-Î¸)=-sinÎ¸]

=-sin(Ï€/3)

= – âˆš3/2

### (ii) sin 17Ï€

Solution:

We have, sin 17Ï€

â‡’sin 17Ï€=sin (34Ã—Ï€/2)

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

=sin 17Ï€                                                [âˆµ 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/âˆš3

### (iv) cos (-25Ï€/4)

Solution:

The cosine function is an even function, Therefore,

cos (-25Ï€/4)=cos (25Ï€/4)

Now, 25Ï€/4=(12Ã—Ï€/2+Ï€/4)

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

cos (25Ï€/4)=cos (12Ã—Ï€/2+Ï€/4)=cos Ï€/4=1/âˆš2

### (v) tan (7Ï€/4)

Solution:

We have, 7Ï€/4=(8Ï€-Ï€)/4 = 2Ï€-Ï€/4

=tan (2Ï€-Ï€/4)                                        [âˆµ tan(2Ï€-Î¸)=-tanÎ¸]

=-tan Ï€/4

=-1

### (vi) sin 17Ï€/6

Solution:

sin 17Ï€/6= sin (3Ï€-Ï€/6)

=sin (2Ï€+(Ï€-Ï€/6))

=sin (Ï€-Ï€/6)                                           [âˆµ sin(2Ï€+Î¸)=sinÎ¸]

=sin Ï€/6                                                [âˆµ sin(Ï€-Î¸)=sinÎ¸]

=1/2

### (vii) cos 19Ï€/6

Solution:

cos 19Ï€/6 = cos (3Ï€+(Ï€+Ï€/6))

= cos (2Ï€+(Ï€+Ï€/6))

=cos (Ï€+Ï€/6)                       [âˆµ cos(2Ï€+Î¸)= cosÎ¸)]

=-cos Ï€/6                            [âˆµ cos(Ï€+Î¸)=-cosÎ¸]

=-âˆš3/2

### (viii) sin (-11Ï€/6)

Solution:

sin (-11Ï€/6) = sin (-(2Ï€-Ï€/6))

=sin (2Ï€-Ï€/6)                                         [âˆµ sin(-Î¸)= -sinÎ¸]

=-(-sin Ï€/6)                                           [âˆµ sin(2Ï€-Î¸)= -sinÎ¸]

=sin Ï€/6

=1/2

### (ix) cosec (-20Ï€/3)

Solution:

cosec (-20Ï€/3)= cosec (-(7Ï€-Ï€/3))

= cosec (7Ï€-Ï€/3)                                     [âˆµ cosec(-Î¸) = -cosecÎ¸]

= – cosec (2Ã—3Ï€+ (Ï€-Ï€/3))

= – cosec (Ï€-Ï€/3)

= – cosec Ï€/3                                           [âˆµ cosec(Ï€-Î¸)= cosecÎ¸]

= – 2/âˆš3

### (x) tan (-13Ï€/4)

Solution:

tan (-13Ï€/4) = -tan (13Ï€/4)                     [âˆµ tan(-Î¸)=-tanÎ¸]

=-tan (3Ï€+Ï€/4)

=- tan (2Ï€+(Ï€+Ï€/4)                                 [âˆµ tan(2Ï€+Î¸)=tanÎ¸]

=-tan Ï€/4                                                [âˆµ tan(Ï€+Î¸)=tanÎ¸]

= -1

### (xi) cos 19Ï€/4

Solution:

cos 19Ï€/4 = cos (5Ï€-Ï€/4))

= cos (2Ã—2Ï€+(Ï€-Ï€/4))                             [âˆµ cos(2nÏ€+Î¸)= cosÎ¸ , n âˆˆ N]

=cos (Ï€-Ï€/4)                                           [âˆµ cos(Ï€-Î¸)= -cosÎ¸]

=-cos Ï€/4

=-1/âˆš2

### (xii) sin (41Ï€/4)

Solution:

sin (41Ï€/4) = sin (10Ï€+Ï€/4)

=sin (2Ã—5Ï€+Ï€/4)                                     [âˆµ sin(-Î¸)= -sinÎ¸]

=sin Ï€/4                                                  [âˆµ sin(2Ï€-Î¸)= -sinÎ¸]

=1/âˆš2

### (xiii) cos 39Ï€/4

Solution:

cos 39Ï€/4 = cos (10Ï€-Ï€/4))

= cos (2Ã—5Ï€-Ï€/4)

=cos Ï€/4                                                 [âˆµ cos(2nÏ€-Î¸)= cosÎ¸ , n âˆˆ N]

=1/âˆš2

### (xiv) sin (151Ï€/6)

Solution:

sin (151Ï€/6) = sin (25Ï€+Ï€/6)

=sin (2Ã—12Ï€+ (Ï€ +Ï€/6))                          [âˆµ sin(2nÏ€+Î¸)= sinÎ¸ , n âˆˆ N]

=sin (Ï€ +Ï€/6)                                          [âˆµ sin(Ï€+Î¸)= -sinÎ¸]

=-sin Ï€/6

=-1/2

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

Solution:

Taking LHS

tan 225Â° cot 405Â°+tan 765Â° cot 675Â°

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

=tan(Ï€/4)Ã—cot(Ï€/4)+tan(Ï€/4)Ã—{-cot(Ï€/4)}                 [âˆµ cot(4Ï€-Ï€/4)=-cot(Ï€/4)]

=1Ã—1+1Ã—(-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)}        [âˆµ sin(6Ï€-Î¸)= -sinÎ¸]

=âˆš3/2Ã—âˆš3/2+1/2Ã—(-1/2)

=3/4-1/4

=2/4

=1/2= RHS        (Hence Proved)

### (iii) cos 24Â° + cos55Â° + cos125Â° + cos204Â° + cos300Â°=1/2

Solution:

Taking LHS

cos 24Â° + cos55Â° + cos125Â° + cos204Â° + cos300Â°

=cos 24Â° – cos ( Ï€+24Â°) + cos 55Â° +cos (Ï€-55Â°) + cos ( 2Ï€-Ï€/3)

=cos 24Â° – cos 24Â° + cos 55Â° – cos 55Â° + cos Ï€/3

= cos Ï€/3

= 1/2 = RHS          (Hence Proved)

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

Solution:

Taking LHS

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

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

=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)Ã—{-cot(Ï€/4)}                      [âˆµ cot(4Ï€-Ï€/4)=-cot(Ï€/4)]

=1Ã—1+1Ã—(-1)

=1-1

=0 = RHS     (Hence Proved)

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

Solution:

Taking LHS

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

=cos (3Ï€+Ï€/6) sin (3Ï€-Ï€/6) – sin 330Â° cos 390Â°            [âˆµ sin(-Î¸)= -sinÎ¸ and cos(-Î¸)= cosÎ¸]

=-cos Ï€/6 sin Ï€/6 + sin Ï€/6 cos Ï€/6                              [âˆµ sin(2Ï€-Î¸)= -sinÎ¸]

=0=RHS (Hence Proved)

### (vi) tan (11Ï€/3)- 2sin (4Ï€/6)-3/4cosec2 (Ï€/4)+4cos2 (17Ï€/6)=(3-4âˆš3)/2

Solution:

Taking LHS

tan (4Ï€-Ï€/3)- 2sin (2Ï€/3)-3/4Ã—(âˆš2)2+4cos2 (3Ï€-Ï€/6)

=-tan Ï€/3- 2sin (Ï€-Ï€/3)-3/4Ã—2+4cos2 Ï€/6                    [âˆµ tan(nÏ€-Î¸)=-tanÎ¸   âˆµcos(2nÏ€-Î¸)= -cosÎ¸ , n âˆˆ N]

=-âˆš3 – 2sin Ï€/3 -3/2+4Ã—(âˆš3/2)2

=-âˆš3 – 2Ã—(âˆš3/2) -3/2+4Ã—(3/4)

=-âˆš3 – âˆš3 -3/2+3

=-2âˆš3+(-3+6)/2

=-2âˆš3+3/2

=(3-4âˆš3)/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)

=3Ã—(1/2)Ã—2- 4sin (Ï€-Ï€/6)Ã—1

=3 – 4sin Ï€/6

=3-4Ã—1/2

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

### Question 3. Prove that:

(i)

Solution:

[âˆµtan(Ï€/2+Î¸)= -cotÎ¸]

=                                                   [âˆµsec(Ï€/2+Î¸)= -cosecÎ¸]

=1

=RHS (Hence Proved)

(ii)=2

Solution:

Taking LHS

=                               [âˆµcot(Ï€/2+Î¸)= -tanÎ¸    âˆµcot(2Ï€+Î¸)= cotÎ¸]]

=(sec x+cot(Ï€/2+x))/(secx-tanx)+1

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

=1+1

=2=RHS (Hence Proved)

{iii}=1

Solution:

Taking LHS

[âˆµ tan(Ï€/2-Î¸)= cotÎ¸     âˆµsin(Ï€/2+Î¸)= -cosÎ¸]

=                                [âˆµ cotÎ¸= cosÎ¸/sinÎ¸ âˆµ cosecÎ¸= 1/sinÎ¸]

=

= 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)}                             [âˆµ sec(Ï€/2+Î¸)= -cosecÎ¸]

={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                                                           [âˆµ 1+cot2 Î¸=cosec2Î¸]

=cosec2 x+2cot x -cosec2 x

=2cot x=RHS (Hence Proved)

(v) =1

Solution:

Taking LHS

=

=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))         [âˆµ sec(-Î¸)= secÎ¸]

=-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                                                      [âˆµ1+cot2 Î¸=cosec2Î¸]

=-1=RHS (Hence Proved)

### (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                                                 [âˆµ cos(Ï€-Î¸)= -cosÎ¸]

=-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)                                                       [âˆµ cos(Ï€/2+Î¸)= sinÎ¸]

=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)                                                     [âˆµ tan(Ï€/2-Î¸)= cotÎ¸]

=cot C/2 = RHS (Hence Proved)

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

=cos(Ï€-(Ï€-B))+cos(Ï€+B)+cos(Ï€+(Ï€-D))-sin(Ï€/2+D)                [âˆµ cos(Ï€+Î¸)= -cosÎ¸]

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

=cos B – cos B +0

=0 =RHS (Hence Proved)

### (i) cosec (Ï€/2+Î¸) + x cos Î¸ cot(Ï€/2+Î¸)=sin(Ï€/2+Î¸)

Solution:

We have,

cosec (Ï€/2+Î¸) + x cos Î¸ cot(Ï€/2+Î¸)=sin(Ï€/2+Î¸)

â‡’ sec Î¸ + x cos Î¸ (-tanÎ¸)=cos Î¸

â‡’1/cosÎ¸ – x cos Î¸ (sinÎ¸/cosÎ¸)=cos Î¸

â‡’1/cosÎ¸ – x sinÎ¸=cos Î¸

â‡’1-x sinÎ¸cosÎ¸/cosÎ¸ =cos Î¸

â‡’1-x sinÎ¸cosÎ¸ =cos2 Î¸

â‡’1-cos2Î¸ =x sinÎ¸cosÎ¸

â‡’sin2Î¸ =x sinÎ¸cosÎ¸

â‡’x=sinÎ¸/cosÎ¸

â‡’x=tanÎ¸

### (ii) x cot (Ï€/2+Î¸) + tan (Ï€/2+Î¸)sin Î¸+ cosec(Ï€/2+Î¸)=0

Solution:

We have,

x cot (Ï€/2+Î¸) + tan (Ï€/2+Î¸)sin Î¸+ cosec(Ï€/2+Î¸)=0

â‡’-x tan Î¸ – cot Î¸ sin Î¸+ sec Î¸=0

â‡’-x sin Î¸/cos Î¸ – (cos Î¸/sin Î¸) sin Î¸+ 1/cos Î¸=0

â‡’-x sin Î¸/cos Î¸ – cos Î¸ + 1/cos Î¸=0

â‡’(-x sin Î¸ – cos2Î¸ + 1)/cos Î¸=0

â‡’-x sin Î¸ +1- cos2Î¸ =0

â‡’-x sin Î¸ + sin2Î¸ =0

â‡’x sin Î¸ = sin2Î¸ =0

â‡’x = sin Î¸

### (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)                    [âˆµ tan nÏ€= 0, âˆ€ nâˆˆ Z ]

=0- cos (Ï€+Ï€/2)-sin (Ï€-Ï€/6)cos(Ï€/2-Ï€/6)

=0- (cos Ï€/2)- (sin Ï€/6)(-sin Ï€/6)

=0-0+sin2 Ï€/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)         [âˆµ sin (4Ï€+Î¸)= sinÎ¸  & sin (3Ï€-Î¸)= sinÎ¸]

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

=(âˆš3/2)Ã—(âˆš3/2)-(1/2)Ã—(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

=(âˆš3/2)Ã—(âˆš3/2)-(1/2)Ã—(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)                           [âˆµ sin (3Ï€+Î¸)= -sinÎ¸  & cos (3Ï€-Î¸)= -cosÎ¸]

=(-âˆš3/2)Ã—(-âˆš3/2)-(1/2)Ã—(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 Ï€/4) (cot Ï€/4) + (tan Ï€/4) (-cot Ï€/4)

=1.1+1.(-1)

=1-1=0 RHS (Hence Proved)

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