# Maths F.Sc Part 1 Exercise 10.1

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```					mathcity.org                          Exercise 10.1 (Solutions)
Textbook of Algebra and Trigonometry for Class XI
Merging man and maths             Available online @ http://www.mathcity.org, Version: 1.0.0

Question # 1
3
(i)   sin(−780) = − sin 780 = − sin ( 4(90) + 60 ) = − sin(60) = −      *
2
(ii) cot(−855o ) = − cot855o = − cot ( 9(90) + 45)
= − ( − tan 45o ) = tan 45o = 1 *            Q 855 is in the IInd quad.
(iii) csc(2040o ) = csc ( 22(90) + 60 ) = − csc(60)        Q 2040o is in the Ist quad.
1            1       2
=−             =−        =−       *
sin ( 60 )      3         3
2
(iv) sec(−960) = sec(960) = sec (10(90) + 60 ) = − sec 60o Q 960o is in the IIIrd quad.
1          1
=−            =−      = −2 *
cos 60o       1
2
1
(v) tan(1110) = tan (12(90) + 30 ) = tan(30) =          *  Q 1110o is in the Ist quad
3
(vi) sin(−300) = − sin(300) = − sin ( 3(90) + 30 )
3
= −(− cos30o ) = cos30o =        *           Q 300o is in the IIIrd quad.
2
Question # 2
(i) sin196o = sin(180 + 16) = sin180 ocos16o + cos180o sin16o
= (0)cos16o + (−1)sin16o = − sin16o *

(ii) cos147o = cos(180 − 33) = cos180 ocos33o + sin180o sin 33o
= (−1)cos33o + (0)sin 33o = − cos33o *

(iii) sin 319o = sin(360 − 41) = sin 360o cos 41o − cos360o sin 41o          Do yourself
(iv) cos 254o = cos(270 − 16)                                                Do yourself
o
sin 294         sin(270 + 24)
(v) tan 294o =           o
=
cos 294 cos(270 + 24)
sin 270o cos 24o + cos 270o sin 24o (−1)cos 24o + (0)sin 24o
=                                          =
cos 270o cos 24o − sin 270o sin 24o (0)cos 24o − (−1)sin 24o
− cos 24o + 0 − cos 24o
=               o
=          o
= − cot 24o *
0 + sin 24          sin 24
Alternative
tan 270o + tan 24o
tan 294o = tan(270 + 24) =
1 − tan 270o tan 24o
o       tan 24o          tan 24o 
tan 270 1 +                       1 + ∞ 
      tan 270o   =             
=
       1                 1          
tan 270o              o
− tan 24o   − tan 24o 
 tan 270                 ∞          
=
(1 + 0 ) = − 1 = − cot 24o *
(
0 − tan 24o    )      tan 24o
FSc-I / 10.1 - 2
(vi) cos 728o = cos(720 + 8)                                                 Do yourself

(vii) sin(−625o ) = − sin 625o = − sin(630 − 5) = − ( sin 630o cos5o − cos630o sin 5o )
= − ( (−1)cos5o − (0)sin 5o ) = − ( − cos5o − 0 ) = cos5o *
(viii) cos(−435o ) = cos 435o = cos(450 − 15)                         Do yourself

Question # 3
(i) L.H.S = sin(180 + α )sin(90 − α )
= ( sin180o cosα + cos180o sin α ) )( sin 90o cosα − cos90o sin α )
= ( (0)cosα + (−1)sin α ) )( (1)cosα − (0)sin α )
= ( 0 − sin α ) )( cosα − 0 ) = − sin α cosα = R.H.S     *
(ii) First we calculate
3
sin 780o = sin(720 + 60) = sin(2 × 360 + 60) = sin 60o =
2
o                         o     o         o
sin 480 = sin(450 + 30) = sin 450 cos30 + cos 450 sin 30o
3
= (1)cos30 + (0)sin 30 = cos30 + 0 =
2
1                         1
cos120 = −        and     sin 30 =
2                         2
So                     o       o         o
L.H.S = sin 780 sin 480 + cos120 sin 30o
 3  3   1  1  3 1 1
=          +  −   = − = = R.H.S                    *
 2  2   2  2  4 4 2
(iii) First we calculate
cos306o = cos(270 + 36) = cos 270o cos36o − sin 270o sin 36o
= (0)cos36o − (−1)sin 36o = 0 + sin 36o = sin 36o
cos 234o = cos(270 − 36) = cos 270cos36 + sin 270cos36
= (0)cos36o + (−1)sin 36o = 0 − sin 36o = − sin 36o
cos162o = cos(180 − 18) = cos180o cos18o + sin180o sin18o
= (−1)cos18 + (0)sin18 = − cos18 + 0 = − cos18
So     L.H.S = sin 306o + cos 234o + cos162o + cos18o
= sin 36o − sin 36o − cos18o + cos18o = 0 = R.H.S *
(iv) First we calculate (Alternative Method)
3
cos330o = cos(360 − 30) = cos(−30o ) = cos(30o ) =
2
3
sin 600 = sin(6 × 90 + 60) = − sin 60 = −               Q 600o is in the IIIrd quad
2
1
cos120o = cos(90 + 30) = − sin 30 = −                   Q 120o is in the IInd quad
2
1
sin150o = sin(90 + 60) = cos60o =         Q 150o is in the IInd quad
2
So                   o        o
L.H.S = cos330 sin 600 + cos120o sin150o
 3     3   1  1       3 1 4
=    −     +  −   = − − = − = −1 = R.H.S *
 2  2   2  2          4 4 4
FSc-I / 10.1 - 3
Question # 4
First we calculate
sin(π + θ ) = sin π cosθ + cos π sin θ = (0)cosθ + (−1)sin θ
= 0 − sinθ = − sinθ
 3π                 π                             3π
tan     + θ  = tan  3 ⋅ + θ  = − cotθ             Q        + θ is in the IVth quad
 2                  2                              2
 3π               π                               3π
cot     − θ  = cot  3 ⋅ − θ  = tanθ               Q        − θ is in the IIIrd quad
 2                2                                2
cos(π − θ ) = cos π cosθ + sin π sin θ = (−1)cosθ + (0)sin θ
= − cosθ + 0 = − cosθ
csc(2π − θ ) = csc(−θ ) = − cscθ
Now
 3π    
sin 2 (π + θ ) tan        +θ 
L.H.S =                                  2     
 3π       
cot       − θ  cos 2 (π − θ ) csc(2π − θ )
 2        
(− sin θ ) 2 ( − cot θ )          sin 2 θ ( − cot θ )
=                                      =
( tan θ ) (− cosθ )2 (− cscθ ) tan θ cos θ (− cscθ )
2                              2        2

cosθ
sin 2 θ
sin θ          sin θ cosθ
=                           =              = cosθ = R.H.S       *
sin 2 θ           1            sin θ
cos 2 θ
cos θ
2
sin θ
Question 4 (ii)
First we calculate
cos(90 + θ ) = − sin θ                          Q 90 + θ is in the IInd quad.
sec(−θ ) = secθ
tan(180 − θ ) = tan ( 2(90) − θ ) = − tanθ      Q 180 − θ is in the IInd quad.
sec(360 − θ ) = sec(−θ ) = secθ
sin(180 + θ ) = sin ( 2(90) + θ ) = − sinθ      Q 180 + θ is in the IIIrd quad.
cot(90 − θ ) = tan θ                            Q 90 − θ is in the Ist quad.
Now
cos(90 + θ ) sec(−θ ) tan(180 − θ )
L.H.S =
sec(360 − θ ) sin(180 + θ ) cot(90 − θ )
(− sin θ )secθ (− tan θ )
=                             = 1 = R.H.S *
secθ (− sin θ ) (− tan θ )

Question # 5 (i)
Since α , β and γ are angels of triangle therefore
α + β + γ = 180      ⇒ α + β = 180 − γ
Now L.H.S = sin(α + β ) = sin(180 − γ )
= sin180 cos γ − cos180 sin γ
= (0)cos γ − (−1)sin γ = 0 + sinγ = sinγ = R.H.S *

Question # 5 (ii)
Since α , β and γ are angels of triangle therefore
α + β + γ = 180
α + β 180 − γ
⇒ α + β = 180 − γ        ⇒        =
2     2
FSc-I / 10.1 - 4
α + β          180 − γ         180 γ 
Now L.H.S = cos         = cos           = cos       − 
 2             2               2     2
      γ               γ              γ
= cos  90 −  = cos90 cos + sin 90 sin
      2               2              2
γ          γ            γ        γ
= (0) cos + (1) sin = 0 + sin = sin = R.H.S *
2         2            2        2
Question # 5 (iii)
Since α , β and γ are angels of triangle therefore
α + β + γ = 180      ⇒ α + β = 180 − γ
Now L.H.S = cos(α + β ) = cos(180 − γ )
= cos180 cos γ + sin180 sin γ
= (−1)cos γ + (0)sin γ = − cos γ + 0 = − cosγ = R.H.S *

Question # 5 (iv)
Since α , β and γ are angels of triangle therefore
α + β + γ = 180        ⇒ α + β = 180 − γ
Now L.H.S = tan(α + β ) + tan γ = tan(180 − γ ) + tan γ
tan180 − tan γ
=                   + tan γ
1 + tan180 tan γ
(0) − tan γ            − tan γ
=               + tan γ =         + tan γ
1 + (0) tan γ             1+ 0
= − tan γ + tan γ = 0 = R.H.S *

Made by: Atiq ur Rehman (mathcity@gmail.com) http://www.mathcity.org

Remember:
● sin (α + β ) = sin α cos β + cosα sin β      ● sin (α − β ) = sin α cos β − cosα sin β
● cos (α + β ) = cosα cos β − sin α sin β           ● cos (α − β ) = cosα cos β + sin α sin β
tan α + tan β                                      tan α − tan β
● tan (α + β ) =                                    ● tan (α − β ) =
1 − tan α tan β                                    1 + tan α tan β
 π         
Three Steps to solve sin  n ⋅ ± θ 
 2         
Step I: First check that n is even or odd
Step II: If n is even then the answer will be in sin and if the n is odd then sin will be
converted to cos and vice virsa (i.e. cos will be converted to sin).
π
Step III: Now check in which quadrant n ⋅ ± θ is lying if it is in Ist or IInd quadrant
2
the answer will be positive as sin is positive in these quadrant and if it is in the IIIrd or
e.g.           sin 667 o = sin ( 7(90) + 37 )
Since n = 7 is odd so answer will be in cos and 667 is in IVth quadrant and sin is
–ive in IVth quadrant therefore answer will be in negative. i.e sin 667o = − cos37
Similar technique is used for other trigonometric rations. i.e tan € cot and sec € csc .

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Description: Mathematic Solved Exercises For 1st Year or F.Sc Part 1