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Chapter 8 – Introduction to Trigonometry

Introduction to Trigonometry

1.   In ?ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
       (i) Sin A, cos A                                          (ii) sin C, cos C
Sol. In right ?ABC, we have:
        p = 24 cm, b = 7 cm
        
2.   In the figure, find tan P – cot R.
Sol. In right ?PQR, using the Pythagoras theorem, we get
        
3.   If sin  calculate cos A and tan A.
Sol. Let us consider, the right ?ABC, we have
        Perp. = BC and Hyp. = AC
        
4.   Given 15 cot A = 8, find sin A and sec A.
Sol. Let in the right ?ABC, we have
        15 cot A = 8
        
        Now, using Pythagoras theorem, we get
        
5.   Given  calculate all other trigonometric ratios.
Sol. Let us have a right ?ABC in which ?B = 90°
        
        
6.   If ?A and ?B are acute angles such that cos A = cos B, then show that ?A = ?B.
Sol. Let us consider a right ?ABC,
        
        
7.   
Sol. Let us have a right ?ABC in which ?B = 90°, and ?A = ?
        
        
8.   If 3 cot A = 4, check whether 
Sol. Let us consider a right angled ?ABC in which ?B = 90°
        ?For ?A, we have:
        Base = AB and Perpendicular = BC. Also Hypotenuse = AC
        3 cot A = 4
        
        
9.   In triangle ABC, right-angled at B, if  find the value of:
        (i) sin A cos C + cos A sin C                        (ii) cos A cos C – sin A sin C
Sol. Let us consider a right ?ABC, in which ?B = 90°
        For ?A, we have
        Base = AB
        Perpendicular = BC
        Hypotenuse = AC
        
        
10.   In ?PQR, right-anlged at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Sol. It is given that PQR is a right ?, such that ?Q = 90°
        PR + QR = 25 cm
        and PQ = 5 cm
        Let QR = x cm
        ?PR = (25 – x)
        ?By Pythagoras theorem, we have
        PR2 = QR2 + PQ2
        ?(25 – x) = x2 + 52
        ?625 – 50x + x2 = x2 + 25
        ?–50x = –600
        
        
11.   State whether the following are true or false. Justify your answer.
        (i) The value of tan A is always less than,1.
        (ii)  for some valued of angle A.
        (iii) cos A is the abbreviation ?used for the cosecant of angle A.
        (iv) cot A is the product of cot and A.
        (v)  for some angle q.
Sol. False [? A tangent of an angle is ratio of sides other than hupotenuse, which may be equal or unequal to each other.]
        (ii) True ? cos A is always less than 1
              
        (iii) False [? ‘cosine A’ is abbreviated as ‘cos A’
        (iv) False [‘cot A’ is a single and meaningful term whereas ‘cot’ alone has no meaning.]
        (v) False [?  is greater than 1 and sin B cannot be greater than 1.]
Exercise 8.2
1.   Evaluate the following:
        
        
        
2.   Choose the correct option and justify your choice:
        
        (iii) When A = 0 then we have:
                sin 2A = sin 2(0°) = sin 0° = 0
                2 sin A = 2 sin 0 = 2 × 0 = 0
                i.e., sin 2A = 2 sin A for A = 0°
                Thus, the option (A) is correct
        
3.   
Sol. From the table, we have tan 
…(1)
        Also tan (A + B) = 
(Given) …(2)
        From (1) and (2), we get
             A + B = 60°
…(3)
        Similarly,
             A – B = 30°
…(4)
        Adding (3) and (4),
             2A = 90°.? A = 45°
        Subtracting (4) from (3), we get
             2B = 30° ? B = 15°.
4.   State whether the following are true or false. Justify your answer.,
        (i) sin (A + B) sin A + sin B.
        (ii) The value of sin? increases as ?increases.
        (iii) The value of cos? increases as ?increases.
        (iv) sin?= cos? for all values of q.
        (v) cot A is not defined for A = 0°.
Sol. (i) Let us take A = 30° and B = 60°
        Then LHS = sin (30° + 60°)
        = sin 90° = 1
        RHS = sin 30° + sin 60°
        
        Since, LHS ? RHS
        ? The statement sin (A + B) = sin A + sin B is false.
        (ii) Since the values of sin?increases from 0 to 1 as the?increases from 0 to 90°.
             ? The given statement
        (iii) Since the value of cos?ecreases from1 to 0 as?increases from 0 to 90°.
             ? The given statement is false.
        (iv) Let us take 0 = 30°
             
             ? sin 30° ? cos 30°
             ? The given statement .is false.
        (iv) From the table, we have:
             cot 0° = not defined.
             ? The given statement is true.
Exercise 8.3
1.   Evaluate:
             
2.   Show that:
        (i) tan 48° tan 23° tan 42° tan 67° = 1
        (ii) cos 38° cos 52° ? sin 38° sin 52° = 0
Sol. (i) tan 48° tan 23° tan 42° tan 67° =1
             L.H.S. = tan 48° tan 23° tan 42° tan 67°
             = tan (90° ? 42°) tan 23° tan.42°.tan (90.° ? 23°)
             = cot 42° tan 23° tan 42° cot 23° [ tan (90 ? A) = cot A]
             
             = R.H.S.
             Thus, tan 48° tan 23° tan 42° tan 67° = 1
        (ii) cos 38° cos 52° – sin 38° sin 52°
             L.H.S. = cos 38° cos 52° – sin 52°
             = cos 38° cos (90°– 38°)- sin 38° sin (90° – 38°)
             = cos(38° sin 38° – sin 38° cos 38°
             [?sin (90° – A) = cos A and cos(90° – A) = sin A]
             = 0 = R.H.S.
             This, cos 38° cos 52° – sin 38° sin 52° = 0
3.   If tan 2A cot (A ?V 18?X), where 2A is an acute angle, find the value of A.
Sol. Since tan 2A = cot (A –18°)
        Also tan (2A)° = cot (90° – 2A) [?tan ?? = cot (90° – ?)]
        ? A – 18 = 90°– 2A
        ? A + 2A = 90° + 18°
        ? 3A = 108°
        
4.   If tan A = cot B, prove that A + B = 90°.
Sol. tan A = cot B (given)
        And cot B = tan (90° – B)
[?tan (90° – ? ) = cot ? )]
        ? A = 90° – B
        ? A + B = 90°.
5.   If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Sol. sec 4A = cosec (A – 20°)
        sec (4A) = cosec (90° – 4A)
[?cosec (90° – ? ) = sec ? )]
        ? A – 20° = 90° – 4A
        ?A + 4A = 90° + 20°
        ? 5A = 110°
        
6.   If A, B and C are interior angles of a triangle ABC, then show that
        
Sol. Since, sum of the angles of ?ABC is A° + B° + C° = 180°
        ? B + C = 180° – A
        Dividing both sides by 2,
        
7.   Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Sol. Since sin 67° = sin(90° – 23°)
        = cos 23°
[?sin (90° – ? ) = cos ? ]
        Also, cos 75° = cos (90° – 15)
        = sin 15°
[?cos(90° – ? ) = sin ? ]]
        ? We have:
        sin 67° + cos 75° = cos 23° + sin 15°.
Exercise 8.4
1.   Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
2.   Write all the other trigonometric ratios of ?A in terms of sec A.
              
3.   Evaluate:
        (ii) sin 25° cos 65° + cos 25° sin 65°
              sin 25° = sin (90° – 65) = cos 65°
[? sin (90° – A = cos A]
              And cos 25° = cos (90° – 65°) = sin 65°
[? cos (90° – A = sin A]
              ? sin 25° cos 65° + cos 25° sin 65°
              = cos 65° cos 65° + sin 65° sin 65°
              = (cos 65°)2 + (sin 65°)2
[? cos2 A + sin2 A = 1]
              = cos2 65° + sin2 65°
              = 1
4.   Choose the correct option. Justify your choice.
        (i) 9 sec2 A – 9 tan2 A = ……………..
              (a) 1               (b) 9              (c) 8              (d) 0
        (ii) (1 + tan ? + sec ? ) (1 + cot ? ?n– cosec ? ) =
              (a) 0              (b) 1              (c) 2              (d) –1
        (iii) (sec A + tan A) (1 – sin A) = ……………..
              (a) sec A              (b) sin A               (c) cosec A              (d) cos A
        
Sol. (i) Since, 9 sec2 A – 9 tan2 A = 9 (sec2 A – tan2 A)
             = 9 (1)
[?tan2 A + 1 = sec2 A ? sec2 A – tan2 A = 1]
             = 9
             ? The option (b) is correct.
        (ii) Here, (1 + tan ? + sec ?) (1 +cot ? – cosec ?)
        
        
5.   Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
        
        
        
        (viii) (sin A + cosec A)2 + (cos A + sec A)2
              = sin2 A + cosec2 A + 2 sin A . cosec A + cos2 A + sec2 A + 2 cos A . sec A
              = (sin2 A + cos2 A) + cosec2 A + sec2 A + 2 + 2
[sin A . cosec A = 1 and sec A . cos A = 1]
              = 1 + cosec2 A + sec2 A + 4
[?sin2 A + cos2 A = 1]
              = 5 + (1 + cot2 A) + (1 tan2 A)
[?cosec2 A = 1 + cot2 A and sec2 A = 1 + tan2 A]
              = 7 + cot2 A + tan2 A
              = R.H.S.
        (ix) L.H.S. = (cosec A – sin A) (sec A – cos A)
        
        
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