Introduction to Trigonometry-NCERT Solutions

Class X Math
NCERT Solution for Intorudction to Trigonometry
1.   In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
       (i) Sin A, cos A                                          (ii) sin C, cos C
Sol. In right ΔABC, we have:
        p = 24 cm, b = 7 cm
        
2.   In the figure, find tan P – cot R.
Sol. In right ΔPQR, using the Pythagoras theorem, we get
        
3.   If sin calculate cos A and tan A.
Sol. Let us consider, the right ΔABC, we have
        Perp. = BC and Hyp. = AC
        
4.   Given 15 cot A = 8, find sin A and sec A.
Sol. Let in the right ΔABC, 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 ΔABC 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 ΔABC,
        
        
7.   
Sol. Let us have a right ΔABC in which ∠B = 90°, and ∠A = θ
        
        
8.   If 3 cot A = 4, check whether
Sol. Let us consider a right angled ΔABC 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 ΔABC, in which ∠B = 90°
        For ∠A, we have
        Base = AB
        Perpendicular = BC
        Hypotenuse = AC
        
        
10.   In ΔPQR, 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 Δ, 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 sinθ increases as θincreases.
        (iii) The value of cosθ increases as θincreases.
        (iv) sinθ= cosθ 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 sinθincreases from 0 to 1 as theθincreases from 0 to 90°.
             ∴ The given statement
        (iii) Since the value of cosθecreases from1 to 0 asθincreases 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° – θ)]
        ⇒ 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° – θ ) = cot θ )]
        ∴ 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° – θ ) = sec θ )]
        ∴ 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 ΔABC 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° – θ ) = cos θ ]
        Also, cos 75° = cos (90° – 15)
        = sin 15°
[∵cos(90° – θ ) = sin θ ]]
        ∴ 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 θ + sec θ ) (1 + cot θ �n– cosec θ ) =
              (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 θ + sec θ) (1 +cot ∵ – cosec θ)
        
        
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)