 # 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)
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)  error: Content is protected !!