Trigonometry
Trigonometry
Trigonometry
Perp. base
D A C D
CONTENTS Hypotenuse
Perp.
Hypo. Hypo.
Right Angle Triangle E F B
Perpendicular
Trigonometric Ratio (T.R.) of some A
Specific Angles
Trigonometric Ratios of
B C
Complementary Angles
Hypotenuse Perpendicular Base
Trigonometric Identities for A AC BC AB
for C AC AB BC
Trigonometry is the branch of mathematics in The trigonometry ratio are
which we study of relationships between the sides sine of cosine of , tangent of ,
& angles of a triangle. cotangent of , secant of , cosecant of .
These ratios are abbreviated as sin , cos , tan ,
Fact : In Greek words :
cot , sec , cosec and the relation with sides
Tri = three are
gon = sides D
(H)
metron = measure (P)
The ratio of sides of a right angle triangle
E F
with respect to acute angles are called (B)
"Trigonometric ratios of the angle". sin = P/H = DE/DF
RIGHT ANGLE TRIANGLE cos = B/H = EF/DF
tan = P/B = DE/EF
1. A having one angle equal to 90º is called
right angle . cot = B/P = EF/DE
2. The sum of other two acute (Less than 90º) sec = H/B = DF/EF
angles is 90º. (or both acute angles are cosec = H/P = DF/DE
complementary) 1 1
By above table sin = , cos = ,
3. The side opposite to 90º, is called hypotenuse, cos ec sec
it is longest side in . 1 sin P/H P
tan = also tan = = =
4. The side opposite to given one acute angle is cot cos B/ H B
perpendicular. we can say ‘‘Trigonometric Ratio’’ represents
5. The rest (IIIrd) side is base. ratio between acute angles & sides of triangle.
EXAMPLES A P
Ex.1 If ABC is right angle triangle, B = 90º,
AB = 12 cm, AC = 13 cm then find sin A and
cos C.
C B R Q
Sol. Using Pythagoras theorem
A AC
We have sin B =
AB
12 13 PR
and sin Q =
PQ
B C
AC PR
BC = AC 2 AB2 = 169 144 = 5 cm Then =
AB PQ
BC 5 AC AB
sin A = = Therefore, = = k, say ....(1)
AC 13 PR PQ
AB 12
cos C = = Ans. Now, using Pythagoras theorem,
AC 13
1 BC = AB 2 AC 2
Ex.2 If sin A = in right triangle ABC, then
2
and QR = PQ 2 PR 2
find value of tan A, cosec A, tan B, cosec B.
A BC AB2 AC 2
So, =
Sol. 2
QR PQ 2 PR 2
1
k 2 PQ 2 k 2 PR 2
C B =
1 PQ 2 PR 2
1 BC
sin A = =
2 AB k PQ 2 PR 2
= = k .....(2)
PQ 2 PR 2
AC = AB 2 BC 2 = ( 2k ) 2 (k ) 2
From (1) and (2), we have
2 2 2
= 2k k = k =k AC AB BC
= =
BC k PR PQ QR
tan A = = =1
AC k
Then, by using Theorem, ACB ~ PRQ and
1 2k therefore, B = Q.
cosec A = = = 2
sin A k
Ex.4 Consider ACB, right-angled at C, in which
AC k AB = 29 units, BC = 21 units and ABC =
tan B = = =1
BC k (see figure). Determine the value of
AB 2k (i) cos2 + sin2,
cosec B = = = 2
AC k
(ii) cos2 – sin2 [NCERT]
Ex.3 If B and Q are acute angles such that sin
A
B = sin Q, then prove that B = Q.
[NCERT] 29
Sol. Let us consider two right triangles ABC and
PQR where sin B = sin Q. C B
21
Sol. In ACB, we have A
25 cm 8 8
1 1
113 113
24 cm =
7 7
1 1
A B 113 113
7 cm
AC = 25 cm, BC = (25 – 1)cm = 24 cm ( 113 8)( 113 8)
=
and AB = 7 cm. ( 113 7)( 113 7)
For T-ratios of C, we have 113 64 49
= = Ans.
base = BC = 24 cm, 113 49 64
perpendicular = AB = 7 cm and 2 2
B 7k 49
hypotenuse = AC = 25 cm. (ii) cot2 = = = Ans.
P 8k 64
AB 7 BC 24
sin C = = and cos C = = . Ex.11 If 3 cot A = 4, check whether
AC 25 AC 25
Ex.9 If A and B are acute angles such that 1 tan 2 A
2
= cos2 A – sin2A or not.
cos A = cos B, then show that A = B. 1 tan A
[NCERT] [NCERT]
Sol. cos A = cos B 4 3
Sol. cot A = tan A =
AC BC 3 4
=
AB AB C
A
3k
B A
4k
C B
AC = AB 2 BC 2 = 16k 2 9k 2
AC = BC
is an isosceles = 25k 2 = 5k
A = B Proved. 3k 3
sin A = =
5k 5
7 4k 4
Ex.10 If cot = , evaluate : [NCERT] cos A = =
8 5k 5
(1 sin )(1 sin ) 1 tan 2 A
(i) , (ii) cot2 LHS =
(1 cos )(1 cos ) 1 tan 2 A
7 Base B 2
Sol. cot = = 3 9
8 Perpendicular P 1 1
= 4 = 16
H= (8k ) 2 (7 k ) 2 = (64 49) k 3
2 9
1 1
4 16
= 113 k
(16 9) / 16 7 P
= = Sol.
(16 9) / 16 25
RHS = cos2A – sin2A
16 9 7
= – = Q R
25 25 25
PR + QR = 25 cm
LHS = RHS
PQ = 5 cm
Let PR = x cm
Ex.12 In triangle ABC, right-angled at B, if
1 QR = (25 – x) cm
tan A = , find the value of : [NCERT] Using Pythagoras theorem
3
PR2 = PQ2 + QR2
(i) sin A cos C + cos A sin C
x2 = 52 + (25 – x)2
(ii) cos A cos C – sin A sin C
x2 = 25 + 625 + x2 – 50x
1 P
Sol. tan A = = 50x = 650
3 B
x = 13 cm = PR
A
QR = 25 – 13 = 12 cm.
QR 12k 12
3k sin P =
PR 13k 13
B C PQ 5k 5
k cos P =
PR 13k 13
AC = ( 3k ) 2 ( k ) 2 = 3k 2 k 2 = 2k QR 12k 12
tan P = Ans.
BC k 1 PQ 5k 5
sin A = = = ;
AC 2k 2
3
Ex.14 If sin A = , find cos A and tan A.
AB 3k 3 5
sin C = ;
AC 2k 2 Perpendicular 3
Sol. Since sin A = = , so
AB 3k 3 Hypotenuse 5
cos A = ;
AC 2k 2 We draw a triangle ABC, right angled at B
BC k 1 such that
cos C = C
AC 2k 2
(i) sin A cos C + cos A sin C
5 3
1 1 3 3
=
2 2 2 2
1 3 A 4 B
= + =1
4 4 Perpendicular = BC = 3 units,
(ii) cos A cos C – sin A sin C and, Hypotenuse = AC = 5 units.
3 1 1 3 By Pythagoras theorem, we have
=
2 2 2 2
– AC2 = AB2 + BC2
52 = AB2 + 32
3 3 AB2 = 52 – 32
= – =0
4 4 AB2 = 16 AB = 4
Ex.13 In PQR, right-angled at Q, PR + QR = 25 cm When we consider the t-ratio of A, we have
and PQ = 5 cm. Determine the values of Base = AB = 4, Perpendicular = BC = 3,
sin P, cos P and tan P. [NCERT] Hypotenuse = AC = 5.
Base 4 B such that Base = AB = 1 and Perpendicular
cos A = =
Hypotenuse 5 = BC = 2 – 1.
Perpendicular 3 By Pythagoras theorem, we have
and, tan A = =
Base 4 AC2 = AB2 + BC2
C
Ex.15 If cosec A = 10 , find other five
trigonometric ratios. 2 1
Hypotenuse 10
Sol. Since cosec A = = , A 1 B
Perpendicular 1
AC2 = 12 + ( 2 1) 2
so we draw a right triangle ABC, right angled
at B such that AC2 = 1 + 2 + 2 – 2 2
Perpendicular = BC = 1 unit. and,
AC2 = 4 – 2 2 AC = 42 2
Hypotenuse = AC = 10 units.
BC 2 1
By Pythagoras theorem, we have Now, sin A = = , and
AC 42 2
AC2 = AB2 + BC2
C AB 1
cos A = =
AC 42 2
10 2 1 1
1
sin A cosA = ×
42 2 42 2
A B 2 1 2 1 1 2
= = = = .
( 10 )2 = AB2 + 12 42 2 2 2 ( 2 1) 2 2 4
AB2 = 10 – 1 = 9
sin2 = (sin )2
AB = 9 =3 cos2 = (cos )2
When we consider the trigonometric ratios of tan2 = (tan )2
A, we have cosec2 = (cosec )2
Base = AB = 3, Perpendicular = BC = 1, and sec2 = (sec )2
Hypotenuse = AC = 10 . cot2 = (cot )2
Perpendicular 1
sin A = = ;
Hypotenuse 10 EXAMPLES
Base 3 1
cos A = = ; Ex.17 In a ABC right angled at C, if tan A =
Hypotenuse 10 3
Perpendicular 1 and tan B = 3 . Show that
tan A = = ;
Base 3 sin A cos B + cos A sin B = 1.
Sol. Let us draw a ABC, right angled at C in
Hypotenuse 10
sec A = = ; 1
Base 3 which tan B = 3 and tan A = .
3
Base 3
and cot A = = =3 B
Perpendicular 1
2
Ex.16 If tan A = 2 – 1, show that sinA cosA= .
4
Perpendicular 2 1
Sol. Since tan A = = , so A C
Base 1
we draw a right triangle ABC, right angled at
1 C
Now, tan A =
3 5
3
BC 1 BC
= tan A AC
AC 3
A 4 B
BC = x and AC = 3x ....(i) By Pythagoras theorem, we have
And, tan B = 3 AC2 = AB2 + BC2
52 = 42 + BC2
AC 3 AC
= tan B BC BC2 = 52 – 42 = 9
BC 1
BC = 9 =3
AC = 3 x and BC = x ....(ii)
BC 3
From (i) and (ii), we have tan = =
AB 4
BC = x, AC = 3 x 3 1
By Pythagoras theorem, we have 1
1 tan 4 = 4 = 1.
AB2 = AC2 + BC2 Now, =
1 tan 1 3 7 7
AB2 = ( 3x ) 2 + x2 4 4
AB2 = 3x2 + x2 12
AB2 = 4x2 Ex.19 If cot B = , prove that
5
AB = 2x tan2 B – sin2B = sin4B. sec2B.
When we find the t-rations of A, we have Base 12
Sol. Since cot B = = , so we
Base = AC = 3 x, Perpendicular = BC = x, Perpendicular 5
and Hypotenuse = AB = 2x. draw a right triangle ABC, right angled at C
BC x 1 such that Base = BC = 12 units.
sin A = = = and
AB 2x 2 Perpendicular = AC = 5 units.
AC 3x 3 A
cos A = = =
AB 2x 2 13
When we consider the t-ratios of B, we have 5
Base = BC = x, Perpendicular = AC = 3 x, B 12 C
and Hypotenuse = AB = 2x.
By Pythagoras theorem, we have
BC x 1 AB2 = BC2 + AC2
cos B = = = and
AB 2x 2 AB2 = 122 + 52 = 169
AC 3x 3 AB = 169 = 13
sin B = = =
AB 2x 2 AC 5 AC 5
Now, sin B = = , tan B = =
AB 13 BC 12
1 1 3 3 AB 13
sinA cosB + cosA sinB = × + × and sec B = =
2 2 2 2 BC 12
1 3 Now, LHS = tan2B – sin2B = (tanB)2 – (sinB)2
= + = 1.
4 4 5
2
5
2
25 25
= – = –
5 1 tan 12 13 144 169
Ex.18 If sec = , evaluate .
4 1 tan 1 1 169 144
Hypotenuse 5 = 25 = 25
Sol. Since sec = = , so we draw 144 169 144 169
Base 4 25 25 25
a right triangle ABC, right angled at B such that = 25 × =
144 169 144 169
Hypotenuse = AC = 5 units, 2 2
5 5
Base = AB = 4 units, and BAC = = 2 ....(i)
12 132
and, RHS = sin4B sec2B Sol. We have,
= (sin B)4 (sec B)2 AB = a
5
4
13
2 AD + DB = a [ AD = DB]
= × AD + AD = a
13 12
a
54 2AD = a AD =
= 2 2
13 12 2 a
52 52 Thus, AD = DB =
= 2 ....(ii) 2
13 12 2 By Pythagoras theorem, we have
From (i) and (ii), we have AC2 = AB2 + BC2
tan2B – sin2B = sin4B sec2B. b2 = a2 + BC2
Ex.20 In a right triangle ABC, right angled at B, the
BC2 = b2 – a2 BC = b2 a 2
ratio of AB to AC is 1 : 2 . Find the values Thus, in BCD, we have
of
2 tan A 2 tan A Base = BC = b2 a 2
(i) and (ii)
2
1 tan A 1 tan 2 A a
and Perpendicular = BD =
AB 1 2
Sol. We have, AB : AC = 1 : 2 i.e. =
AC 2 Applying Pyhthagoras theorem in BCD, we
have
AB = x and AC = 2 x.
BC2 + BD2 = CD2
C 2
a
( b 2 a 2 ) 2 + = CD2
2
a2
CD2 = b2 – a2 +
4
A B 4 b 2 4a 2 a 2
By Pythagoras theorem, we have CD2 =
4
AC2 = AB2 + BC2
4b 2 3a 2
( 2 x)2 = x2 + BC2 CD =
BC2 = 2x2 – x2 = x2 2
BC = x Now,
BC x BD
tan A = = =1 (i) sin =
AB x CD
2 tan A 2 1 2 a/2 a
Now, = = =1 sin = =
2 2
2
1 tan A 11 2
2 4b 3a 4b 3a 2
2
2 tan A 2 1 2 2
Now, = = , which is
2
1 tan A 11 0 BC
(ii) cos =
undefined. CD
Ex.21 In fig. AD = DB and B is a right angle. b2 a 2 2 b2 a 2
Determine cos = =
4b 2 3a 2 4b 2 3a 2
(i) sin (ii) cos
(iii) tan (iv) sin2 + cos2 2
A BD
(iii) tan =
CD
a/2 a
tan = = , and
2 2
b Da b a 2 b a22
Q.11 If tan = 8/15 and 0º < < 90º, find sin. Q.18 If 4 tan = 3, find the value of
4 sin 2 cos
Q.12 If sin = 8/17 and 0º < < 90º, find tan. .
4 sin 3 cos
24
Q.13 If sinA = , find the value of tanA + secA, 13
25 Q.19 If cosec = , find the value of
12
where 0º < A < 90º.
2 sin 3 cos
.
Q.14 If 5 tan = 12, find the value of 4 sin 9 cos
2 cos sin
.
sin cos
3 1 cos
Q.15 If tan = , find the value of .
4 1 cos
12 1 sin
Q.16 If tan = , find the value of .
5 1 sin
ANSWER KEY
ANSWER KEY
c
5 5 13
1. 108° 2. 3. and
6 12 12
x 2 y2 1 1
4. 5. 6.
x 2 y2 6 7
12 7 3
7. 8. 9.
25 17 4
3 1 271
10. 11. 12.
5 3 979
7
13. 14. 2 3 15. cos
25
16. 9 17. 2
18. sec A + tan A 19. sec A – tan A 20. cosec x – cot x
21. cosec x + cot x 22. sec x – tan x 23. 1
3 1
24. 0 25. 26. 0
2 2
27. (A) 28. 1
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