“Mathematics Class 11 – Trigonometric Ratios and Identities (Handwritten Notes with Solutions)”

3 Trigonometric Ratios & Identities


Note
BASIC OF TRIGONOMETRY
(1) 90°, 180°, 270°... ⇒ Allied angles

# Point-01: Angles (2) If α + β = π ⇒ α & β Supplementary angles

Measurement of Angles π
(3) If α + β = ⇒ α & β complementary angles
2
⇒ Degree system & Radian (circular system)

⇒ π radian = 180° RAR
180° π Ξ 3.14, π Ξ 1.57, π Ξ 0.78
⇒ 1 radian = 2 4
π
180° Note
1 radian = = 57°16' approximately
π
Angle measured in anticlockwise direcn. ≡ (+ve)
 π 
⇒ 1° =   radian
 180  (+x)
(0, 0)
π
1° = radian = 0.01746 radian approximately
180 Angle measured in clockwise direcn. ≡ (–ve)

# Point-02: Quadrant
r
π π
 ( 4n − 3) or ( 4n + 1)
O θ (in radian) = or  = rθ 2 2
r 90°/–270°

II I
Sin/cosec → +ve All → +ve
RAR ([kwu esa clk yks) 180°/–180° 0°/360°/–360
( 2n ± 1)π, n ∈ I O ( 2nπ), n ∈ I
π π π III IV
30° = 45° = 60° = cos/sec → +ve
6 4 3 tan/cot → +ve

π 2π 3π 270°/–90°
90° = 120° = 135° = π π
2 3 4 ( 4n − 1) or ( 4n + 3)
2 3
5π 7π Point-03: Purana Pyaar (Class X wala)
150° = 180° = p 210° =
6 6
Sine Cosine Tangent
5π 4π 3π
225° = 240° = 270° =
4 3 2 P B P
H H B
5π 7π 11π
300° = 315° = 330° =
3 4 6 Cosecant Secant Cotangent

, Acidic Question
TRIGONOMETRIC IDENTITIES
Q. The value of
log(sin2x + cos4x + 2)(cos2x + sin4x + 2) is
(1) (sin2θ + cos2θ) = 1
(A) 1    (B) –1 (C) 0    (D) 2
(2) (sec2θ - tan2θ) = 1
Sol. cos2x + sin4x + 2 = cos2x + (1 – cos2x)2 + 2
⇒ (secθ + tanθ) (secθ – tanθ) = 1

= cos2x + 1 + cos4x – 2cos2x + 2
(3) (cosec2θ – cot2θ) = 1
= 1 – cos2x + cos4x + 2

⇒ (cosecθ + cotθ) (cosecθ – cotθ) = 1 = sin2x + cos4x + 2

⇒ Req. ans = 1 fnf.
Question
Note
3
Q. If q ∈ I quadrant & secq + tanq = . Then
2 Dealing with –ve sign
value of cos q is _____________.
sin(–θ) = –sinθ cot(–θ) = -cot(θ)
3
Sol. Given: secq + tanq =  ...(i) cosec(-θ) = -cosecθ cos(-θ) = cosθ
2
tan(-θ) = -tanθ sec(–θ) = secθ
2
⇒ secq – tanq =  ...(ii)
3
add ______________________________ Note
13 13 "How to reduce angle"

2 sec=
θ ⇒ sec=
θ
6 12 (1) Angle ka quadrant pta kro (θ acute
12 imagine krte hue)

cos θ = fnf.
13 (2) Angle = a.(90°) ± θ
Question [JEE Main-2019] a Ξ even → function nahi badlega
a Ξ odd → function will change
1
Q. Let fk = (sinkx + coskx) for k = 1, 2, 3, ... sin ↔ cos
k
tan ↔ cot
then for all x ∈ R, the value of f4(x) – f6(x) cosec ↔ sec
is equal to

(A)
1
(B)
1 REDUCTION FORMULA
12 4
1 5
(C) − (D) π  π 
12 12 sin  − θ  = + cos θ sin  + θ  = + co s θ
Sol. ⇒ sin2x + cos2x = 1 2  2 

SBS: sin4x + cos4x + 2sin2xcos2x = 1
...(i) π  π 
cos  − θ  = + sin θ cos  + θ  = − sin θ

⇒ sin6x + cos6x = (sin2x)3 + (cos2x)3 2  2 
= (sin2x + cos2x)(sin4x + cos4x – sin2xcos2x) π 
π 
tan  − θ  = + cot θ tan  + θ  = − cot θ
sin6x + cos6x = 1 – 3sin2xcos2x...(ii) 2   2 
1
now; f4(x) = (1 – 2sin2xcos2x) π  π 
4 cot  − θ  = + tan θ cot  + θ  = − tan θ
2  2 
1
and, f6(x) = (1 – 3sin2xcos2x)
6 π  π 
cosec  − θ  = + sec θ cosec  + θ  = + sec θ
1 1 1 2  2 

⇒ f4(x) – f6(x) = − = fnf.
4 6 12

2
Mathematics

, π  π  0° 30° 45° 60° 90°
sec  − θ  = + co sec θ sec  + θ  = −co sec θ
2  2  not 1
cot 3 1 0
defined 3
sin(π – θ) = +sinθ sin(π + θ) = –sinθ 2 not
sec 1 2 2
3 defined
cos(π – θ) = –cosθ cos(π + θ) = –cosθ
not 2
cosec 2 2 1
tan(π – θ) = –tanθ tan(π + θ) = tanθ defined 3


cot(π – θ) = –cotθ cot(π + θ) = +cotθ Q. Find value of following:

(1) cos(240°) = cos(2 × 90° + 60°)
cosec(π – θ) = +cosecθ cosec(π + θ) = –cosecθ
−1
= –(cos60°) =
sec(π – θ) = –secθ sec(π + θ) = –secθ 2

(2) tan(–300°) = –tan(300°)
Examples
= –tan(3 × 90° + 30°)
 4π 
(1) sin(2p – θ) = sin  − θ  = − sin θ
 2          = −( − cot 30°) = + 3
 3π 
(2) tan  + θ  = − cot θ  11π  11π  π
 2  (3) sin  −  = − sin = − sin  4 π − 
3 3  3
 10 π 
(3) cosec(5p – θ) = cos ec  − θ  = + cosec θ  π 3
 2  = −  − sin  =
(4) sec(2p + θ) = sec θ  3 2

 5π  19ππ
19  π  π
(5) cos  − θ  = + sin θ (4) tan
tan  = tan  6π +  = + tan   = 3
 2   33  3 3

 3π   3π  (5) cot(–1470°) = –cot(1470°)
(6) sin  θ −  = − sin  − θ  = −( − cos θ) = + cos θ
 2   2  = –cot(16×90° + 30°)
 π  π  π 
(7) cos  θ −  = co s  −  − θ   = cos  − θ  = + sin θ = –(+cot30°) = − 3
 2   2    2 
Question

VALUE OF Q. Find value of following:

TRIGONOMETRIC FUNCTION (i) a = tan1° tan2° tan3° ... tan89° = 1

(ii) a = cos15° cos16° cos17° ... cos175° = 0
0° 30° 45° 60° 90°
19
sin 1
(iii) g = sin25° + sin210° + sin215° ... sin290° =
1 3 2
0 1
2 2 2 π 3π 5π 7π
(iv) δ = cos2 + cos2 + cos2 + cos2 =2
1 16 16 16 16
3 1
cos 1 0
2 2 2
Question
1 not
tan 0 1 3 Q. If cosecq – sinq = m and secq – cosq = n,
3 defined
then (m2n)2/3 + (mn2)2/3 =


3
Trigonometric Ratios & Identities