﻿ Class 11 NCERT Math Solution
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TOPICS
Unit-3(Examples)

Example-1 :-  Convert 40° 20′ into radian measure.

Solution :-
```  We know that 180° = π radian.
Hence,
40° 20′ = 40 + 1/3 degree
= π/180 x 121/3 radian
= 121π/540 radian
Therefore, 40° 20′ = 121π/540 radian
```

Example-2 :-  Convert 6 radians into degree measure.

Solution :-
```  We know that π radian = 180°.
Hence,
6 radians = 180/π x 6 degree
= (1080 x 7)/22 degree
= 343° + 7/11 degree
= 343° + (7 x 60)/11 minute    [1° = 60']
= 343° + 420/11 minute
= 343° + 38' + 2/11 minute     [1' = 60"]
= 343° + 38' + 10.9"
6 radians = 343° 38' 11" approximately.
```

Example-3 :- Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm. (use π = 22/7)

Solution :-
```   Given that l = 37.4 cm and θ = 60° = 60π/180 radian = π/3
By using formula,
r = l/θ
= (37.4 x 3)/π
= (112.2 x 7)/22
= 785.4/22
= 35.7 cm
```

Example-4 :-  The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14).

Solution :-
```  In 60 minutes, the minute hand of a watch completes one revolution.
Therefore,in 40 minutes, the minute hand turns through 2/3 of a revolution.
Therefore, θ = 2/3 x 360° = 4π/3 radian and r = 1.5 cm
Hence, the required distance travelled is given
By formula,
l = r θ
= 1.5 x 4π/3
= 2π
= 2 x 3.14
= 6.28 cm
```

Example-5 :-  If the arcs of the same lengths in two circles subtend angles 65°and 110° at the centre, find the ratio of their radii.

Solution :-
```  Let r₁  and r₂ be the radii of the two circles. Given that
θ₁  = 65° = π/180 x 65 = 13π/36 radian
θ₂ = 110° = π/180 x 110 = 22π/36 radian
Let l be the length of each of the arc.
Then l =  r₁ θ₁  =  r₂θ₂, which gives
13π/36 x r₁  = 22π/36 x r₂ i.e.
r₁ /r₂ = 22/13
Hence, r₁  : r₂ = 22 : 13
```

Example-6 :-  If cos x = –3/5 , x lies in the third quadrant, find the values of other five trigonometric functions.

Solution :-
```  Given that cos x = –3/5
(i) sec x = 1/cos x = -5/3
(ii) sin x = √1 - cos²x
= √1 - (-3/5)²x
= √1 - 9/25
= √16/25
= ±4/5
Since x lies in second quadrant, sin x will be negative. Therefore
sin x = -4/5
(iii) cosec x = 1/sin x = -5/4
(iv) tan x = sin x/cos x
= -4/5 ÷ (-3)/5
= -4/5 x (-5)/3
= 4/3
(v) cot x = 1/tan x = 3/4
```

Example-7 :-  If cot x = –5/12, x lies in second quadrant, find the values of other five trigonometric functions.

Solution :-
```  Given that cot x = –5/12
(i) tan x = 1/cot x = -12/5
(ii) sec x = √1 + tan²x
= √1 + (-12/5)²x
= √1 + 144/25
= √169/25
= ±13/5
Since x lies in second quadrant, sec x will be negative. Therefore
sec x = -13/5
(iii) cos x = 1/sec x = -5/13
(iv) sin x = √1 - cos²x
= √1 - (-5/13)²x
= √1 - 25/169
= √144/169
= ±12/13
Since x lies in second quadrant, sin x will be positive. Therefore
sin x = 12/13
(v) cosec x = 1/sin x = 13/12
```

Example-8 :-  Find the value of sin 31π/3.

Solution :-
```  We know that values of sin x repeats after an interval of 2π.
Therefore,
sin 31π/3 = sin(10π + π/3)
= sin π/3
= √3/2
```

Example-9 :-  Find the value of cos (–1710°).

Solution :-
```  We know that values of cos x repeats after an interval of 2π or 360°.
Therefore,
cos (–1710°) = cos (–1710° + 5 x 360°)
= cos (–1710° + 1800°)
= cos 90°
= 0
```

Example-10 :-  Prove that: 3sin π/6 . sec π/3 - 4sin 5π/6 . cot π/4 = 1

Solution :-
```  L.H.S.
3sin π/6 . sec π/3 - 4sin 5π/6 . cot π/4
= 3 x 1/2 x 2 - 4sin(π - π/6) x 1
= 3 - 4sin π/6
= 3 - 4 x 1/2
= 3 - 2
= 1
= R.H.S
```

Example-11 :-  Find the value of sin 15°.

Solution :-
```  sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
= 1/√2 x √3/2 - 1/√2 x 1/2
= √3/2√2 - 1/2√2
= (√3 - 1)/√2
```

Example-12 :-  Find the value of tan 13π/12.

Solution :-
```  tan 13π/12 = tan(π + π/12)
= tan π/12
= tan(π/4 - π/6)
= (tan π/4 - tan π/6)/(1 + tan π/4 . tan π/6)
= (1 - 1/√3)/(1 + 1/√3)
= (√3 - 1)/(√3 + 1)
By rationalising,
= (√3 - 1)/(√3 + 1) x (√3 - 1)/(√3 - 1)
= (√3 - 1)²/[(√3)² - (1)²]
= 2 - √3
```

Example-13 :-  Prove that:

Solution :-
```
```

Example-14 :-  Show that: tan 3x . tan 2x . tan x = tan 3x – tan 2x – tan x

Solution :-
```  We know that 3x = 2x + x.
Therefore, tan 3x = tan (2x + x)
tan 3x =  (tan 2x + tan x)/(1 - tan 2x . tan x)
tan 3x – tan 3x tan 2x tan x = tan 2x + tan x
tan 3x – tan 2x – tan x = tan 3x tan 2x tan x
tan 3x tan 2x tan x = tan 3x – tan 2x – tan x
```

Example-15 :-  Prove that: cos(π/4 + x) + cos(π/4 - x) = √2 cos x

Solution :-
```
```

Example-16 :-  Prove that:

Solution :-
```
```

Example-17 :-  Prove that:

Solution :-
```
```

Example-18 :-  Find the principal solutions of the equation sin x = √3/2

Solution :-
```  We know that sin π/3 = √3/2 and sin 2π/3 = (π - π/3) = √3/2
Therefore, principal solutions are π/3 and 2π/3.
```

Example-19 :-  Find the principal solutions of the equation tan x = -1/√3

Solution :-
```  We know that tan π/6 = 1/√3
tan(π - π/6) = -tan π/6 = -1/√3
and tan(2π - π/6) = -tan π/6 = -1/√3
Thus, tan 5π/6 = tan 11π/6 = -1/√3
Therefore, principal solutions are 5π/6 and 11π/6.
```

Example-20 :-  Find the solution of sin x = -√3/2

Solution :-
```  Given:
sin x = -√3/2
sin x = -sin π/3
sin x = sin(π + π/3)
sin x = sin 4π/3
which gives,
x = nπ + (-1)ⁿ 4π/6, where  n ∈ Z.
```

Example-21 :-  Solve cos x = 1/2

Solution :-
```  Given:
cos x = 1/2
cos x = cos π/3
Therefore,
x = 2nπ ± π/3, where  n ∈ Z.
```

Example-22 :-  Solve tan 2x = -cot(x + π/3).

Solution :-
```  Given:
tan 2x = -cot(x + π/3)
tan 2x = tan(π/2 + x + π/3)
tan 2x = tan(x + 5π/6)
Therefore,
2x = nπ + x + 5π/6 where  n ∈ Z. or
x = nπ + 5π/6 where  n ∈ Z.
```

Example-23 :-  Solve sin 2x – sin 4x + sin 6x = 0.

Solution :-
```  Given:
sin 2x – sin 4x + sin 6x = 0
(sin 6x + sin 2x) – sin 4x = 0
2sin 4x cos 2x - sin 4x = 0
sin 4x (2cos 2x - 1) = 0
Therefore,
sin 4x = 0 or 2cos 2x = 1
sin 4x = 0 or  cos 2x = 1/2
4x = nπ or 2x = 2nπ ± π/3, where  n ∈ Z.
x = nπ/4 or x = nπ ± π/6, where  n ∈ Z.
```

Example-24 :-  Solve 2 cos² x + 3 sin x = 0

Solution :-
```  Given:
2 cos² x + 3 sin x = 0
2 (1 - sin²x) + 3 sin x = 0
2 - 2 sin²x + 3 sin x = 0
2 sin²x - 3 sin x - 2 = 0
(2sin x + 1)(sin x - 2) = 0
Hence, sin x = -1/2 or sin x = 2
But, sin x = 2 is not possible.
Therefore,
sin x = -1/2 = sin 7π/6
x = nπ + (-1)ⁿ 7π/6, where  n ∈ Z.
```
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