Class 11: Maths Chapter 5 solutions. Complete Class 11 Maths Chapter 5 Notes.
Contents
RD Sharma Solutions for Class 11 Maths Chapter 5–Trigonometric Functions
RD Sharma 11th Maths Chapter 5, Class 11 Maths Chapter 5 solutions
EXERCISE 5.1 PAGE NO: 5.18
Prove the following identities:
1. sec4 x – sec2 x = tan4 x + tan2 x
Solution:
Let us consider LHS: sec4 x – sec2 x
(sec2 x)2 – sec2 x
By using the formula, sec2 θ = 1 + tan2 θ.
(1 + tan2 x) 2 – (1 + tan2 x)
1 + 2tan2 x + tan4 x – 1 – tan2 x
tan4 x + tan2 x
= RHS
∴ LHS = RHS
Hence proved.
2. sin6 x + cos6 x = 1 – 3 sin2 x cos2 x
Solution:
Let us consider LHS: sin6 x + cos6 x
(sin2 x) 3 + (cos2 x) 3
By using the formula, a3 + b3 = (a + b) (a2 + b2 – ab)
(sin2 x + cos2 x) [(sin2 x) 2 + (cos2 x) 2 – sin2 x cos2 x]
By using the formula, sin2 x + cos2 x = 1 and a2 + b2 = (a + b) 2 – 2ab
1 × [(sin2 x + cos2 x) 2 – 2sin2 x cos2 x – sin2 x cos2 x
12 – 3sin2 x cos2 x
1 – 3sin2 x cos2 x
= RHS
∴ LHS = RHS
Hence proved.
3. (cosec x – sin x) (sec x – cos x) (tan x + cot x) = 1
Solution:
Let us consider LHS: (cosec x – sin x) (sec x – cos x) (tan x + cot x)
By using the formulas
cosec θ = 1/sin θ;
sec θ = 1/cos θ;
tan θ = sin θ / cos θ;
cot θ = cos θ / sin θ
Now,

1 = RHS
∴ LHS = RHS
Hence proved.
4. cosec x (sec x – 1) – cot x (1 – cos x) = tan x – sin x
Solution:
Let us consider LHS: cosec x (sec x – 1) – cot x (1 – cos x)
By using the formulas
cosec θ = 1/sin θ;
sec θ = 1/cos θ;
tan θ = sin θ / cos θ;
cot θ = cos θ / sin θ
Now,

By using the formula, 1 – cos2x = sin2x;

= RHS
∴ LHS = RHS
Hence Proved.
5.

Solution:
Let us consider the LHS:

By using the formula,
cosec θ = 1/sin θ;
sec θ = 1/cos θ;
Now,


sin x
= RHS
∴ LHS = RHS
Hence Proved.
6.

Solution:
Let us consider the LHS:

By using the formula,
tan θ = sin θ / cos θ;
cot θ = cos θ / sin θ
Now,

By using the formula, a3 – b3 = (a – b) (a2 + b2 + ab)
By using the formula,
cosec θ = 1/sin θ,
sec θ = 1/cos θ;
cosec x × sec x + 1
sec x cosec x + 1
=RHS
∴ LHS = RHS
Hence Proved.
7.

Solution:
Let us consider LHS:
By using the formula a3 ± b3 = (a ± b) (a2 + b2∓ ab)

We know, sin2x + cos2x = 1.
1 – sinx cosx + 1 + sinx cosx
2
= RHS
∴ LHS = RHS
Hence Proved.
8. (sec x sec y + tan x tan y)2 – (sec x tan y + tan x sec y)2 = 1
Solution:
Let us consider LHS:
(sec x sec y + tan x tan y)2 – (sec x tan y + tan x sec y)2
Expanding the above equation we get,[(sec x sec y)2 + (tan x tan y)2 + 2 (sec x sec y) (tan x tan y)] – [(sec x tan y)2 + (tan x sec y)2 + 2 (sec x tan y) (tan x sec y)] [sec2 x sec2 y + tan2 x tan2 y + 2 (sec x sec y) (tan x tan y)] – [sec2 x tan2 y + tan2 x sec2 y + 2 (sec2 x tan2 y) (tan x sec y)]
sec2 x sec2 y – sec2 x tan2 y + tan2 x tan2 y – tan2 x sec2 y
sec2 x (sec2 y – tan2 y) + tan2 x (tan2 y – sec2 y)
sec2 x (sec2 y – tan2 y) – tan2 x (sec2 y – tan2 y)
We know, sec2 x – tan2 x = 1.
sec2 x × 1 – tan2 x × 1
sec2 x – tan2 x
1 = RHS
∴ LHS = RHS
Hence proved.
9.

Solution:
Let us Consider RHS:




= LHS
∴ LHS = RHS
Hence Proved.
10.

Solution:
Let us consider LHS:

By using the formulas,
1 + tan2x = sec2x and 1 + cot2x = cosec2x


= RHS
∴ LHS = RHS
Hence Proved.
11.

Solution:
Let us consider LHS:

By using the formula,
tan θ = sin θ / cos θ;
cot θ = cos θ / sin θ
Now,

By using the formula, a3 + b3 = (a + b) (a2 + b2– ab)

We know, sin2 x + cos2 x = 1.
1 – 1 + sin x cos x
Sin x cos x
= RHS
∴ LHS = RHS
Hence proved.
12.

Solution:
Let us consider LHS:

By using the formula,
cosec θ = 1/sin θ,
sec θ = 1/cos θ;


= RHS
∴ LHS = RHS
Hence proved.
13. (1 + tan α tan β) 2 + (tan α – tan β) 2 = sec2 α sec2 β
Solution:
Let us consider LHS: (1 + tan α tan β) 2 + (tan α – tan β) 2
1+ tan2 α tan2 β + 2 tan α tan β + tan2 α + tan2 β – 2 tan α tan β
1 + tan2 α tan2 β + tan2 α + tan2 β
tan2 α (tan2 β + 1) + 1 (1 + tan2 β)
(1 + tan2 β) (1 + tan2 α)
We know, 1 + tan2 θ = sec2 θ
So,
sec2 α sec2 β
= RHS
∴ LHS = RHS
Hence proved.
EXERCISE 5.2 PAGE NO: 5.25
1. Find the values of the other five trigonometric functions in each of the following:
(i) cot x = 12/5, x in quadrant III
(ii) cos x = -1/2, x in quadrant II
(iii) tan x = 3/4, x in quadrant III
(iv) sin x = 3/5, x in quadrant I
Solution:
(i) cot x = 12/5, x in quadrant III
In third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative.
By using the formulas,
tan x = 1/cot x
= 1/(12/5)
= 5/12
cosec x = –√(1 + cot2 x)
= –√(1 + (12/5)2)
= –√(25+144)/25
= –√(169/25)
= -13/5
sin x = 1/cosec x
= 1/(-13/5)
= -5/13
cos x = – √(1 – sin2 x)
= – √(1 – (-5/13)2)
= – √(169-25)/169
= – √(144/169)
= -12/13
sec x = 1/cos x
= 1/(-12/13)
= -13/12
∴ sin x = -5/13, cos x = -12/13, tan x = 5/12, cosec x = -13/5, sec x = -13/12
(ii) cos x = -1/2, x in quadrant II
In second quadrant, sin x and cosec x are positive. tan x, cot x, cos x, sec x are negative.
By using the formulas,
sin x = √(1 – cos2 x)
= √(1 – (-1/2)2)
= √(4-1)/4
= √(3/4)
= √3/2
tan x = sin x/cos x
= (√3/2)/(-1/2)
= -√3
cot x = 1/tan x
= 1/-√3
= -1/√3
cosec x = 1/sin x
= 1/(√3/2)
= 2/√3
sec x = 1/cos x
= 1/(-1/2)
= -2
∴ sin x = √3/2, tan x = -√3, cosec x = 2/√3, cot x = -1/√3 sec x = -2
(iii) tan x = 3/4, x in quadrant III
In third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative.
By using the formulas,
sin x = √(1 – cos2 x)
= – √(1-(-4/5)2)
= – √(25-16)/25
= – √(9/25)
= – 3/5
cos x = 1/sec x
= 1/(-5/4)
= -4/5
cot x = 1/tan x
= 1/(3/4)
= 4/3
cosec x = 1/sin x
= 1/(-3/5)
= -5/3
sec x = -√(1 + tan2 x)
= – √(1+(3/4)2)
= – √(16+9)/16
= – √ (25/16)
= -5/4
∴ sin x = -3/5, cos x = -4/5, cosec x = -5/3, sec x = -5/4, cot x = 4/3
(iv) sin x = 3/5, x in quadrant I
In first quadrant, all trigonometric ratios are positive.
So, by using the formulas,
tan x = sin x/cos x
= (3/5)/(4/5)
= 3/4
cosec x = 1/sin x
= 1/(3/5)
= 5/3
cos x = √(1-sin2 x)
= √(1 – (-3/5)2)
= √(25-9)/25
= √(16/25)
= 4/5
sec x = 1/cos x
= 1/(4/5)
= 5/4
cot x = 1/tan x
= 1/(3/4)
= 4/3
∴ cos x = 4/5, tan x = 3/4, cosec x = 5/3, sec x = 5/4, cot x = 4/3
2. If sin x = 12/13 and lies in the second quadrant, find the value of sec x + tan x.
Solution:
Given:
Sin x = 12/13 and x lies in the second quadrant.
We know, in second quadrant, sin x and cosec x are positive and all other ratios are negative.
By using the formulas,
Cos x = √(1-sin2 x)
= – √(1-(12/13)2)
= – √(1- (144/169))
= – √(169-144)/169
= -√(25/169)
= – 5/13
We know,
tan x = sin x/cos x
sec x = 1/cos x
Now,
tan x = (12/13)/(-5/13)
= -12/5
sec x = 1/(-5/13)
= -13/5
Sec x + tan x = -13/5 + (-12/5)
= (-13-12)/5
= -25/5
= -5
∴ Sec x + tan x = -5
3. If sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2 find the value of 8 tan x -√5 sec y.
Solution:
Given:
sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2
We know that, x is in second quadrant and y is in third quadrant.
In second quadrant, cos x and tan x are negative.
In third quadrant, sec y is negative.
By using the formula,
cos x = – √(1-sin2 x)
tan x = sin x/cos x
Now,
cos x = – √(1-sin2 x)
= – √(1 – (3/5)2)
= – √(1 – 9/25)
= – √((25-9)/25)
= – √(16/25)
= – 4/5
tan x = sin x/cos x
= (3/5)/(-4/5)
= 3/5 × -5/4
= -3/4
We know that sec y = – √(1+tan2 y)
= – √(1 + (1/2)2)
= – √(1 + 1/4)
= – √((4+1)/4)
= – √(5/4)
= – √5/2
Now, 8 tan x – √5 sec y = 8(-3/4) – √5(-√5/2)
= -6 + 5/2
= (-12+5)/2
= -7/2
∴ 8 tan x – √5 sec y = -7/2
4. If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x.
Solution:
Given:
Sin x + cos x = 0 and x lies in fourth quadrant.
Sin x = -cos x
Sin x/cos x = -1
So, tan x = -1 (since, tan x = sin x/cos x)
We know that, in fourth quadrant, cos x and sec x are positive and all other ratios are negative.
By using the formulas,
Sec x = √(1 + tan2 x)
Cos x = 1/sec x
Sin x = – √(1- cos2 x)
Now,
Sec x = √(1 + tan2 x)
= √(1 + (-1)2)
= √2
Cos x = 1/sec x
= 1/√2
Sin x = – √(1 – cos2 x)
= – √(1 – (1/√2)2)
= – √(1 – (1/2))
= – √((2-1)/2)
= – √(1/2)
= -1/√2
∴ sin x = -1/√2 and cos x = 1/√2
5. If cos x = -3/5 and π<x<3π/2 find the values of other five trigonometric functions and hence evaluate
Solution:
Given:
cos x= -3/5 and π <x < 3π/2
We know that in the third quadrant, tan x and cot x are positive and all other rations are negative.
By using the formulas,
Sin x = – √(1-cos2 x)
Tan x = sin x/cos x
Cot x = 1/tan x
Sec x = 1/cos x
Cosec x = 1/sin x
Now,
Sin x = – √(1-cos2 x)
= – √(1-(-3/5)2)
= – √(1-9/25)
= – √((25-9)/25)
= – √(16/25)
= – 4/5
Tan x = sin x/cos x
= (-4/5)/(-3/5)
= -4/5 × -5/3
= 4/3
Cot x = 1/tan x
= 1/(4/3)
= 3/4
Sec x = 1/cos x
= 1/(-3/5)
= -5/3
Cosec x = 1/sin x
= 1/(-4/5)
= -5/4
∴
= [(-5/4) + (3/4)] / [(-5/3) – (4/3)]
= [(-5+3)/4] / [(-5-4)/3]
= [-2/4] / [-9/3]
= [-1/2] / [-3]
= 1/6
EXERCISE 5.3 PAGE NO: 5.39
1. Find the values of the following trigonometric ratios:
(i) sin 5π/3
(ii) sin 17π
(iii) tan 11π/6
(iv) cos (-25π/4)
(v) tan 7π/4
(vi) sin 17π/6
(vii) cos 19π/6
(viii) sin (-11π/6)
(ix) cosec (-20π/3)
(x) tan (-13π/4)
(xi) cos 19π/4
(xii) sin 41π/4
(xiii) cos 39π/4
(xiv) sin 151π/6
Solution:
(i) sin 5π/3
5π/3 = (5π/3 × 180)o
= 300o
= (90×3 + 30)o
Since, 300o lies in IV quadrant in which sine function is negative.
sin 5π/3 = sin (300)o
= sin (90×3 + 30)o
= – cos 30o
= – √3/2
(ii) sin 17π
Sin 17π = sin 3060o
= sin (90×34 + 0)o
Since, 3060o lies in the negative direction of x-axis i.e., on boundary line of II and III quadrants.
Sin 17π = sin (90×34 + 0)o
= – sin 0o
= 0
(iii) tan 11π/6
tan 11π/6 = (11/6 × 180)o
= 330o
Since, 330o lies in the IV quadrant in which tangent function is negative.
tan 11π/6 = tan (300)o
= tan (90×3 + 60)o
= – cot 60o
= – 1/√3
(iv) cos (-25π/4)
cos (-25π/4) = cos (-1125)o
= cos (1125)o
Since, 1125o lies in the I quadrant in which cosine function is positive.
cos (1125)o = cos (90×12 + 45)o
= cos 45o
= 1/√2
(v) tan 7π/4
tan 7π/4 = tan 315o
= tan (90×3 + 45)o
Since, 315o lies in the IV quadrant in which tangent function is negative.
tan 315o = tan (90×3 + 45)o
= – cot 45o
= -1
(vi) sin 17π/6
sin 17π/6 = sin 510o
= sin (90×5 + 60)o
Since, 510o lies in the II quadrant in which sine function is positive.
sin 510o = sin (90×5 + 60)o
= cos 60o
= 1/2
(vii) cos 19π/6
cos 19π/6 = cos 570o
= cos (90×6 + 30)o
Since, 570o lies in III quadrant in which cosine function is negative.
cos 570o = cos (90×6 + 30)o
= – cos 30o
= – √3/2
(viii) sin (-11π/6)
sin (-11π/6) = sin (-330o)
= – sin (90×3 + 60)o
Since, 330o lies in the IV quadrant in which the sine function is negative.
sin (-330o) = – sin (90×3 + 60)o
= – (-cos 60o)
= – (-1/2)
= 1/2
(ix) cosec (-20π/3)
cosec (-20π/3) = cosec (-1200)o
= – cosec (1200)o
= – cosec (90×13 + 30)o
Since, 1200o lies in the II quadrant in which cosec function is positive.
cosec (-1200)o = – cosec (90×13 + 30)o
= – sec 30o
= -2/√3
(x) tan (-13π/4)
tan (-13π/4) = tan (-585)o
= – tan (90×6 + 45)o
Since, 585o lies in the III quadrant in which the tangent function is positive.
tan (-585)o = – tan (90×6 + 45)o
= – tan 45o
= -1
(xi) cos 19π/4
cos 19π/4 = cos 855o
= cos (90×9 + 45)o
Since, 855o lies in the II quadrant in which the cosine function is negative.
cos 855o = cos (90×9 + 45)o
= – sin 45o
= – 1/√2
(xii) sin 41π/4
sin 41π/4 = sin 1845o
= sin (90×20 + 45)o
Since, 1845o lies in the I quadrant in which the sine function is positive.
sin 1845o = sin (90×20 + 45)o
= sin 45o
= 1/√2
(xiii) cos 39π/4
cos 39π/4 = cos 1755o
= cos (90×19 + 45)o
Since, 1755o lies in the IV quadrant in which the cosine function is positive.
cos 1755o = cos (90×19 + 45)o
= sin 45o
= 1/√2
(xiv) sin 151π/6
sin 151π/6 = sin 4530o
= sin (90×50 + 30)o
Since, 4530o lies in the III quadrant in which the sine function is negative.
sin 4530o = sin (90×50 + 30)o
= – sin 30o
= -1/2
2. prove that:
(i) tan 225o cot 405o + tan 765o cot 675o = 0
(ii) sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2
(iii) cos 24o + cos 55o + cos 125o + cos 204o + cos 300o = 1/2
(iv) tan (-125o) cot (-405o) – tan (-765o) cot (675o) = 0
(v) cos 570o sin 510o + sin (-330o) cos (-390o) = 0
(vi) tan 11π/3 – 2 sin 4π/6 – 3/4 cosec2 π/4 + 4 cos2 17π/6 = (3 – 4√3)/2
(vii) 3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1
Solution:
(i) tan 225o cot 405o + tan 765o cot 675o = 0
Let us consider LHS:
tan 225° cot 405° + tan 765° cot 675°
tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8 + 45°) cot (90° × 7 + 45°)
We know that when n is odd, cot → tan.
tan 45° cot 45° + tan 45° [-tan 45°]
tan 45° cot 45° – tan 45° tan 45°
1 × 1 – 1 × 1
1 – 1
0 = RHS
∴ LHS = RHS
Hence proved.
(ii) sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2
Let us consider LHS:
sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6
sin 480° cos 690° + cos 780° sin 1050°
sin (90° × 5 + 30°) cos (90° × 7 + 60°) + cos (90° × 8 + 60°) sin (90° × 11 + 60°)
We know that when n is odd, sin → cos and cos → sin.
cos 30° sin 60° + cos 60° [-cos 60°]
√3/2 × √3/2 – 1/2 × 1/2
3/4 – 1/4
2/4
1/2
= RHS
∴ LHS = RHS
Hence proved.
(iii) cos 24o + cos 55o + cos 125o + cos 204o + cos 300o = 1/2
Let us consider LHS:
cos 24o + cos 55o + cos 125o + cos 204o + cos 300o
cos 24° + cos (90° × 1 – 35°) + cos (90° × 1 + 35°) + cos (90° × 2 + 24°) + cos (90° × 3 + 30°)
We know that when n is odd, cos → sin.
cos 24° + sin 35° – sin 35° – cos 24° + sin 30°
0 + 0 + 1/2
1/2
= RHS
∴ LHS = RHS
Hence proved.
(iv) tan (-125o) cot (-405o) – tan (-765o) cot (675o) = 0
Let us consider LHS:
tan (-125o) cot (-405o) – tan (-765o) cot (675o)
We know that tan (-x) = -tan (x) and cot (-x) = -cot (x).[-tan (225°)] [-cot (405°)] – [-tan (765°)] cot (675°)
tan (225°) cot (405°) + tan (765°) cot (675°)
tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8 + 45°) cot (90° × 7 + 45°)
tan 45° cot 45° + tan 45° [-tan 45°]
1 × 1 + 1 × (-1)
1 – 1
0
= RHS
∴ LHS = RHS
Hence proved.
(v) cos 570o sin 510o + sin (-330o) cos (-390o) = 0
Let us consider LHS:
cos 570o sin 510o + sin (-330o) cos (-390o)
We know that sin (-x) = -sin (x) and cos (-x) = +cos (x).
cos 570o sin 510o + [-sin (330o)] cos (390o)
cos 570o sin 510o – sin (330o) cos (390o)
cos (90° × 6 + 30°) sin (90° × 5 + 60°) – sin (90° × 3 + 60°) cos (90° × 4 + 30°)
We know that cos is negative at 90° + θ i.e. in Q2 and when n is odd, sin → cos and cos → sin.
-cos 30° cos 60° – [-cos 60°] cos 30°
-cos 30° cos 60° + cos 60° cos 30°
0
= RHS
∴ LHS = RHS
Hence proved.
(vi) tan 11π/3 – 2 sin 4π/6 – 3/4 cosec2 π/4 + 4 cos2 17π/6 = (3 – 4√3)/2
Let us consider LHS:
tan 11π/3 – 2 sin 4π/6 – 3/4 cosec2 π/4 + 4 cos2 17π/6
tan (11 × 180o)/3 – 2 sin (4 × 180o)/6 – 3/4 cosec2 180o/4 + 4 cos2 (17 × 180o)/6
tan 660o – 2 sin 120o – 3/4 (cosec 45o)2 + 4 (cos 510o)2
tan (90° × 7 + 30°) – 2 sin (90° × 1 + 30°) – 3/4 [cosec 45°]2 + 4 [cos (90° × 5 + 60°)]2
We know that tan and cos is negative at 90° + θ i.e. in Q2 and when n is odd, tan → cot, sin → cos and cos → sin.[-cot 30°] – 2 cos 30° – 3/4 [cosec 45°]2 + [-sin 60°]2
– cot 30° – 2 cos 30° – 3/4 [cosec 45°]2 + [sin 60°]2
-√3 – 2√3/2 – 3/4 (√2)2 + 4 (√3/2)2
-√3 – √3 – 6/4 + 12/4
(3 – 4√3)/2
= RHS
∴ LHS = RHS
Hence proved.
(vii) 3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1
Let us consider LHS:
3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4
3 sin 180o/6 sec 180o/3 – 4 sin 5(180o)/6 cot 180o/4
3 sin 30° sec 60° – 4 sin 150° cot 45°
3 sin 30° sec 60° – 4 sin (90° × 1 + 60°) cot 45°
We know that when n is odd, sin → cos.
3 sin 30° sec 60° – 4 cos 60° cot 45°
3 (1/2) (2) – 4 (1/2) (1)
3 – 2
1
= RHS
∴ LHS = RHS
Hence proved.
3. Prove that:
(i)

(ii)

(iii)

(iv)

(v)

Solution:
(i)

1 = RHS
∴ LHS = RHS
Hence proved.
(ii)

1 + 1
2 = RHS
∴ LHS = RHS
Hence proved.
(iii)


1 = RHS
∴ LHS = RHS
Hence proved.
(iv)

{1 + cot x – (-cosec x)} {1 + cot x + (-cosec x)}
{1 + cot x + cosec x} {1 + cot x – cosec x}
{(1 + cot x) + (cosec x)} {(1 + cot x) – (cosec x)}
By using the formula, (a + b) (a – b) = a2 – b2
(1 + cot x)2 – (cosec x)2
1 + cot2 x + 2 cot x – cosec2 x
We know that 1 + cot2 x = cosec2 x
cosec2 x + 2 cot x – cosec2 x
2 cot x = RHS
∴ LHS = RHS
Hence proved.
(v)

1 = RHS
∴ LHS = RHS
Hence proved.
4. Prove that: sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 = 2
Solution:
Let us consider LHS:
sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9
sin2 π/18 + sin2 2π/18 + sin2 7π/18 + sin2 8π/18
sin2 π/18 + sin2 2π/18 + sin2 (π/2 – 2π/18) + sin2 (π/2 – π/18)
We know that when n is odd, sin → cos.
sin2 π/18 + sin2 2π/18 + cos2 2π/18 + cos2 2π/18
when rearranged,
sin2 π/18 + cos2 2π/18 + sin2 π/18 + cos2 2π/18
We know that sin2 + cos2x = 1.
So,
1 + 1
2 = RHS
∴ LHS = RHS
Hence proved.
RD Sharma Solutions for Class 11 Maths Chapter 5: Download PDF
RD Sharma Solutions for Class 11 Maths Chapter 5–Trigonometric Functions
Download PDF: RD Sharma Solutions for Class 11 Maths Chapter 5–Trigonometric Functions PDF
Chapterwise RD Sharma Solutions for Class 11 Maths :
- Chapter 1–Sets
- Chapter 2–Relations
- Chapter 3–Functions
- Chapter 4–Measurement of Angles
- Chapter 5–Trigonometric Functions
- Chapter 6–Graphs of Trigonometric Functions
- Chapter 7–Values of Trigonometric Functions at Sum or Difference of Angles
- Chapter 8–Transformation Formulae
- Chapter 9–Values of Trigonometric Functions at Multiples and Submultiples of an Angle
- Chapter 10–Sine and Cosine Formulae and their Applications
- Chapter 11–Trigonometric Equations
- Chapter 12–Mathematical Induction
- Chapter 13–Complex Numbers
- Chapter 14–Quadratic Equations
- Chapter 15–Linear Inequations
- Chapter 16–Permutations
- Chapter 17–Combinations
- Chapter 18–Binomial Theorem
- Chapter 19–Arithmetic Progressions
- Chapter 20–Geometric Progressions
- Chapter 21–Some Special Series
- Chapter 22–Brief review of Cartesian System of Rectangular Coordinates
- Chapter 23–The Straight Lines
- Chapter 24–The Circle
- Chapter 25–Parabola
- Chapter 26–Ellipse
- Chapter 27–Hyperbola
- Chapter 28–Introduction to Three Dimensional Coordinate Geometry
- Chapter 29–Limits
- Chapter 30–Derivatives
- Chapter 31–Mathematical Reasoning
- Chapter 32–Statistics
- Chapter 33–Probability
About RD Sharma
RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.
R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.