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RD Sharma Class 11 Solutions Chapter 9 Exercise 9.1
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Prove the following identities:
1. √[(1 – cos 2x) / (1 + cos 2x)] = tan x
Solution:
Let us consider LHS:
√[(1 – cos 2x) / (1 + cos 2x)]
We know that cos 2x = 1 – 2 sin2 x
= 2 cos2 x – 1
So,
√[(1 – cos 2x) / (1 + cos 2x)] = √[(1 – (1 – 2sin2 x)) / (1 + (2cos2x – 1))]
= √[(1 – 1 + 2sin2 x) / (1 + 2cos2 x – 1)]
= √[2 sin2 x / 2 cos2 x]
= sin x/cos x
= tan x
= RHS
Hence proved.
2. sin 2x / (1 – cos 2x) = cot x
Solution:
Let us consider LHS:
sin 2x / (1 – cos 2x)
We know that cos 2x = 1 – 2 sin2 x
Sin 2x = 2 sin x cos x
So,
sin 2x / (1 – cos 2x) = (2 sin x cos x) / (1 – (1 – 2sin2 x))
= (2 sin x cos x) / (1 – 1 + 2sin2 x)]
= [2 sin x cos x / 2 sin2 x]
= cos x/sin x
= cot x
= RHS
Hence proved.
3. sin 2x / (1 + cos 2x) = tan x
Solution:
Let us consider LHS:
sin 2x / (1 + cos 2x)
We know that cos 2x = 1 – 2 sin2 x
= 2 cos2 x – 1
Sin 2x = 2 sin x cos x
So,
sin 2x / (1 + cos 2x) = [2 sin x cos x / (1 + (2cos2x – 1))]
= [2 sin x cos x / (1 + 2cos2 x – 1)]
= [2 sin x cos x / 2 cos2 x]
= sin x/cos x
= tan x
= RHS
Hence proved.
5. [1 – cos 2x + sin 2x] / [1 + cos 2x + sin 2x] = tan x
Solution:
Let us consider LHS:
[1 – cos 2x + sin 2x] / [1 + cos 2x + sin 2x]We know that, cos 2x = 1 – 2 sin2 x
= 2 cos2 x – 1
Sin 2x = 2 sin x cos x
So,
6. [sin x + sin 2x] / [1 + cos x + cos 2x] = tan x
Solution:
Let us consider LHS:
[sin x + sin 2x] / [1 + cos x + cos 2x]We know that, cos 2x = cos2 x – sin2 x
Sin 2x = 2 sin x cos x
So,
= RHS
Hence proved.
7. cos 2x / (1 + sin 2x) = tan (π/4 – x)
Solution:
Let us consider LHS:
cos 2x / (1 + sin 2x)
We know that, cos 2x = cos2 x – sin2 x
Sin 2x = 2 sin x cos x
So,
8. cos x / (1 – sin x) = tan (π/4 + x/2)
Solution:
Let us consider LHS:
cos x / (1 – sin x)
We know that, cos 2x = cos2 x – sin2 x
Cos x = cos2 x/2 – sin2 x/2
Sin 2x = 2 sin x cos x
Sin x = 2 sin x/2 cos x/2
So,
11. (cos α + cos β) 2 + (sin α + sin β) 2 = 4 cos2 (α – β)/2
Solution:
Let us consider LHS:
(cos α + cos β)2 + (sin α + sin β)2
Upon expansion, we get,
(cos α + cos β)2 + (sin α + sin β)2 =
= cos2 α + cos2 β + 2 cos α cos β + sin2 α + sin2 β + 2 sin α sin β
= 2 + 2 cos α cos β + 2 sin α sin β
= 2 (1 + cos α cos β + sin α sin β)
= 2 (1 + cos (α – β)) [since, cos (A – B) = cos A cos B + sin A sin B]
= 2 (1 + 2 cos2 (α – β)/2 – 1) [since, cos2x = 2cos2 x – 1]
= 2 (2 cos2 (α – β)/2)
= 4 cos2 (α – β)/2
= RHS
Hence Proved.
12. sin2 (π/8 + x/2) – sin2 (π/8 – x/2) = 1/√2 sin x
Solution:
Let us consider LHS:
sin2 (π/8 + x/2) – sin2 (π/8 – x/2)
we know, sin2 A – sin2 B = sin (A+B) sin (A-B)
so,
sin2 (π/8 + x/2) – sin2 (π/8 – x/2) = sin (π/8 + x/2 + π/8 – x/2) sin (π/8 + x/2 – (π/8 – x/2))
= sin (π/8 + π/8) sin (π/8 + x/2 – π/8 + x/2)
= sin π/4 sin x
= 1/√2 sin x [since, since π/4 = 1/√2]
= RHS
Hence proved.
13. 1 + cos2 2x = 2 (cos4 x + sin4 x)
Solution:
Let us consider LHS:
1 + cos2 2x
We know, cos2x = cos2 x – sin2 x
cos2 x + sin2 x = 1
so,
1 + cos2 2x = (cos2 x + sin2 x) 2 + (cos2 x – sin2 x) 2
= (cos4 x + sin4 x + 2 cos2 x sin2 x) + (cos4 x + sin4 x – 2 cos2 x sin2 x)
= cos4 x + sin4 x + cos4 x + sin4 x
= 2 cos4 x + 2 sin4 x
= 2 (cos4 x + sin4 x)
= RHS
Hence proved.
14. cos3 2x + 3 cos 2x = 4 (cos6 x – sin6 x)
Solution:
Let us consider RHS:
4 (cos6 x – sin6 x)
Upon expansion, we get,
4 (cos6 x – sin6 x) = 4 [(cos2 x)3 – (sin2 x)3]
= 4 (cos2 x – sin2 x) (cos4 x + sin4 x + cos2 x sin2 x)
By using the formula,
a3 – b3 = (a-b) (a2 + b2 + ab)
= 4 cos 2x (cos4 x + sin4 x + cos2 x sin2 x + cos2 x sin2 x – cos2 x sin
We know, cos 2x = cos2 x – sin2 x
So,
= 4 cos 2x (cos4 x + sin4 x + 2 cos2 x sin2 x – cos2 x sin2 x)
= 4 cos 2x [(cos2 x)2 + (sin2 x)2 + 2 cos2 x sin2 x – cos2 x sin2 x]
We know, a2 + b2 + 2ab = (a + b)2
= 4 cos 2x [(1)2 – 1/4 (4 cos2 x sin2 x)]
= 4 cos 2x [(1)2 – 1/4 (2 cos x sin x)2]
We know, sin 2x = 2sin x cos x
= 4 cos 2x [(12) – 1/4 (sin 2x)2]
= 4 cos 2x (1 – 1/4 sin2 2x)
We know, sin2 x = 1 – cos2 x
= 4 cos 2x [1 – 1/4 (1 – cos2 2x)]
= 4 cos 2x [1 – 1/4 + 1/4 cos2 2x]
= 4 cos 2x [3/4 + 1/4 cos2 2x]
= 4 (3/4 cos 2x + 1/4 cos3 2x)
= 3 cos 2x + cos3 2x
= cos3 2x + 3 cos 2x
= LHS
Hence proved.
15. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
Solution:
Let us consider LHS:
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= (sin 3x) (sin x) + sin2 x + (cos 3x) (cos x) – cos2 x
= [(sin 3x) (sin x) + (cos 3x) (cos x)] + (sin2 x – cos2 x)
= [(sin 3x) (sin x) + (cos 3x) (cos x)] – (cos2 x – sin2 x)
= cos (3x – x) – cos 2x
We know, cos 2x = cos2 x – sin2 x
cos A cos B + sin A sin B = cos(A – B)
So,
= cos 2x – cos 2x
= 0
= RHS
Hence Proved.
16. cos2 (π/4 – x) – sin2 (π/4 – x) = sin 2x
Solution:
Let us consider LHS:
cos2 (π/4 – x) – sin2 (π/4 – x)
We know, cos2 A – sin2 A = cos 2A
So,
cos2 (π/4 – x) – sin2 (π/4 – x) = cos 2 (π/4 – x)
= cos (π/2 – 2x)
= sin 2x [since, cos (π/2 – A) = sin A]
= RHS
Hence proved.
17. cos 4x = 1 – 8 cos2 x + 8 cos4 x
Solution:
Let us consider LHS:
cos 4x
We know, cos 2x = 2 cos2 x – 1
So,
cos 4x = 2 cos2 2x – 1
= 2(2 cos2 2x – 1)2 – 1
= 2[(2 cos2 2x) 2 + 12 – 2×2 cos2 x] – 1
= 2(4 cos4 2x + 1 – 4 cos2 x) – 1
= 8 cos4 2x + 2 – 8 cos2 x – 1
= 8 cos4 2x + 1 – 8 cos2 x
= RHS
Hence Proved.
18. sin 4x = 4 sin x cos3 x – 4 cos x sin3 x
Solution:
Let us consider LHS:
sin 4x
We know, sin 2x = 2 sin x cos x
cos 2x = cos2 x – sin2 x
So,
sin 4x = 2 sin 2x cos 2x
= 2 (2 sin x cos x) (cos2 x – sin2 x)
= 4 sin x cos x (cos2 x – sin2 x)
= 4 sin x cos3 x – 4 sin3 x cos x
= RHS
Hence proved.
19. 3(sin x – cos x) 4 + 6 (sin x + cos x) 2 + 4 (sin6 x + cos6 x) = 13
Solution:
Let us consider LHS:
3(sin x – cos x) 4 + 6 (sin x + cos x) 2 + 4 (sin6 x + cos6 x)
We know, (a + b)2 = a2 + b2 + 2ab
(a – b)2 = a2 + b2 – 2ab
a3 + b3 = (a + b) (a2 + b2 – ab)
So,
3(sin x – cos x) 4 + 6 (sin x + cos x) 2 + 4 (sin6 x + cos6 x) = 3{(sin x – cos x) 2}2 + 6 {(sin x)2 + (cos x)2 + 2 sin x cos x)} + 4 {(sin2 x)3 + (cos2 x)3}
= 3{(sin x) 2 + (cos x)2 – 2 sin x cos x)}2 + 6 (sin2 x + cos2 x + 2 sin x cos x) + 4{(sin2 x + cos2 x) (sin4 x + cos4 x – sin2 x cos2 x)}
= 3(1 – 2 sin x cos x) 2 + 6 (1 + 2 sin x cos x) + 4{(1) (sin4 x + cos4 x – sin2 x cos2 x)}
We know, sin2 x + cos2 x = 1
So,
= 3{12 + (2 sin x cos x) 2 – 4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin2 x)2 + (cos2 x)2 + 2 sin2 x cos2 x – 3 sin2 x cos2 x)}
= 3{1 + 4 sin2 x cos2 x – 4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin2 x + cos2 x) 2 – 3 sin2 x cos2 x)}
= 3 + 12 sin2 x cos2 x – 12 sin x cos x + 6 + 12 sin x cos x + 4{(1)2 – 3 sin2 x cos2 x)}
= 9 + 12 sin2 x cos2 x + 4(1 – 3 sin2 x cos2 x)
= 9 + 12 sin2 x cos2 x + 4 – 12 sin2 x cos2 x
= 13
= RHS
Hence proved.
20. 2(sin6 x + cos6 x) – 3(sin4 x + cos4 x) + 1 = 0
Solution:
Let us consider LHS:
2(sin6 x + cos6 x) – 3(sin4 x + cos4 x) + 1
We know, (a + b)2 = a2 + b2 + 2ab
a3 + b3 = (a + b) (a2 + b2 – ab)
So,
2(sin6 x + cos6 x) – 3(sin4 x + cos4 x) + 1 = 2{(sin2 x) 3 + (cos2 x) 3} – 3{(sin2 x) 2 + (cos2 x) 2} + 1
= 2{(sin2 x + cos2 x) (sin4 x + cos4 x – sin2 x cos2 x} – 3{(sin2 x) 2 + (cos2 x) 2 + 2sin2 x cos2 x – 2sin2 x cos2 x} + 1
= 2{(1) (sin4 x + cos4 x + 2 sin2 x cos2 x – 3 sin2 x cos2 x} – 3{(sin2 x + cos2 x) 2 – 2sin2 x cos2 x} + 1
We know, sin2 x + cos2 x = 1
= 2{(sin2 x + cos2 x) 2 – 3 sin2 x cos2 x} – 3{(1)2 – 2sin2 x cos2 x} + 1
= 2{(1)2 – 3 sin2 x cos2 x} – 3(1 – 2sin2 x cos2 x) + 1
= 2(1 – 3 sin2 x cos2 x) – 3 + 6 sin2 x cos2 x + 1
= 2 – 6 sin2 x cos2 x – 2 + 6 sin2 x cos2 x
= 0
= RHS
Hence proved.
21. cos6 x – sin6 x = cos 2x (1 – 1/4 sin2 2x)
Solution:
Let us consider LHS:
cos6 x – sin6 x
We know, (a + b) 2 = a2 + b2 + 2ab
a3 – b3 = (a – b) (a2 + b2 + ab)
So,
cos6 x – sin6 x = (cos2 x)3 – (sin2 x)3
= (cos2 x – sin2 x) (cos4 x + sin4 x + cos2 x sin2 x)
We know, cos 2x = cos2 x – sin2 x
So,
= cos 2x [(cos2 x) 2 + (sin2 x) 2 + 2 cos2 x sin2 x – cos2 x sin2 x]
= cos 2x [(cos2 x) 2 + (sin2 x) 2 – 1/4 × 4 cos2 x sin2 x]
We know, sin2 x + cos2 x = 1
So,
= cos 2x [(1)2 – 1/4 × (2 cos x sin x) 2]
We know, sin 2x = 2 sin x cos x
So,
= cos 2x [1 – 1/4 × (sin 2x) 2]
= cos 2x [1 – 1/4 × sin2 2x]
= RHS
Hence proved.
22. tan (π/4 + x) + tan (π/4 – x) = 2 sec 2x
Solution:
Let us consider LHS:
tan (π/4 + x) + tan (π/4 – x)
We know,
tan (A+B) = (tan A + tan B)/(1- tan A tan B)
tan (A-B) = (tan A – tan B)/(1+ tan A tan B)
So,
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