# ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities

1. If A is an acute angle and sin A = 3/5, find out all other trigonometric ratios of angle A (using trigonometric identities).
Solution:
As we have Given that ,
sin A = 3/5 and A is an acute angle
Then, in ∆ABC
As we have ∠B = 90°
And, AC = 5 and BC = 3
By The Pythagoras theorem,
AB = √(AC2 – BC2
= √(52 - 32 )
= √(25 - 9)
= √16
Then ,

2. If A is an acute angle and sec A =17/8, find out  all other trigonometric ratios of angle A (using trigonometric identities).
Solution:
As we have Given,

sec A =17/8 and A is an acute angle
Therefore, in ∆ ABC
AS  we have Given
∠B = 90o And, AC = 17 and AB = 8
By using the  Pythagoras theorem,
BC = √(AC2 – AB2
= √(172 - 8 2 )
= √(289 - 64)
= √225
Hence , BC = 15
Therefore,

3. Express the ratios cos A, tan A and sec A in terms of sin A.
Solution:
As we know that,
sin2 A + cos2 A = 1
So,
cos A = √(1 – sin2 A)
tan A = sin A/cos A = sin A/ √(1 – sin2 A)
sec A = 1/cos A = 1/ (√1 – sin2 A)

4. If tan A = 1/√3, find out all other trigonometric ratios of angle A.
Solution:
As we have Given that ,
tan A = 1/√3
In the right ∆ ABC,
Then, BC = 1 and AB = √3
By using the Pythagoras theorem,
AC = √(AB2 + BC2
= √[(√3)2 + (1)2
= √(3 + 1)
= √4
Hence AC = 2
Therefore,

5. If 12 cosec θ = 13, find the value of (2 sin θ – 3 cos θ)/ (4 sin θ – 9 cos θ)
Solution:
As we have Given,
12 cosec θ = 13
⇒ cosec θ = 13/12
In the right ∆ ABC,
∠A = θ
So, cosec θ = AC/BC = 13/12
AC = 13 and BC = 12
By using the  Pythagoras theorem,
AB = √(AC2 – BC2
= √[(13)2 - (12)2
= √(169 - 144)
= √25
Hence AC= 5
Now,
sin θ = BC/AC = 12/13
cos θ = AB/AC = 5/13
Therefore,

6. Without using trigonometric tables, evaluate the following (6 to 10):
(i) cos2 26o + cos 64o sin 26o + (tan 36o / cot 54o
(ii) (sec 17o / cosec 73o ) + (tan 68o / cot 22o ) + cos2 44o + cos2 46
Solution:
As  we have Given that,
(i) cos2 26o + cos 64o sin 26o + (tan 36o / cot 54o
= cos2 26o + cos (90o - 16o ) sin 26o + [tan 36o / cot (90o - 54o )]
= [cos2 26o + sin2 26o ] + (tan 36o / tan 36o
= 1 + 1 = 2
(ii) (sec 17o / cosec 73o ) + (tan 68o / cot 22o ) + cos2 44o + cos2 46o
= [sec 17o / cosec (90o - 73o )] + [(tan 90o – 22o )/ cot 22o ] + cos2 (90o - 44o ) + cos2 46o
= [sec 17o / sec 17o ] + [cot 22o / cot 22o ] + [sin2 46o + cos2 46o
= 1 + 1 + 1
= 3

(ii) L.H.S = tan θ/ tan (90o - θ) + sin (90o - θ)/ cos θ
= tan θ/ cot θ + cos θ/ cos θ
= tan θ/ (1/tan θ) + 1
= tan2 θ + 1
= sec2 θ = R.H.S.
(iii) L.H.S. = (cos (90o - θ) cos θ)/ tan θ + cos2 (90o - θ)
= (sin θ cos θ)/ tan θ + sin2 θ
= (sin θ cos θ)/ (sin θ/ cos θ) + sin2 θ = cos2 θ + sin2 θ
= 1 = R.H.S.

Prove that following (12 to 30) identities, where the angles involved are acute angles for which the trigonometric ratios as defined:

12.
(i) (sec A + tan A) (1 – sin A) = cos A
(ii) (1 + tan2 A) (1 – sin A) (1 + sin A) = 1.
Solution:
As we have Given that,
(i) L.H.S = (sec⁡A+tan⁡A )(1 –sin⁡A )

13. (i) tan A + cot A = sec A cosec A
(ii) (1 – cos A) (1 + sec A) = tan A sin A.
Solution:
(i) L.H.S. = tan A + cot A
= sin A/cos A + cos A/sin A
= (sin2 A + cos2 A)/ (sin A cos A)
= 1/ (sin A cos A)
= sec A cosec A
= R.H.S

14. (i) 1/ (1 + cos A) + 1/ (1 – cos A) = 2 cosec2
(ii) 1/(sec A + tan A) + 1/(sec A – tan A) = 2 sec A

Solution:
(i) L.H.S =1/ (1 + cos A) + 1/ (1 – cos A)

15. (i) sin A/ (1 + cos A) = (1 – cos A)/ sin A
(ii) (1 – tan2 A)/ (cot2 A - 1) = tan2 A
(iii) sin A/ (1 + cos A) = cosec A – cot A
Solution:
(i)=L.H.S=sinA / (1+cosA)
When multiplying and dividing by (1 – cos A), we have

16. (i) (sec A – 1)/(sec A + 1) = (1 – cos A)/(1 + cos A)
(ii) tan2 θ/ (sec θ – 1)2 = (1 + cos θ)/ (1 – cos θ)
(iii) (1 + tan A)2 + (1 – tan A)2 = 2 sec2
(iv) sec2 A + cosec2 A = sec2 A. cosec2
Solution:

(iii) L.H.S. = (1 + tan A)2 + (1 – tan A)
= 1 + 2 tan A + tan2 A + 1 – 2 tan A + tan2
= 2 + 2 tan2 A = 2(1 + tan2 A)               [As 1 + tan2 A = sec2 A]
= 2 sec2 A = R.H.S.
(iv) L.H.S = sec2 A + cosec2 A
= 1/cos2 A + 1/sin2
= (sin2 A + cos2 A)/ (sin2 A cos2 A)
= 1/ (sin2 A cos2 A)
= sec2 A cosec2 A = R.H.S

17.(i) (1 + sin A)/ cos A + cos A/ (1 + sin A) = 2 sec A
(ii) tan A/ (sec A - 1) + tan A/ (sec A + 1) = 2cosec A
Solution:
(i) L.H.S. = (1 + sin A)/ cos A + cos A/ (1 + sin A)

18. (i) cosec A/ (cosec A - 1) + cosec A/ (cosec A + 1) = 2 sec2 A
(ii) cot A – tan A = (2cos2 A - 1)/ (sin A – cos A)
(iii) (cot A – 1)/ (2 – sec2 A) = cot A/ (1 + tan A)
Solution:

19. (i) tan2 θ – sin2 θ = tan2 θ sin2 θ
(ii) cos θ/ (1 – tan θ) – sin2 θ/ (cos θ - sin θ) = cos θ + sin θ
Solution:
L.H.S= tan2 θ- sin2 θ

20. (i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ
(ii) 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ = cot4 θ – tan4 θ.
Solution:
(i) L.H.S. = cosec4 θ – cosec2 θ
= cosec2 θ (cosec2 θ - 1)
= cosec2 θ cot2 θ            [cosec2 θ – 1 = cot2 θ]
= (cot2 θ + 1) cot2 θ
= cot4 θ + cot2 θ = R.H.S.

= tan2 θ+tan θ+1=tan2 θ+1+tanθ
=sec2 θ+tan θ                {∴ sec2 θ=tan2 θ+1}
=R.H.S

25. (i) (1 + tan A)/ sin A + (1 + cot A)/ cos A = 2 (sec A + cosec A)
(ii) sec4 A – tan4 A = 1 + 2 tan2
Solution:

=2(secA+cosecA)=R.H.S
(ii) sec4 A – tan4 A = 1 + 2 tan2
L.H.S. = sec4 A – tan4 A = (sec2 A – tan2 A) (sec2 A + tan2 A)
= (1 + tan4 A – tan4 A) (1 + tan4 A + tan4 A)            [As sec2 A = tan4 A + 1]
= 1 (1 + 2 tan2 A)
= 1 + 2 tan2 A = R.H.S.

28. (i) (sin θ + cos θ) (sec θ + cosec θ) = 2 + sec θ cosec θ
(ii) (cosec A – sin A) (sec A – cos A) sec2A = tan A
Solution:
(i) (sin θ + cos θ) (sec θ + cosec θ) = 2 + sec θ cosec θ
L.H.S. = (sin θ + cos θ) (sec θ + cosec θ)

30. (i) 1/ (sec A + tan A) – 1/cos A = 1/cos A – 1/(sec A – tan A)
(ii) (sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A cosec A)2
(iii) (tan A + sin A)/ (tan A – sin A) = (sec A + 1)/ (sec A – 1)
Solution:

31. If sin θ + cos θ = √2 sin (90° – θ), show that cot θ = √2 + 1
Solution:
As we have Given that,
sin θ + cos θ = √2 sin (90° – θ)
sin θ + cos θ = √2 cos θ
On dividing by sin θ,
we have
1 + cot θ = √2 cot θ
1 = √2 cot θ - cot θ
(√2 - 1) cot θ = 1

32. If 7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°,  then find out the value of θ.
Solution:
As we have Given that ,
7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°
3 sin2 θ + 3 cos2 θ + 4 sin2 θ = 4
3 (sin2 θ + 3 cos2 θ) + 4 sin2 θ = 4
3 (1) + 4 sin2 θ = 4
4 sin2 θ = 4 – 3
sin2 θ = 1/4
Taking square-root on both sides,
Then we get
sin θ = 1/2
Thus, θ = 30o

33. If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.
Solution:
As we have  Given that ,
sec θ + tan θ = m
sec θ – tan θ = n
Now,
mn = (sec θ + tan θ) (sec θ – tan θ)
= sec2 θ – tan2 θ = 1
Thus, mn = 1

35. If x = h + a cos θ and y = k + a sin θ, prove that (x – h)2 + (y – k)2 = a2
Solution:
As we have  Given that ,
x = h + a cos θ
y = k + a sin θ
Now,
x – h = a cos θ
y – k = a sin θ
On squaring and adding we get
(x – h)2 + (y – k)2 = a2 cos2 θ + a2 sin2 θ
= a2 (sin2 θ + cos2 θ)
= a2 (1)                [Since, sin2 θ + cos2 θ = 1]
Hence proved

Chapter Test

1. (i) If θ is an acute angle and cosec θ = √5, find out the value of cot θ – cos θ.
(ii) If θ is an acute angle and tan θ = 8/15, find the value of sec θ + cosec θ.
Solution:
As we have Given that , θ is an acute angle and cosec θ = √5
So,
sin θ = 1/√5 And, cos θ = √(1 – sin2 θ)
cos θ = √(1 – (1/√5)2 )
= √(1 – (1/5))
= √(4/5)
cos θ = 2/√5
Now,
cot θ – cos θ = (cos θ/sin θ) – cos θ

In fig. we have
tan θ = BC/AB = 8/15
So, BC = 8 and AB = 15
Using By the Pythagoras theorem,
we have AC = √(AB2 + BC2 ) = √(52 + 82 ) = √(25 + 64) = √289
⇒ AC = 17
Now,
sec θ = AC/AB = 17/15
cosec θ = AC/BC = 17/8
So,
sec θ + cosec θ = 17/15 + 17/8
= (136 + 255)/ 120
= 391/120
=3 31/120

4. (i) cos A/ (1 – sin A) + cos A/ (1 + sin A) = 2 sec A
(ii) cos A/ (cosec A + 1) + cos A/ (cosec A - 1) = 2 tan A
Solution:

14. If tan A = n tan B and sin A = m sin B, prove that cos2 A = (m2 – 1)/ (n2 - 1)
Solution:
As we have Given, tan A = n tan B and sin A = m sin B
n = tan A/ tan B
m = sin A/ sin B

15. If sec A = x + 1/4x, then prove that sec A + tan A = 2x or 1/2x
Solution:
As we have Given that,
sec A = x + 1/4x
We know that,

16. When 0° < θ < 90°, solve the following equations:
(i) 2 cos2 θ + sin θ – 2 = 0
(ii) 3 cos θ = 2 sin2 θ
(iii) sec2 θ – 2 tan θ = 0
(iv) tan2 θ = 3 (sec θ – 1).
Solution:
As we have Given that , 0° < θ < 90°

(i) 2 cos2 θ + sin θ – 2 = 0
2 (1 - sin2 θ) + sin θ – 2 = 0
2 – 2 sin2 θ + sin θ – 2 = 0
– 2 sin2 θ + sin θ = 0
sin θ (1 – 2 sin θ) = 0
So, either sin θ = 0 or 1 – 2 sin θ = 0
If sin θ = 0
⇒ θ = 0o
And, if 1 – 2 sin θ = 0
sin θ = ½
⇒ θ = 30o
Thus, θ = 0o or 30o

(ii) 3 cos θ = 2 sin2 θ
3 cos θ = 2 (1 - cos2 θ)
3 cos θ = 2 – 2 cos2 θ
2 cos2 θ + 3 cos θ – 2 = 0
2 cos2 θ + 4 cos θ – cos θ – 2 = 0
2 cos θ (cos θ + 2) – 1(cos θ + 2)
(2 cos θ – 1) (cos θ + 2) = 0
So, either 2 cos θ – 1 = 0 or cos θ + 2 = 0
If 2 cos θ – 1 = 0
cos θ = ½
⇒ θ = 60o
And, for cos θ + 2 = 0
⇒ cos θ = -2 which is not possible being out of range.
Thus, θ = 60o

(iii) sec2 θ – 2 tan θ = 0
(1 + tan2 θ) – 2 tan θ = 0
tan2 θ – 2 tan θ + 1 = 0
(tan θ - 1)2 = 0
tan θ – 1 = 0
⇒ tan θ = 1
Thus, θ = 45o

(iv). tan2 θ = 3 (sec θ – 1)
(sec2 θ – 1) = 3 sec θ – 3
sec2 θ – 1 – 3 sec θ + 3 = 0
sec2 θ – 3 sec θ + 2 = 0
sec2 θ – 2 sec θ – sec θ + 2 = 0
sec θ (sec θ – 2) – 1 (sec θ = 2) = 0
(sec θ - 1) (sec θ - 2) = 0
So, either sec θ – 1 = 0 or sec θ – 2 = 0
If sec θ – 1 = 0
sec θ = 1
⇒ θ = 0o
And, if sec θ – 2 = 0
sec θ = 2
⇒ θ = 60
Thus, θ = 0o or 60o