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,
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities
AB = √(AC2 – BC2
= √(52 - 32 )
= √(25 - 9)
 = √16
Hence  AD  = 4
Then , 
 
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-
 
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, 
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-1
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, 
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-2
 
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,
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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,  
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-4
 
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,
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-5
 
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
 
 
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(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.
 
 ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-15
 
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 )
 
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ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-17
 
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
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-18
 
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)
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-19
 
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
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-20
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-21
 
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:

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(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)
 
 ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-23
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-24
 
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:

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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 
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-29
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-30
 
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.
 
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= tan2 θ+tan θ+1=tan2 θ+1+tanθ 
=sec2 θ+tan θ                {∴ sec2 θ=tan2 θ+1}
=R.H.S 
 
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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:

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=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.
 
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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)
 
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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:
 
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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 
 
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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 

ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-55

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
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-56
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 
 
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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:

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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

ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-88

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,
 
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities-89
 
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
ML Aggarwal Solutions Class 10 Maths Chapter 1 Goods and Service Tax (GST)
ML Aggarwal Solutions Class 10 Maths Chapter 2 Banking
ML Aggarwal Solutions Class 10 Maths Chapter 3 Shares and Dividends
ML Aggarwal Solutions Class 10 Maths Chapter 4 Linear Inequations
ML Aggarwal Solutions Class 10 Maths Chapter 5 Quadratic Equations in One Variable
ML Aggarwal Solutions Class 10 Maths Chapter 6 Factorization
ML Aggarwal Solutions Class 10 Maths Chapter 7 Ratio and Proportion
ML Aggarwal Solutions Class 10 Maths Chapter 7 Ratio and Proportion
ML Aggarwal Solutions Class 10 Maths Chapter 8 Matrices
ML Aggarwal Solutions Class 10 Maths Chapter 9 Arithmetic and Geometric Progression
ML Aggarwal Solutions Class 10 Maths Chapter 10 Reflection
ML Aggarwal Solutions Class 10 Maths Chapter 11 Section Formula
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line
ML Aggarwal Solutions Class 10 Maths Chapter 13 Similarity
ML Aggarwal Solutions Class 10 Maths Chapter 14 Locus
ML Aggarwal Solutions Class 10 Maths Chapter 15 Circles
ML Aggarwal Solutions Class 10 Maths Chapter 16 Constructions
ML Aggarwal Solutions Class 10 Maths Chapter 17 Mensuration
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities
ML Aggarwal Solutions Class 10 Maths Chapter 19 Trigonometric Tables
ML Aggarwal Solutions Class 10 Maths Chapter 20 Heights and Distances
ML Aggarwal Solutions Class 10 Maths Chapter 21 Measures Of Central Tendency
ML Aggarwal Solutions Class 10 Maths Chapter 22 Probability
NCERT Exemplar Solutions Class 10 Maths Areas related to Circles
NCERT Exemplar Solutions Class 10 Maths Arithmetic Progression
NCERT Exemplar Solutions Class 10 Maths Circles
NCERT Exemplar Solutions Class 10 Maths Construction
NCERT Exemplar Solutions Class 10 Maths Coordinate Geometry
NCERT Exemplar Solutions Class 10 Maths Linear Equations
NCERT Exemplar Solutions Class 10 Maths Polynomials
NCERT Exemplar Solutions Class 10 Maths Quadratic Equation
NCERT Exemplar Solutions Class 10 Maths Real Numbers
NCERT Exemplar Solutions Class 10 Maths Surface Area and Volume
NCERT Exemplar Solutions Class 10 Maths Triangles
NCERT Exemplar Solutions Class 10 Maths Trigonometry

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