Trigonometric Identities ML Aggarwal Solutions ICSE Class-10

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Trigonometric Identities ML Aggarwal Solutions ICSE Class-10 Chapter-18 . We Provide Step by Step Answer of Exercise-18 of Trigonometric Identities with  MCQs and Chapter-Test Questions  / Problems related  for ICSE Class-10 APC Understanding Mathematics  .  Visit official Website CISCE  for detail information about ICSE Board Class-10.

Trigonometric Identities ML Aggarwal Solutions ICSE Class-10 Chapter-18


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Before viewing Answer of Chapter-18 Trigonometric Identities of ML Aggarwal Solutions. Read the Chapter Carefully with formula and then solve all example of Exe-17 given in  your text book . For more practice on Trigonometric Identities  related problems /Questions / Exercise try to solve Trigonometric Identities  exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE)  / Concise Selina Publications ICSE  Mathematics. Get the Formula of Trigonometric Identities    for ICSE Class 10 Maths  to understand the topic more clearly in effective way.

 


 

 Chapter 18 Trigonometric Identities ML Aggarwal Solutions ICSE Class 10th

If A is an acute angle and sin A = \\ \frac { 3 }{ 5 }  find all other trigonometric ratios of angle A (using trigonometric identities).

Answer 1


sin A = \\ \frac { 3 }{ 5 }
In ∆ABC, ∠B = 90°
AC = 5 and BC = 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 1

Question 2


If A is an acute angle and sec A = \\ \frac { 17 }{ 8 } , find all other trigonometric ratios of angle A (using trigonometric identities).


Answer 2


sec A = \\ \frac { 17 }{ 8 }  (A is an acute angle)
In right ∆ABC
sec A = \\ \frac { AC }{ AB }  = \\ \frac { 17 }{ 8 }
AC = 17, AB = 8
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 2

Question 3


Express the ratios cos A, tan A and sec A in terms of sin A.


Answer 3


cos A = \sqrt { { 1-sin }^{ 2 }A }
tan A = \frac { SinA }{ CosA } =\frac { sinA }{ \sqrt { { 1-sin }^{ 2 }A } }
sec A = \frac { 1 }{ cosA } =\frac { 1 }{ \sqrt { { 1-sin }^{ 2 }A } }

Question 4


If tan A = \frac { 1 }{ \sqrt { 3 } } , find all other trigonometric ratios of angle A.


Answer 4


tan A = \frac { 1 }{ \sqrt { 3 } }
In right ∆ABC,
tan A = \\ \frac { BC }{ AB }  = \frac { 1 }{ \sqrt { 3 } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 4

Question 5


If 12 cosec θ = 13, find the value of \frac { 2sin\theta -3cos\theta }{ 4sin\theta -9cos\theta }


Answer 5


12 cosec θ = 13
⇒ cosec θ = \\ \frac { 13 }{ 12 }
In right ∆ABC,
∠A = θ
cosec θ = \\ \frac { AC }{ BC }  = \\ \frac { 13 }{ 12 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 5

Without using trigonometric tables, evaluate the following (6 to 10) :

Question 6


(i) cos² 26° + cos 64° sin 26° + \frac { tan{ 36 }^{ O } }{ { cot54 }^{ O } }
(ii) \frac { sec{ 17 }^{ O } }{ { cosec73 }^{ O } } +\frac { tan68^{ O } }{ cot22^{ O } }  + cos² 44° + cos² 46°


Answer 6


Given that
(i) cos² 26° + cos 64° sin 26° + \frac { tan{ 36 }^{ O } }{ { cot54 }^{ O } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 6

Question 7


(i) \frac { sin65^{ O } }{ { cos25 }^{ O } } +\frac { cos32^{ O } }{ sin58^{ O } }  – sin 28° sec 62° + cosec² 30° (2015)
(ii) \frac { sin29^{ O } }{ { cosec61 }^{ O } }  + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).


Answer 7


given that
(i) \frac { sin65^{ O } }{ { cos25 }^{ O } } +\frac { cos32^{ O } }{ sin58^{ O } }  – sin 28° sec 62° + cosec² 30°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 7.1

Question 8


(i) \frac { { sin }35^{ O }{ cos55 }^{ O }+{ cos35 }^{ O }{ sin }55^{ O } }{ { cosec }^{ 2 }{ 10 }^{ O }-{ tan }^{ 2 }{ 80 }^{ O } }
(ii) { sin }^{ 2 }{ 34 }^{ O }+{ sin }^{ 2 }{ 56 }^{ O }+2tan{ 18 }^{ O }{ tan72 }^{ O }-{ cot }^{ 2 }{ 30 }^{ O }


Answer 8


Given that
(i) \frac { { sin }35^{ O }{ cos55 }^{ O }+{ cos35 }^{ O }{ sin }55^{ O } }{ { cosec }^{ 2 }{ 10 }^{ O }-{ tan }^{ 2 }{ 80 }^{ O } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 8

Question 9


(i) { \left( \frac { { tan25 }^{ O } }{ { cosec }65^{ O } } \right) }^{ 2 }+{ \left( \frac { { cot25 }^{ O } }{ { sec65 }^{ O } } \right) }^{ 2 }+{ 2tan18 }^{ O }{ tan }45^{ O }{ tan72 }^{ O }
(ii) \left( { cos }^{ 2 }25+{ cos }^{ 2 }65 \right) +cosec\theta sec\left( { 90 }^{ O }-\theta \right) -cot\theta tan\left( { 90 }^{ O }-\theta \right)


Answer 9


(i) { \left( \frac { { tan25 }^{ O } }{ { cosec }65^{ O } } \right) }^{ 2 }+{ \left( \frac { { cot25 }^{ O } }{ { sec65 }^{ O } } \right) }^{ 2 }+{ 2tan18 }^{ O }{ tan }45^{ O }{ tan72 }^{ O }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 9
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 9.1

Question 10


(i) 2(sec² 35° – cot² 55°) – \frac { { cos28 }^{ O }cosec{ 62 }^{ O } }{ { tan18 }^{ O }tan{ 36 }^{ O }{ tan30 }^{ O }{ tan54 }^{ O }{ tan72 }^{ O } }
(ii) \frac { { cosec }^{ 2 }(90-\theta )-{ tan }^{ 2 }\theta }{ 2({ cos }^{ 2 }{ 48 }^{ O }+{ cos }^{ 2 }{ 42 }^{ O }) } -\frac { { 2tan }^{ 2 }{ 30 }^{ O }{ sec }^{ 2 }{ 52 }^{ O }{ sin }^{ 2 }{ 38 }^{ O } }{ { cosec }^{ 2 }{ 70 }^{ O }-{ tan }^{ 2 }{ 20 }^{ O } }


Answer 10


(i) 2(sec² 35° – cot² 55°) – \frac { { cos28 }^{ O }cosec{ 62 }^{ O } }{ { tan18 }^{ O }tan{ 36 }^{ O }{ tan30 }^{ O }{ tan54 }^{ O }{ tan72 }^{ O } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 10
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 10.1

Question 11


Prove that following:
(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
(ii) \frac { tan\theta }{ tan({ 90 }^{ O }-\theta ) } +\frac { sin({ 90 }^{ O }-\theta ) }{ cos\theta } ={ sec }^{ 2 }\theta
(iii) \frac { cos({ 90 }^{ O }-\theta )cos\theta }{ tan\theta } +{ cos }^{ 2 }({ 90 }^{ O }-\theta )=1
(iv) sin (90° – θ) cos (90° – θ) = \frac { tan\theta }{ { 1+tan }^{ 2 }\theta }


Answer 11


(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
L.H.S. = cos θ sin (90° – θ) + sin θ cos (90° – θ)
= cos θ . cos θ + sin θ . sin θ
= cos2 θ + sin2 θ = 1 = R.H.S.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 11
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 11.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 11.2

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

Question 12


(i) (sec A + tan A) (1 – sin A) = cos A
(ii) (1 + tan2 A) (1 – sin A) (1 + sin A) = 1.


Answer 12


(i) (sec A + tan A) (1 – sin A) = cos A
L.H.S. = (sec A + tan A) (1 – sin A)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 12

Question 13


(i) tan A + cot A = sec A cosec A
(ii) (1 – cos A)(1 + sec A) = tan A sin A.


Answer 13


(i) tan A + cot A = sec A cosec A
L.H.S. = tan A + cot A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 13
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 13.1

Question 14


(i) \frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA } =2{ cosec }^{ 2 }A
(ii) \frac { 1 }{ secA+tanA } +\frac { 1 }{ secA-tanA } =2{ sec }A


Answer 14


(i) \frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA } =2{ cosec }^{ 2 }A
L.H.S = \frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 14

Question 15


(i) \frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA }
(ii) \frac { 1-{ tan }^{ 2 }A }{ { cot }^{ 2 }A-1 } ={ tan }^{ 2 }A
(iii) \frac { sinA }{ 1+cosA } =cosecA-cotA


Answer 15


(i) \frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA }
L.H.S = \frac { sinA }{ 1+cosA }
(multiplying and dividing by (1 – cosA))
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 15
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 15.1

Question 16


(i) \frac { secA-1 }{ secA+1 } =\frac { 1-cosA }{ 1+cosA }
(ii) \frac { { tan }^{ 2 }\theta }{ { (sec\theta -1) }^{ 2 } } =\frac { 1+cos\theta }{ 1-cos\theta }
(iii) { (1+tanA) }^{ 2 }+{ (1-tanA) }^{ 2 }=2{ sec }^{ 2 }A
(iv) { sec }^{ 2 }A+{ cosec }^{ 2 }A={ sec }^{ 2 }A{ .cosec }^{ 2 }A


Answer 16


(i) \frac { secA-1 }{ secA+1 } =\frac { 1-cosA }{ 1+cosA }
L.H.S = \frac { secA-1 }{ secA+1 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 16
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 16.1

Question 17


(i) \frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } =2secA
(ii) \frac { tanA }{ secA-1 } +\frac { tanA }{ secA+1 } =2cosecA


Answer 17


(i) \frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } =2secA
L.H.S = \frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 17

Question 18


(i) \frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } =2{ sec }^{ 2 }A
(ii) cotA-tanA=\frac { { 2cos }^{ 2 }A-1 }{ sinA-cosA }
(iii) \frac { cotA-1 }{ 2-{ sec }^{ 2 }A } =\frac { cotA }{ 1+tanA }


Answer 18


(i) \frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } =2{ sec }^{ 2 }A
L.H.S = \frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 18
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 18.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 18.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 18.3

Question 19


(i) { tan }^{ 2 }\theta -{ sin }^{ 2 }\theta ={ tan }^{ 2 }\theta { sin }^{ 2 }\theta
(ii) \frac { cos\theta }{ 1-tan\theta } -\frac { { sin }^{ 2 }\theta }{ cos\theta -sin\theta } =cos\theta +sin\theta


Answer 19


(i) { tan }^{ 2 }\theta -{ sin }^{ 2 }\theta ={ tan }^{ 2 }\theta { sin }^{ 2 }\theta
L.H.S = { tan }^{ 2 }\theta -{ sin }^{ 2 }\theta
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 19
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 19.1

Question 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.


Answer 20


(i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ
L.H.S = cosec4 θ – cosec2 θ
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 20
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 20.1

Question 21


(i) \frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) } =cot\theta
(ii) \frac { { tan }^{ 3 }\theta -1 }{ tan\theta -1 } ={ sec }^{ 2 }\theta +tan\theta


Answer 21


(i) \frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) } =cot\theta
L.H.S = \frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 21

Question 22


(i) \frac { 1+cosecA }{ cosecA } =\frac { { cos }^{ 2 }A }{ 1-sinA }
(ii) \sqrt { \frac { 1-cosA }{ 1+cosA } } =\frac { sinA }{ 1+cosA }


Answer 22


(i) \frac { 1+cosecA }{ cosecA } =\frac { { cos }^{ 2 }A }{ 1-sinA }
L.H.S = \frac { 1+cosecA }{ cosecA }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 22
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 22.1

Question 23


(i) \sqrt { \frac { 1+sinA }{ 1-sinA } } =tanA+secA
(ii) \sqrt { \frac { 1-cosA }{ 1+cosA } } =cosecA-cotA


Answer 23


(i) \sqrt { \frac { 1+sinA }{ 1-sinA } } =tanA+secA
L.H.S = \sqrt { \frac { 1+sinA }{ 1-sinA } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 23
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 23.1

Question 24


(i) \sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA
(ii) \frac { cotAcotA }{ 1-sinA } =1+cosecA


Answer 24


(i) \sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA
L.H.S = \sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 24
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 24.1

Question 25


(i) \frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } =2(secA+cosecA)
(ii) { sec }^{ 4 }A-{ tan }^{ 4 }A=1+2{ tan }^{ 2 }A


Answer 25


(i) \frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } =2(secA+cosecA)
L.H.S = \frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 25
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 25.1

Question 26


(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1
(ii) sec6 A – tan6 A = 1 + 3 tan2 A + 3 tan4 A


Answer 26


(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1
L.H.S = cosec6 A – cot6 A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 26

Question 27


(i) \frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 } =\frac { 1+cos\theta }{ sin\theta }
(ii) \frac { sin\theta }{ cot\theta +cosec\theta } =2+\frac { sin\theta }{ cot\theta -cosec\theta }


Answer 27


(i) \frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 } =\frac { 1+cos\theta }{ sin\theta }
L.H.S = \frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 27
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 27.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 27.2

Question 28


(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
(ii) (cosecA – sinA)(secA – cosA) sec2A = tanA


Answer 28


(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
L.H.S = (sinθ + cosθ)(secθ + cosecθ)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 28

Question 29

 

(i) \frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } =2
(ii) \frac { { tan }^{ 2 }A }{ { 1+tan }^{ 2 }A } +\frac { cot^{ 2 }A }{ 1+{ cot }^{ 2 }A } =1


Answer 29:


(i) \frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } =2
L.H.S = \frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 29
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 29.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 29.2

Question 30


(i) \frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } =\frac { 1 }{ cosA } -\frac { 1 }{ secA-tanA }
(ii) { (sinA+secA) }^{ 2 }+{ (cosA+cosecA) }^{ 2 }={ (1+secA\quad cosecA) }^{ 2 }
(iii) \frac { tanA+sinA }{ tanA-sinA } =\frac { secA+1 }{ secA-1 }


Answer 30


(i) \frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } =\frac { 1 }{ cosA } -\frac { 1 }{ secA-tanA }
L.H.S = \frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 30
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 30.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 30.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 30.3

Question 31


If sin θ + cos θ = √2 sin (90° – θ), show that cot θ = √2 + 1


Answer 31


sin θ + cos θ = √2 sin (90° – θ)
sin θ + cos θ = √2 cos θ
dividing by sin θ
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 31

Question 32


If 7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°, then find the value of θ.


Answer 32


7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°
3 sin2 θ + 3 cos2 θ + 4 sin2 θ = 4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 32

Question 33


If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.


Answer 33


sec θ + tan θ = m and sec θ – tan θ = n
mn = (sec θ + tan θ) (sec θ – tan θ) = sec2 θ – tan2 θ = 1
(∴ sec2 θ – tan2 θ = 1)
Hence proved.

Question 34


If x – a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 – y2 = a2 – b2.


Answer 34


x – a sec θ + b tan θ and y = a tan θ + b sec θ
To prove that x2 – y2 = a2 – b2.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 34

Question 35


If x = h + a cos θ and y = k + a sin θ, prove that (x – h)2 + (y – k)2 = a2.


Answer 35


x = h + a cos θ and y = k + a sin θ
To prove that (x – h)2 + (y – k)2 = a2.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 35


MCQ Solutions of ML Aggarwal Trigonometric Identities Chapter 18 

Choose the correct answer from the given four options (1 to 12) :

Question 1


{ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta }  is equal to
(a) 1
(b) -1
(c) sin2 θ
(d) sec2 θ
Answer 1
{ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta }
\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta } -\frac { 1 }{ { sin }^{ 2 }\theta }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 1

Question 2


(sec2 θ – 1) (1 – cosec2 θ) is equal to
(a) – 1
(b) 1
(c) 0
(d) 2


Answer 2


(sec2 θ – 1) (1 – cosec2 θ)
\left( \frac { 1 }{ { cos }^{ 2 }\theta } -1 \right) \left( 1-\frac { 1 }{ { sin }^{ 2 }\theta } \right)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 2

Question 3


\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta }  is equal to
(a) 2 sin2 θ
(b) 2 cos2 θ
(c) sin2 θ
(d) cos2 θ
Answer 3
\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 3
=\frac { { sin }^{ 2 }\theta }{ 1 } ={ sin }^{ 2 }\theta  (c)

Question 4


(cos θ + sin θ)2 + (cos θ – sin θ)2 is equal to
(a) – 2
(b) 0
(c) 1
(d) 2


Answer 4


(cos θ + sin θ)2 + (cos θ – sin θ)2
= cos2 θ + sin2 θ + 2 sin θ cos θ + cos2 θ + sin2 θ – 2 sin θ cos θ
so = 2(sin2 θ + cos2 θ)
Hence = 2 × 1 = 2 (d)
(∵ sin2 θ + cos2 θ = 1)

 

Question 5


(sec A + tan A) (1 – sin A) is equal to
(a) sec A
(b) sin A
(c) cosec A
(d) cos A


Answer 5


(sec A + tan A) (1 – sin A)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 5

Question 6


\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A }  is equal to
(a) sec2 A
(b) – 1
(c) cot2 A
(d) tan2 A


Answer 6


\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 6
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 6.1

Question 7


If sec θ – tan θ = k, then the value of sec θ + tan θ is
(a) 1-\frac { 1 }{ k }
(b) 1 – k
(c) 1 + k
(d) \\ \frac { 1 }{ k }


Answer 7


sec θ – tan θ = k
\frac { 1 }{ cos\theta } -\frac { sin\theta }{ cos\theta } =k
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 7.1

Question 8


Which of the following is true for all values of θ (0° < θ < 90°):
(a) cos2 θ – sin2 θ = 1
(b) cosec2 θ – sec2 θ = 1
(c) sec2 θ – tan2 θ = 1
(d) cot2 θ – tan2 θ = 1.


Answer 8


∴ sec2 θ – tan2 θ = 1 is true for all values of θ as it is an identity.
(0° < θ < 90°) (c)

Question 9


If θ is an acute angle of a right triangle, then the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is
(a) 0
(b) 2 sin θ cos θ
(c) 1
(d) 2 sin2 θ


Answer 9


sin θ cos (90° – θ) + cos θ sin (90° – θ)
= sin θ sin θ + cos θ cos θ
{ ∵ sin(90° – θ) = cosθ, cos (90° – θ) = sin θ }
= sin2 θ + cos2 θ = 1 (c)

Question 10


The value of cos 65° sin 25° + sin 65° cos 25° is
(a) 0
(b) 1
(b) 2
(d) 4


Answer 10


cos 65° sin 25° + sin 65° cos 25°
so = cos (90° – 25°) sin 25° + sin (90° – 25°) cos 25°
and = sin 25° . sin 25° + cos 25° . cos 25°
therefore = sin2 25° + cos2 25°
( ∵ sin2 θ + cos2 θ = 1)
hence = 1 (b)

Question 11


The value of 3 tan2 26° – 3 cosec2 64° is
(a) 0
(b) 3
(c) – 3
(d) – 1


Answer 11


3 tan2 26° – 3 cosec2 64°
= 3 tan2 26° – 3 cosec (90° – 26°)
= 3 tan2 26° – 3 sec2 26°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 11

Question 12


The value of \frac { sin({ 90 }^{ O }-\theta )sin\theta }{ tan\theta } -1  is
(a) – cot θ
(b) – sin2 θ
(c) – cos2 θ
(d) – cosec2 θ

Answer 12

\frac { sin({ 90 }^{ O }-\theta )sin\theta }{ tan\theta } -1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS 12


Chapter- Test ML Aggarwal Solutions Trigonometric Identities

Question 1


(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.
(ii) If θ is an acute angle and tan θ = \\ \frac { 8 }{ 15 } , find the value of sec θ + cosec θ.

 

Answer 1


(i) θ is an acute angle.
cosec θ = √5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 1.2

Question 2


Evaluate the following:
(i) 2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
(ii) \frac { { sin }^{ 2 }{ 22 }^{ O }+{ sin }^{ 2 }{ 68 }^{ O } }{ { cos }^{ 2 }{ 22 }^{ O }+{ cos }^{ 2 }{ 68 }^{ O } }  + sin2 63° + cos 63° sin 27°


Answer 2


(i) 2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right)  – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

(ii) \frac { { sin }^{ 2 }{ 22 }^{ O }+{ sin }^{ 2 }{ 68 }^{ O } }{ { cos }^{ 2 }{ 22 }^{ O }+{ cos }^{ 2 }{ 68 }^{ O } }  + sin2 63° + cos 63° sin 27°

Question 3


If \\ \frac { 4 }{ 3 }  (sec2 59° – cot2 31°) – \\ \frac { 2 }{ 2 }  sin 90° + 3tan2 56° tan2 34° = \\ \frac { x }{ 2 } , then find the value of x.


Answer 3


Given
\\ \frac { 4 }{ 3 }  (sec2 59° – cot2 31°) – \\ \frac { 2 }{ 2 }  sin 90° + 3tan2 56° tan2 34° = \\ \frac { x }{ 2 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 3.1

Prove the following (4 to 11) identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

Question 4


(i) \frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA
(ii) \frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA


Answer 4


(i) \frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA
L.H.S = \frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 4.1

Question 5


(i) \frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta
(ii) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.


Answer 5


(i) \frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta
L.H.S = \frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 5.1

Question 6


(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ
(ii) \frac { cot\theta }{ cosec\theta +1 } +\frac { cosec\theta +1 }{ cot\theta } =2sec\theta


Answer 6


L.H.S. = sin2 θ + cos4 θ
so = (1 – cos2 θ + cos4 θ
threfore = 1 – cos2 θ + cos4 θ
hence = 1 – cos2 θ (1 – cos2 θ)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 6
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 6.1

Question 7


(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
(ii) \frac { 1 }{ sinA+cosA+1 } +\frac { 1 }{ sinA+cosA-1 } =secA+cosecA


Answer 7


(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
L.H.S = sec4 A (1 – sin4 A) – 2 tan2 A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 7.1

Question 8


(i) \frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1
(ii) (sec A – tan A)2 (1 + sin A) = 1 – sin A.


Answer 8


(i) \frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1
L.H.S = \frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 8
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 8.1

Question 9


(i) \frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA
(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
(iii) \frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }


Answer 9


(i) \frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA
L.H.S = \frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 9
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 9.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 9.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 9.3

Question 10


\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }
Answer 10
\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }
L.H.S = \frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 10
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 10.1

Question 11


2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ


Answer 11


2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
L.H.S = 2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 11
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 11.1

Question 12


If cot θ + cos θ = m, cot θ – cos θ = n, then prove that (m2 – n2)2 = 16 run.


Answer 12


cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii)
Adding (i)&(ii) we get
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 12

Question 13


If sec θ + tan θ = p, prove that sin θ = \frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }


Answer 13


sec θ + tan θ = p,
prove that sin θ = \frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }
\frac { 1 }{ cos\theta } +\frac { sin\theta }{ cos\theta } =p
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 13

Question 14


If tan A = n tan B and sin A = m sin B, prove that cos2 A = \frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 }

 

Answer 14


m = \\ \frac { sinA }{ sinB }
n = \\ \frac { tanA }{ tanB }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 14
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 14.1

Question 15


If sec A = x+ \frac { 1 }{ 4x } , then prove that sec A + tan A = 2x or \\ \frac { 1 }{ 2x }


Answer 15


sec A = x+ \frac { 1 }{ 4x }
To prove that sec A + tan A = 2x or \\ \frac { 1 }{ 2x }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 15
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 15.1

Question 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).


Answer 16


0° < θ < 90°
(i) 2 cos2 θ + sin θ – 2 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 16
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test 16.1

–: End of Trigonometric Identities ML Aggarwal Solutions  :–


Return to  ML Aggarwal Solutions for ICSE Class-10

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