# Trigonometric Identities ML Aggarwal Solutions ICSE Class-10

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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### How to Solve Trigonometric Identities Problems/Questions / Exercise of ICSE Class-10 Mathematics

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).

sin A = $\\ \frac { 3 }{ 5 }$
In ∆ABC, ∠B = 90°
AC = 5 and BC = 3

#### 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).

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

#### Question 3

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

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.

tan A = $\frac { 1 }{ \sqrt { 3 } }$
In right ∆ABC,
tan A = $\\ \frac { BC }{ AB }$ = $\frac { 1 }{ \sqrt { 3 } }$

#### Question 5

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

12 cosec θ = 13
⇒ cosec θ = $\\ \frac { 13 }{ 12 }$
In right ∆ABC,
∠A = θ
cosec θ = $\\ \frac { AC }{ BC }$ = $\\ \frac { 13 }{ 12 }$

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°

Given that
(i) cos² 26° + cos 64° sin 26° + $\frac { tan{ 36 }^{ O } }{ { cot54 }^{ O } }$

#### 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°).

given that
(i) $\frac { sin65^{ O } }{ { cos25 }^{ O } } +\frac { cos32^{ O } }{ sin58^{ O } }$ – sin 28° sec 62° + cosec² 30°

#### Question 8

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

Given that
(i) $\frac { { sin }35^{ O }{ cos55 }^{ O }+{ cos35 }^{ O }{ sin }55^{ O } }{ { cosec }^{ 2 }{ 10 }^{ O }-{ tan }^{ 2 }{ 80 }^{ O } }$

#### 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) $\left( { cos }^{ 2 }25+{ cos }^{ 2 }65 \right) +cosec\theta sec\left( { 90 }^{ O }-\theta \right) -cot\theta tan\left( { 90 }^{ O }-\theta \right)$

(i) ${ \left( \frac { { tan25 }^{ O } }{ { cosec }65^{ O } } \right) }^{ 2 }+{ \left( \frac { { cot25 }^{ O } }{ { sec65 }^{ O } } \right) }^{ 2 }+{ 2tan18 }^{ O }{ tan }45^{ O }{ tan72 }^{ O }$

#### 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) $\frac { { cosec }^{ 2 }(90-\theta )-{ tan }^{ 2 }\theta }{ 2({ cos }^{ 2 }{ 48 }^{ O }+{ cos }^{ 2 }{ 42 }^{ O }) } -\frac { { 2tan }^{ 2 }{ 30 }^{ O }{ sec }^{ 2 }{ 52 }^{ O }{ sin }^{ 2 }{ 38 }^{ O } }{ { cosec }^{ 2 }{ 70 }^{ O }-{ tan }^{ 2 }{ 20 }^{ O } }$

(i) 2(sec² 35° – cot² 55°) – $\frac { { cos28 }^{ O }cosec{ 62 }^{ O } }{ { tan18 }^{ O }tan{ 36 }^{ O }{ tan30 }^{ O }{ tan54 }^{ O }{ tan72 }^{ O } }$

#### 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) = $\frac { tan\theta }{ { 1+tan }^{ 2 }\theta }$

(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.

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.

(i) (sec A + tan A) (1 – sin A) = cos A
L.H.S. = (sec A + tan A) (1 – sin A)

#### Question 13

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

(i) tan A + cot A = sec A cosec A
L.H.S. = tan A + cot A

#### Question 14

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

(i) $\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA } =2{ cosec }^{ 2 }A$
L.H.S = $\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA }$

#### 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) $\frac { sinA }{ 1+cosA } =cosecA-cotA$

(i) $\frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA }$
L.H.S = $\frac { sinA }{ 1+cosA }$
(multiplying and dividing by (1 – cosA))

#### 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) ${ sec }^{ 2 }A+{ cosec }^{ 2 }A={ sec }^{ 2 }A{ .cosec }^{ 2 }A$

(i) $\frac { secA-1 }{ secA+1 } =\frac { 1-cosA }{ 1+cosA }$
L.H.S = $\frac { secA-1 }{ secA+1 }$

#### Question 17

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

(i) $\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } =2secA$
L.H.S = $\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA }$

#### 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) $\frac { cotA-1 }{ 2-{ sec }^{ 2 }A } =\frac { cotA }{ 1+tanA }$

(i) $\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } =2{ sec }^{ 2 }A$
L.H.S = $\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 }$

#### Question 19

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

(i) ${ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta ={ tan }^{ 2 }\theta { sin }^{ 2 }\theta$
L.H.S = ${ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta$

#### 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.

(i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ
L.H.S = cosec4 θ – cosec2 θ

#### Question 21

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

(i) $\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) } =cot\theta$
L.H.S = $\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) }$

#### Question 22

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

(i) $\frac { 1+cosecA }{ cosecA } =\frac { { cos }^{ 2 }A }{ 1-sinA }$
L.H.S = $\frac { 1+cosecA }{ cosecA }$

#### Question 23

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

(i) $\sqrt { \frac { 1+sinA }{ 1-sinA } } =tanA+secA$
L.H.S = $\sqrt { \frac { 1+sinA }{ 1-sinA } }$

#### Question 24

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

(i) $\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA$
L.H.S = $\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } }$

#### Question 25

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

(i) $\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } =2(secA+cosecA)$
L.H.S = $\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA }$

#### 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

(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1
L.H.S = cosec6 A – cot6 A

#### Question 27

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

(i) $\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 } =\frac { 1+cos\theta }{ sin\theta }$
L.H.S = $\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 }$

#### Question 28

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

(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
L.H.S = (sinθ + cosθ)(secθ + cosecθ)

#### Question 29

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

(i) $\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } =2$
L.H.S = $\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA }$

#### 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) $\frac { tanA+sinA }{ tanA-sinA } =\frac { secA+1 }{ secA-1 }$

(i) $\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } =\frac { 1 }{ cosA } -\frac { 1 }{ secA-tanA }$
L.H.S = $\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA }$

#### Question 31

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

sin θ + cos θ = √2 sin (90° – θ)
sin θ + cos θ = √2 cos θ
dividing by sin θ

#### Question 32

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

7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°
3 sin2 θ + 3 cos2 θ + 4 sin2 θ = 4

#### Question 33

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

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.

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

#### Question 35

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

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

### 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
${ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta }$
$\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta } -\frac { 1 }{ { sin }^{ 2 }\theta }$

#### Question 2

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

(sec2 θ – 1) (1 – cosec2 θ)
$\left( \frac { 1 }{ { cos }^{ 2 }\theta } -1 \right) \left( 1-\frac { 1 }{ { sin }^{ 2 }\theta } \right)$

#### 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
$\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta }$

$=\frac { { sin }^{ 2 }\theta }{ 1 } ={ sin }^{ 2 }\theta$ (c)

#### Question 4

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

(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

(sec A + tan A) (1 – sin A)

#### 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

$\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A }$

#### 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) $\\ \frac { 1 }{ k }$

sec θ – tan θ = k
$\frac { 1 }{ cos\theta } -\frac { sin\theta }{ cos\theta } =k$

#### 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.

∴ 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

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

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

3 tan2 26° – 3 cosec2 64°
= 3 tan2 26° – 3 cosec (90° – 26°)
= 3 tan2 26° – 3 sec2 26°

#### 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

### 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.

(i) θ is an acute angle.
cosec θ = √5

#### 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°

(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.

Given
$\\ \frac { 4 }{ 3 }$ (sec2 59° – cot2 31°) – $\\ \frac { 2 }{ 2 }$ sin 90° + 3tan2 56° tan2 34° = $\\ \frac { x }{ 2 }$

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) $\frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA$

(i) $\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA$
L.H.S = $\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA }$

#### 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.

(i) $\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta$
L.H.S = $\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 }$

#### Question 6

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

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)

#### Question 7

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

(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
L.H.S = sec4 A (1 – sin4 A) – 2 tan2 A

#### 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.

(i) $\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1$
L.H.S = $\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta$

#### 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) $\frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }$

(i) $\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA$
L.H.S = $\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA }$

#### Question 10

$\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }$
$\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }$
L.H.S = $\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA }$

#### Question 11

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

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

#### Question 12

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

cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii)

#### Question 13

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

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

#### Question 14

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

m = $\\ \frac { sinA }{ sinB }$
n = $\\ \frac { tanA }{ tanB }$

#### Question 15

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

sec A = $x+ \frac { 1 }{ 4x }$
To prove that sec A + tan A = 2x or $\\ \frac { 1 }{ 2x }$

#### 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).

0° < θ < 90°
(i) 2 cos2 θ + sin θ – 2 = 0

–: End of Trigonometric Identities ML Aggarwal Solutions  :–

Thanks