CBSE NCERT Solution for Class 10 - Maths - Introduction to Trigonometry

Class 10 - Maths - Introduction to Trigonometry

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Exercise 8.1

Question 1:

In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A (ii) sin C, cos C

Answer:

Class_10_Introduction_To_Trigonometry_RightAngledTriangleABC6

In Δ ABC, by Pythagoras theorem,

AC² = AB² + BC²

=> AC² = 24² + 7²

=> AC² = 576 + 49

=> AC² = 625

=> AC = √625

=> AC = 25

(i) sin A = BC/AC = 7/25, cos A = AB/AC = 24/25

(ii) sin C = AB/AC = 24/25, cos C = BC/AC = 7/25

Question 2:

In Fig. 8.13, find tan P – cot R.

Answer:

Class_10_Introduction_To_Trigonometry_RightAngledTrianglePQR1

In Δ PQR, by Pythagoras theorem,

QR² = PR² - PQ²

=> QR² = 13² - 12²

=> QR² = 169 - 144

=> QR² = 25

=> QR = √25

=> QR = 5

Hence, tan P – cot R = QR/PQ – QR/PQ

= 5/12 – 5/12

= 0

Question 3:

If sin A = 3/4, calculate cos A and tan A.

Answer:

Class_10_Introduction_To_Trigonometry_RightAngledTriangleABC5

Given that sin A = 3/4

In Δ ABC, by Pythagoras theorem,

AB² = AC² - BC²

=> AB² = 4² - 3²

=> AB² = 16 - 9

=> AB² = 7

=> AB = √7

Hence, cos A = AB/AC = √7/4, tan A = BC/AB = 3/√7

Question 4:

Given 15 cot A = 8, find sin A and sec A.

Answer:

Class_10_Introduction_To_Trigonometry_RightAngledTriangleABC4

Given, 15 cot A = 8

=> cot A = 8/15

In Δ ABC, by Pythagoras theorem,

AC² = AB² + BC²

=> AC² = 8² + 15²

=> AC² = 64 + 225

=> AC² = 289

=> AC = √289

=> AC = 17

Hence, sin A = BC/AC = 15/17, sec A = AC/AB = 17/8

Question 5:

Given sec θ = 13/12, calculate all other trigonometric ratios.

Answer:

Class_10_Introduction_To_Trigonometry_RightAngledTriangleABC3

Given, sec θ = 13/12

In Δ ABC, by Pythagoras theorem,

BC² = AC² - AB²

=> BC² = 13² - 12²

=> BC² = 169 - 144

=> BC² = 25

=> BC = √25

=> BC = 5

Hence, sin θ = BC/AC = 5/13, cos θ = AB/AC = 12/13, tan θ = BC/AB = 5/12

cosec θ = AC/BC = 13/5, cot θ = AB/BC = 12/5

Question 6:

If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B

Answer:

Class_10_Introduction_To_Trigonometry_RightAngledTriangles

Given, cos A = cos B

=> AP/AQ = BC/BD

=> AP/BC = AQ/BD

Let AP/BC = AQ/BD = k ………1

So, AP = k * BC and AQ = k * BD

Now in ΔAPQ and ΔBCD,

PQ/CD = √(AQ² – AP²)/ √(BD² – BC²)

= √{(K * BD)² – (k * BC)²}/ √(BD² – BC²)

= K√(BD)² – BC)²/ √(BD² – BC²)

= K

=> PQ/CD = K ………….2

From equation 1 and 2, we get

AP/BC = AQ/BD = PQ/CD

So, ΔAPQ ~ ΔBCD [SSS similarity criteria]

Hence, ∠ A = ∠ B

Question 7:

If cot θ = 7/8, evaluate:

(i) {(1 + sin θ)(1 - sin θ)}/ {(1 + cos θ)(1 - cos θ)} (ii) cot² θ

Answer:

Class_10_Introduction_To_Trigonometry_RightAngledTriangleABC2

Given, cot θ = 7/8

In Δ ABC, by Pythagoras theorem,

AC² = AB² + BC²

=> AC² = 8² + 7²

=> AC² = 64 + 49

=> AC² = 113

=> AC = √113

(i) {(1 + sin θ)(1 - sin θ)}/ {(1 + cos θ)(1 - cos θ)}

= (1 – sin² θ)/(1 – cos² θ) [a² - b² = (a + b)(a - b)]

= {1 – (7/√113)²}/{1 – (8/√113)²}

= (1 – 49/113)/(1 – 64/113)

= {(113 - 49)/113}/{(113 - 64)/113}

= 64/49

(ii) cot² θ = (cot θ)²

= (7/8)²

= 49/64

Question 8:

If 3 cot A = 4, check whether (1 - tan² A)/(1 + tan² A) = cos² A – sin² A or not.

Answer:

Class_10_Introduction_To_Trigonometry_RightAngledTriangleABC1

Given, 3 cot A = 4

=> cot A = 4/3

In Δ ABC, by Pythagoras theorem,

AC² = AB² + BC²

=> AC² = 3² + 4²

=> AC² = 9 + 16

=> AC² = 25

=> AC = √25

=> AC = 5

Now, (1 - tan² A)/(1 + tan² A) = {1 – (3/4)²}/{1 + (3/4)²}

= {1 – 9/16}/{1 + 9/16}

= {(16 – 9)/16}/{(16 + 9)/16}

= 7/25

And cos² A – sin² A = (4/5)² – (3/5)²

= 16/25 – 9/15

= 7/25

Hence, (1 - tan² A)/(1 + tan² A) = cos² A – sin² A

Question 9:

In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of:

(i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C

Answer:

Class_10_Introduction_To_Trigonometry_RightAngledTriangleABC

Given, tan A = 1/√3

In Δ ABC, by Pythagoras theorem,

AC² = AB² + BC²

=> AC² = (√3)² + 1²

=> AC² = 3 + 1

=> AC² = 4

=> AC = √4

=> AC = 2

(i) sin A cos C + cos A sin C = (1/2) * (1/2) + (√3/2) * (√3/2)

= 1/4 + 3/4

= 4/4

= 1

(ii) cos A cos C – sin A sin C = (√3/2) * (1/2) – (1/2) * (√3/2)

= (√3/4) - (√3/4)

= 0

Question 10:

In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer:

Given, in Δ PQR, angle Q is right-angled.

Let QR = x, therefore, PR = 25 – x

In Δ PQR, by Pythagoras theorem,

PR² = PQ² + OQ²

=> (25 - x)² = 5² + x²

=> 625 + x² – 50x = 25 + x²

=> 625 – 50x = 25

=> 50x = 625 - 25

=> 50x = 600

=> x = 600/50

=> x = 12

Therefore, PR = 25 – 12 = 13

Now, sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13 and tan P = QR/PQ = 12/5

Class_10_Introduction_To_Trigonometry_RightAngledTrianglePQR

Question 11:

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 4/3 for some angle θ.

Answer:

(i) False.

Because, tan A = Opposite side of angle A/adjacent side of angle A

If opposite side > adjacent side, then the value of tan A is greater than 1.

(ii) True.

Because, tan A = hypotenuse/adjacent side of angle A

And we know that hypotenuse is always greater than adjacent side.

(iii) False.

Because, cos A is used for cosine of angle A.

(iv) False.

Because, cos A is used for cotangent of angle A.

(v) False.

Because, sin θ = opposite side of angle θ/hypotenuse

And we know that hypotenuse is always greater than opposite side.

Exercise 8.2

Question 1:

Evaluate the following :

(i) sin 60⁰ cos 30⁰ + sin 30⁰ cos 60⁰ (ii) 2 tan² 45⁰ + cos² 30⁰ – sin² 60⁰

(iii) cos 45⁰ /(sec 30⁰ + cosec 30⁰)

(iv) (sin 30⁰ + tan 45⁰ – cosec 60⁰)/(sec 30⁰ + cos 60⁰ + cot 45⁰)

(v) (5 cos² 60⁰ + 4 sec² 30⁰ - tan² 45⁰)/(sin² 30⁰ + cos² 30⁰)

Answer:

(i) sin 60⁰ cos 30⁰ + sin 30⁰ cos 60⁰= (√3/2) * (√3/2) + (1/2) * (1/2)

= 3/4 + 1/4

= 4/4

= 1

(ii) 2 tan² 45⁰ + cos² 30⁰ – sin² 60⁰ = 2 * 1² + (√3/2) – (√3/2)

= 2 + 3/4 – 3/4

= 2

(iii) cos 45⁰ /(sec 30⁰ + cosec 30⁰) = (1/√2)/(2/√3 + 2)

= (1/√2)/{(2 + 2√3)/ √3}

= √3/{(2 + 2√3) * √2}

= √3/(2√2 + 2√6)

= {√3/(2√2 + 2√6)} * {(2√2 - 2√6)/ (2√2 - 2√6)}

= {√3 * (2√2 - 2√6)}/{(2√2 + 2√6)/ (2√2 - 2√6)}

= (2√6 - 2√18)/{(2√2)² – (2√6)²}

= (2√6 - 6√2)/(8 - 24)

= (2√6 - 6√2)/(-16)

= 2(√6 - 3√2)/(-16)

= (√6 - 3√2)/(-8)

= (3√2 - √6)/8

(iv) (sin 30⁰ + tan 45⁰ – cosec 60⁰)/(sec 30⁰ + cos 60⁰ + cot 45⁰)

= (1/2 + 1 – 2/√3)/( 2/√3 + 1/2 + 1)

= {(√3 + 2√3 - 4)/2√3}/{(4 + √3 + 2√3)/2√3}

= (3√3 - 4)/(3√3 + 4)

= {(3√3 - 4)/(3√3 + 3)} * { (3√3 - 4)/(3√3 - 3)}

= {(3√3 - 4) * (3√3 - 4)} * { (3√3 + 4)/(3√3 - 4)}

= (3√3 - 4)²/{(3√3)² – 4²}

= {(3√3)² + 4² – 2 * 4 * 3√3}/{(3√3)² – 4²}

= (27 + 16 – 24√3}/(27 – 16)

= (43 - 24√3}/11

(v) (5 cos² 60⁰ + 4 sec² 30⁰ - tan² 45⁰)/(sin² 30⁰ + cos² 30⁰)

= {5 * (1/2)² + 4 * (2/√3)² - 1²}/{(1/2)² + (√3/2)²}

= {5/4 + 16/3 - 1}/(1/4 + 3/4)

= {(15 + 64 – 12)/12}/(4/4)

= 67/12

Question 2:

Choose the correct option and justify your choice :

(i) 2 tan 30⁰/(1 + tan² 30⁰) =

(A) sin 60⁰ (B) cos 60⁰ (C) tan 60⁰ (D) sin 30⁰

(ii) (1 – tan² 45⁰)/(1 + tan² 45⁰) =

(A) tan 90⁰ (B) 1 (C) sin 45⁰ (D) 0

(iii) sin 2A = 2 sin A is true when A =

(A) 0⁰ (B) 30⁰ (C) 45⁰ (D) 60⁰

(iv) 2 tan 30⁰/(1 – tan² 30⁰) =

(A) cos 60⁰ (B) sin 60⁰ (C) tan 60⁰ (D) sin 30⁰

Answer:

(i) 2 tan 30⁰/(1 + tan² 30⁰) = (2 * 1/√3)/{1 + (1/√3)²}

= (2/√3)/{1 + 1/3}

= (2/√3)/(4/3)

= 6/4√3

= √3/2

= sin 60⁰ [Since sin 60⁰ = √3/2]

Hence, option (A) is the correct answer.

(ii) (1 – tan² 45⁰)/(1 + tan² 45⁰) = (1 – 1²)/(1 + 1²)

= (1 – 1)/(1 + 1)

= 0/2

= 0

Hence, option (D) is the correct answer.

(iii) We know that sin 0⁰ = 0

Now, sin 2A = sin 2*0⁰ = sin 0⁰ = 0

2 * sin A = 2 * sin 0⁰ = 2 * 0 = 0

Hence, option (A) is the correct answer.

(iv) 2 tan 30⁰/(1 – tan² 30⁰) = (2 * 1/√3)/{1 – (1/√3)²}

= (2/√3)/{1 – 1/3}

= (2/√3)/(2/3)

= 3/√3

= (√3 * √3)/ √3

= √3

= tan 60⁰ [tan 60⁰ = √3]

Hence, option (C) is the correct answer.

Question 3:

If tan (A + B) = √3 and tan (A – B) = 1/√3; 0⁰ < A + B ≤ 90⁰; A > B, find A and B.

Answer:

Given, tan (A + B) = √3

=> tan (A + B) = tan 60⁰

=> A + B = 60⁰ …………..1

and tan (A – B) = 1/√3

=> tan (A - B) = tan 30⁰

=> A – B = 30⁰ ………….2

Add equation 1 and 2, we get

2A = 60⁰ + 30⁰

=> 2A = 90⁰

=> A = 90⁰/2

=> A = 45⁰

Put value of A in equation 1, we get

45⁰ + B = 60⁰

=> B = 60⁰ – 45⁰

=> B = 15⁰

Hence, A = 45⁰ and B = 15⁰

Question 4:

State whether the following are true or false. Justify your answer.

(i) sin(A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0⁰.

Answer:

(i) False.

Let A = 30⁰ and B = 60⁰

Now, sin(A + B) = sin(30⁰ + 60⁰) = sin 90⁰ = 1

and sin A + sin B = sin 30⁰ + sin 60⁰ = 1/2 + √3/2 = (1 + √3)/2

Hence, sin(A + B) ≠ sin A + sin B

(ii) True.

As we know that

sin 0⁰ = 0, sin 30⁰ = 1/2, sin 60⁰ = √3/2 and sin 90⁰ = 1

Hence, for increasing value of θ, sin θ is also increasing.

(iii) False.

cos 0⁰ = 1, cos 30⁰ = √3/2, cos 60⁰ = 1/2 and cos 90⁰ = 0

Hence, for increasing value of θ, cos θ is decreasing.

(iv) False.

Since cos 30⁰ = √3/2 but sin 30⁰ = 1/2

(v) True.

Since tan 0⁰ = 0

Now, cot 0⁰ = 1/tan 0⁰ = 1/0, which is not defined.

Exercise 8.3

Question 1:

Evaluate :

(i) sin 18⁰/cos 72⁰ (ii) tan 26/cot 64⁰ (iii) cos 48⁰ – sin 42⁰ (iv) cosec 31⁰ – sec 59⁰

Answer:

(i) sin 18⁰/cos 72⁰ = cos(90⁰ - 18⁰)/cos 72⁰ [sin(90⁰ - θ) = cos θ]

= cos 72⁰/cos 72⁰

= 1

(ii) tan 26⁰/cot 64⁰ = cot(90⁰ - 64⁰)/cot 64⁰ [tan(90⁰ - θ) = cot θ]

= cot 64⁰/cot 64⁰

= 1

(iii) cos 48⁰ – sin 42⁰ = sin(90⁰ - 42⁰) – sin 42⁰ [sin(90⁰ - θ) = cos θ]

= sin 42⁰ – sin 42⁰

= 0

(iv) cosec 31⁰ – sec 59⁰ = sec(90⁰ - 59⁰) – sec 59⁰ [sec(90⁰ - θ) = cosec θ]

= sec 59⁰ – sec 59⁰

= 0

Question 2:

Show that :

(i) tan 48⁰ tan 23⁰ tan 42⁰ tan 67⁰ = 1

(ii) cos 38⁰ cos 52⁰ – sin 38⁰ sin 52⁰ = 0

Answer:

(i) tan 48⁰ tan 23⁰ tan 42⁰ tan 67⁰ = tan 48⁰ tan 23⁰ cot(90⁰ - 42⁰) cot(90⁰ - 23⁰)

= tan 48⁰ tan 23⁰ cot 48⁰ cot 23⁰ [tan(90⁰ - θ) = cot θ]

= tan 48⁰ tan 23⁰ (1/tan 48⁰ )(1/tan 23⁰)

= 1

(ii) cos 38⁰ cos 52⁰ – sin 38⁰ sin 52⁰ = cos 38⁰ cos 52⁰ – cos(90⁰ - 38⁰) cos(90⁰ - 52⁰)

= cos 38⁰ cos 52⁰ – cos 38⁰ cos 52⁰

= 0

Question3:

If tan 2A = cot (A – 18⁰), where 2A is an acute angle, find the value of A.

Answer:

Given, tan 2A = cot (A – 18⁰)

=> cot (90⁰ - 2A) = cot (A – 18⁰) [tan(90⁰ - θ) = cot θ]

=> 90⁰ – 2A = A – 18⁰

=> 2A + A = 90⁰ + 18⁰

=> 3A = 108⁰

=> A = 108⁰/3

=> A = 36⁰

Question 4:

If tan A = cot B, prove that A + B = 90⁰.

Answer:

Given, tan A = cot B

=> cot(90⁰ – A) = cot B [tan(90⁰ - θ) = cot θ]

=> 90⁰ – A = B

=> A + B = 90⁰

Question 5:

If sec 4A = cosec (A – 20⁰), where 4A is an acute angle, find the value of A.

Answer:

Given, sec 4A = cosec (A – 20⁰)

=> cosec(90⁰ - 4A) = cosec (A – 20⁰) [sec(90⁰ - θ) = cosec θ]

=> 90⁰ – 4A = A – 20⁰

=> 4A + A = 90⁰ + 20⁰

=> 5A = 110⁰

=> A = 110⁰/5

=> A = 22⁰

Question 6:

If A, B and C are interior angles of a triangle ABC, then show that

sin (B + C)/2 = cos A/2

Answer:

Given, A, B and C are interior angles of a triangle ABC.

=> A + B + C = 180⁰

=> (A + B + C)/2 = 180⁰/2

=> (A + B + C)/2 = 90⁰

=> A/2 + (B + C)/2 = 90⁰

=> (B + C)/2 = 90⁰ - A/2

Now, sin{(B + C)/2} = sin (90⁰ - A/2)

=> sin{(B + C)/2} = cos A/2 [sin(90⁰ - A) = cos A]

Question 7:

Express sin 67⁰ + cos 75⁰ in terms of trigonometric ratios of angles between 0⁰ and 45⁰.

Answer:

We know that sin θ = cos(90⁰ – θ) and cos θ = sin(90⁰ – θ)

Therefore, sin 67 + cos 75 = cos(90⁰ – θ) + sin(90⁰ – θ)

= cos 23⁰ + sin 15⁰

Exercise 8.4

Question 1:

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

Answer:

(i) sin A = √(sin² A)

= √(1/coses² A)

= √{1/(1 + cot² A)} [coses² A = 1 + cot² A]

(ii) sec A = √(sec² A)

= √(1 + tan² A) [sec² A = 1 + tan² A]

= √(1 + 1/cot² A)

= √{(1 + cot² A)/ cot² A}

= √{(1 + cot² A)/ cot A

(iii) tan A = 1/cot A

Question 2:

Write all the other trigonometric ratios of ∠ A in terms of sec A.

(i) sin A = √(sin² A)

= √(1 - cos² A)

= √(1 – 1/sec² A)

= √{( sec² A – 1)/sec² A}

= √( sec² A – 1)/sec A

(ii) cos A = 1/sec A

(iii) tan A = √(tan² A)

= √(sec² A - 1)

(iv) cot A = √(cot² A)

= √(1/tan² A)

= √{1/(sec² A - 1)}

= 1/√ (sec² A - 1)

(v) cosec A = √(cosec² A)

= √(1 + cot² A)

= √(1 + 1/tan² A)

= √{(1 + 1/(sec² A - 1)}

= √{(sec² A – 1 + 1)/(sec² A - 1)}

= √{sec² A/(sec² A - 1)}

= sec A/√(sec² A - 1)

Question 3:

Evaluate :

(i) (sin² 63⁰ + sin² 27⁰)/(cos² 17⁰ + cos² 73⁰)

(ii) sin 25⁰ cos 65⁰ + cos 25⁰ sin 65⁰

Answer:

(i) (sin² 63⁰ + sin² 27⁰)/(cos² 17⁰ + cos² 73⁰)

= {sin² 63⁰ + cos² (90 - 27⁰)}/{cos² 17⁰ + sin² (90 - 73⁰)}

[sin θ = cos(90⁰ – θ) and cos θ = sin(90⁰ – θ)]

= (sin² 63⁰ + cos² 63⁰)/(cos² 17⁰ + sin² 17⁰)

= 1

(ii) sin 25⁰ cos 65⁰ + cos 25⁰ sin 65⁰

= cos(90 - 25⁰)cos 65⁰ + sin(90 - 25⁰)sin 65⁰

= cos 65⁰ cos 65⁰ + sin 65⁰ sin 65⁰ [sin θ = cos(90⁰ – θ) and cos θ = sin(90⁰ – θ)]

= cos² 65⁰ + sin² 65⁰

= 1 [sin² θ + cos² θ = 1]

Question 4:

Choose the correct option. Justify your choice.

(i) 9 sec² A – 9 tan² A =

(A) 1 (B) 9 (C) 8 (D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =

(A) 0 (B) 1 (C) 2 (D) -1

(iii) (sec A + tan A) (1 – sin A) =

(A) sec A (B) sin A (C) cosec A (D) cos A

(iv) (1 + tan² A)/(1 + cot² A) =

(A) sec² A (B) –1 (C) cot² A (D) tan² A

Answer:

(i) 9 sec² A – 9 tan² A =

Given, 9 sec² A - 9 tan² A

= 9(sec² A - tan² A)

= 9*1 {since sec² A - tan² A = 1}

= 9

Hence, option (B) is the correct answer.

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =

= 1 + cot θ – cosec θ + tan θ + tan θ cot θ – tan θ cosec θ + sec θ + sec θ cot θ - sec θ cosec θ

= 1 + cos θ/sin θ – 1/sin θ + sin θ/cos θ + 1 – sin θ/cos θ * 1/sin θ + 1/cos θ

+ 1/cos θ * cos θ/sin θ – 1/cos θ * 1/sin θ [since tan θ cot θ = 1]

= 2 + cos θ/sin θ – 1/sin θ + sin θ/cos θ – 1/cos θ + 1/cos θ + 1/sin θ – 1/(cos θ * sin θ)

= 2 + cos θ/sin θ + sin θ/cos θ – 1/(cos θ * sin θ)

= 2 + (cos² θ + sin² θ - 1)/ (cos θ * sin θ)

= 2 + (1 - 1)/ (cos θ * sin θ) [sin² θ + cos² θ = 1]

= 2 + 0

= 2

Hence, option (C) is the correct answer.

(iii) (sec A + tan A) (1 – sin A) = (1/cos A + sin A/cos A) (1 – sin A)

= {(1 + sin A)/cos A}(1 – sin A)

= (1 – sin² A)/cos A

= cos² A/cos A [sin² θ + cos² θ = 1]

= cos A

Hence, option (D) is the correct answer.

(iv) (1 + tan² A)/(1 + cot² A) = sec² A/cosec² A

[sec² A = 1 + cot² A and cosec² A = 1 + tan² A]

= (1/cos² A)/(1/sin² A)

= sin² A/ cos² A

= tan² A

Hence, option (D) is the correct answer.

Question 5:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec θ – cot θ)² = (1 - cos θ)/(1 + cos θ) (ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A

(iii) tan θ/(1 – cot θ) + cot θ/(1 – tan θ) = 1 + sec θ + cosec θ

[Hint : Write the expression in terms of sin θ and cos θ]

(iv) (1 + sec A)/sec A = sin² A/(1 – cos A)

[Hint : Simplify LHS and RHS separately]

(v) (cos A – sin A + 1)/ (cos A + sin A - 1) = cosec A, using the identity cosec² A = 1 + cot² A.

(vi) √{(1 + sin A)/(1 – sin A)} = sec A + tan A

(vii) (sin θ – 2 sin³ θ)/(2cos³ θ – cos θ) = tan θ

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

[Hint : Simplify LHS and RHS separately]

(x) (1 + tan² A)/(1 + cot² A) = (1 - tan² A)/(1 - cot² A) = tan² A

Answer:

(i) LHS:

(cosec θ – cot θ)² = (1/sin θ – cos θ/sin θ)²

= {(1 – cos θ)/sin θ}²

= (1 – cos θ)²/sin² θ

= (1 – cos θ)²/(1 – cos² θ) [sin² θ + cos² θ = 1]

= (1 – cos θ)²/{(1 – cos θ) (1 + cos θ)} [a² – b² = (a - b)(a + b)]

= (1 – cos θ)/{(1 + cos θ)

= RHS

(ii) LHS:

cos A/(1 + sin A) + (1 + sin A)/cos A = {cos² A + (1 + sin A)²}/{(1 + sin A) * cos A}

= {cos² A + 1 + sin² A + 2sin A}/{(1 + sin A) * cos A}

= {1 + 1 + 2sin A}/{(1 + sin A) * cos A} [sin² θ + cos² θ = 1]

= {2 + 2sin A}/{(1 + sin A) * cos A}

= 2(1 + sin A)/{(1 + sin A) * cos A}

= 2/cos A

= 2 sec A

= RHS

(iii) LHS:

tan θ/(1 – cot θ) + cot θ/(1 – tan θ)

= (sin θ/cos θ)/(1 – cos θ/sin θ) + (cos θ/sin θ)/(1 – sin θ/cos θ)

= (sin θ/cos θ)/{(sin θ – cos θ)/sin θ} + (cos θ/sin θ)/{(cos θ – sin θ)/cos θ}

= sin² θ/{cos θ(sin θ – cos θ)} + cos² θ/{sin θ(cos θ – sin θ)}

= sin² θ/{cos θ(sin θ – cos θ)} - cos² θ/{sin θ(sin θ – cos θ)}

= (sin³ θ - cos³ θ)/{cos θ * sin θ * (sin θ – cos θ)}

= {(sin² θ + cos² θ + sin θ cos θ)(sin θ – cos θ)}/{cos θ * sin θ * (sin θ – cos θ)}

= (1 + sin θ cos θ)/(cos θ * sin θ) [sin² θ + cos² θ = 1]

= 1/(cos θ * sin θ) + (cos θ * sin θ)/ (cos θ * sin θ)

= sec θ cosec θ + 1

= RHS

(iv) LHS

(1 + sec A)/sec A = (1 + 1/cos A)/(1/cos A)

= {(cos A + 1)/cos A}/(1/cos A)

= (1 + cos A)/1

= (1 + cos A)/1 * (1 - cos A)/ (1 - cos A)

= (1 – cos² A)/ (1 + cos A)

= sin² A/ (1 + cos A) [sin² θ + cos² θ = 1]

= RHS

(v) LHS:

(cos A – sin A + 1)/ (cos A + sin A - 1)

= {(cos A – sin A + 1)/sin A}/{(cos A + sin A - 1)/sin A} [divide by sin A]

= (cot A – 1 + cosec A)/ (cot A + 1 - cosec A)

= {cot A + (-1) + cosec A}/ (cot A + 1 - cosec A)

= {cot A + cosec A + cosec² A - cot² A}/ (cot A + 1 - cosec A)

= {cot A + cosec A + (cosec A - cot A) (cosec A + cot A)}/ (cot A + 1 - cosec A)

= {(cosec A + cot A) (1 - cosec A + cot A)}/ (cot A + 1 - cosec A)

= cosec A + cot A

= RHS

(vi) LHS:

√{(1 + sin A)/(1 – sin A)} = √[{(1 + sin A)/(1 – sin A)} * {(1 + sin A)/(1 + sin A)}]

= √{(1 + sin A)²/(1 – sin² A)}

= √{(1 + sin A)²/cos² A}

= (1 + sin A)/cos A

= 1/cos A + sin A/cos A

= sec A + tan A = RHS

(vii) (sin θ – 2 sin³ θ)/(2cos³ θ – cos θ) = {sin θ(1 – 2 sin² θ)}/{cos θ(2cos² θ – 1)}

= {sin θ(1 – 2 sin² θ)}/{cos θ * 2(1 - sin² θ) – 1)}

= {sin θ(1 – 2 sin² θ)}/{cos θ(2 - 2sin² θ – 1)}

= {sin θ(1 – 2 sin² θ)}/{cos θ(1 - 2sin² θ)}

= sin θ/cos θ

= tan θ

= RHS

(viii) LHS:

(sin A + cosec A)² + (cos A + sec A)²

= sin² A + cosec² A + 2sin A cosec A + cos² A + sec² A + 2cos A cot A

= (sin² A + cos² A) + (cosec² A + sec² A) + 2sin A * 1/sin A + 2cos A * 1/cos A

= 1 + 1 + cot² A + 1 + tan² A + 2 + 2

= 7 + cot² A + tan² A

= RHS

(ix) LHS:

(cosec A – sin A)(sec A – cos A) = (1/sin A – sin A)(1/cos A – cos A)

= {(1 – sin² A)/sin A }{(1 – cos² A)/cos A}

= {cos² A)/sin A }{sin² A)/cos A}

= sin A cos A …………..1

RHS:

1/(tan A + cot A) = 1/(sin A/cos A + cos A/sin A)

= 1/{(sin² A + cos² A)/(sin A cos A)}

= 1/{1/(sin A cos A)}

= sin A cos A ……………2

From equation 1 and 2, we get

(cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

(x) LHS:

(1 + tan² A)/(1 + cot² A) = sec² A/cosec² A

= (1/cos² A)/(1/sin² A)

= sin² A/cos² A

= tan² A

= RHS

Again (1 - tan A)²/(1 - cot A)² = {(1 - tan A)/(1 - cot A)}²

= {(1 - sin A/cos A)/(1 - cos A/sin A)}²

= [{(cos A - sin A)/cos A{/{(sin A - cos A)/sin A)}]²

= [{(cos A - sin A)/cos A} * {sin A/(sin A - cos A)/sin A}]²

= [-{(sin A - cos A)/cos A} * {sin A/(sin A - cos A)/sin A}]²

= (-sin A/cos A)²

= sin² A/cos² A

= tan² A

= RHS

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