**N****CERT Solutions Class ****10** **Maths**** Chapter ****8** **Introduction To Trigonometry**

**NCERT Solutions Class ****10** **Maths**** Chapter ****8** **Introduction To Trigonometry****:** **National Council of Educational Research and Training Class 10 Maths Chapter 8 Solutions – Introduction To Trigonometry. NCERT Solutions Class 10 Maths Chapter 8 PDF Download.**

**NCERT Solutions Class ****10** **Maths**** Chapter ****8****: Overview**

Board |
NCERT |

Class |
10 |

Subject |
Maths |

Chapter |
8 |

Chapter Name |
Introduction To Trigonometry |

Topic |
Exercise Solutions |

**NCERT Solutions **

**Class ****10** **Maths**

**Chapter ****8**** – ****Introduction To Trigonometry**

__EXERCISE 8.1__

__EXERCISE 8.1__

**(1) In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:**

**(i) Sin A, cos A**

**(ii) Sin C, cos C**

**Solution:**

In ∆ABC, given ∠B = 90°

AB = 24 cm, BC = 7 cm

We know that,

AC^{2} = AB^{2} + BC^{2} [Pythagoras Theorem]

AC^{2} = (24)^{2} + (7)^{2}

AC^{2 }= 576 + 49

AC^{2} = 625 cm^{2}

AC = √625 = 2

Hence AC = 25cm

(i) Sin A, Cos A

Sin A = Opposite side/hypotenuse = BC/AC = 7/25

Cos A = Adjacent side/Hypotenuse = AB/AC = 24/25

(ii) Similarity, we have to find = Sin C, Cos C

Cos C = Adjacent side/Hypotenuse = BC/AC = 7/25

Sin C = Opposite side/Hypotenuse = AB/AC = 24/25

**(2) In Fig. 8.13, find tan P – cot R.**

**Solution:**

In ∆PQR, given ∠B = 90°, PQ = 12 PR = 13.

Since ∆PQR is right angle triangle applying Pythagoras theorem,

PR^{2} = PQ^{2} + OR^{2}

13^{2} = 12^{2} + QR^{2} [Putting the value of PA & PR]

QR^{2} = 169 – 144

QR^{2} = 25

Q = 5 cm

∴ Tan P = Opposite side/adjacent side = QR/PQ = 5/12

Cot R = adjacent side/opposite side = 5/12

Than tan P – cot R

= 5/12 – 5/12

= 0

** **

**(3) If sin A = ¾, calculate cos A and tan A. **

**Solution: **

Sin A = 3/4

Opposite side (Perpendicular)/Hypotenuse = 3/4

Let perpendicular = P

Hypotenuse = H

P/1+ = 3/4

Let P = 3x, H = 4x

Applying Pythagoras theorem,

AC^{2 }= AB^{2} + BC^{2}

(4K)^{2} = (3x)^{2} + BC^{2} (Putting the value)

BC^{2} = 16x^{2} – 9x^{2}

BC^{2} = 7x^{2}

BC = √7 x

∴ Cos A = Adjacent side (base)/Hypotenuse = AB/AC

= √7 x/4x = √7/4

Tan A = Opposite side (Perpendicular)/Adjacent side (Base)

= BC/AB = 3/√7

**(4) Given 15 cot A = 8, find sin A and sec A.**

**Solution: **

Let ∆ABC is a right angled triangle, ∠B = 90°

Given, 15 cot A = 8

Or, Cot A = 8/15

Adjacent side (Base)/Opposite side (perpendicular) = AB/BC = 8/15

Let, AB = 8x

BC = 15

Applying Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (8x)^{2} + (15x)^{2}

AC^{2} = 64x^{2} + 225x^{2}

AC = √289x^{2} = 17x

∴ Sin A = Opposite side (Perpendicular)/Hypotenuse

= BC/AC = 15x/17x = 15/17

Sec A = Hypotenuse/Adjacent side (Base)

= AC/BC = 17K/8K = 17/8

** **

**(5) Given sec θ = 13/12, calculate all other trigonometric ratios.**

**Solution: **

Let, ∆ABC is a right angled triangle, ∠B = 90°

Given, Sec θ = 13/12

Hypotenuse/Adjacent side (Base) = 13/12

AC/BC = 13/12

Let, AC = 13x, BC = 12x

Applying Pythagoras theorem, we get

AC^{2} = AB^{2} + BC^{2}, BC = 12x

AC^{2} = AB^{2} + BC^{2}

(13x)^{2} = AB^{2} + (12x)^{2} [Putting the value]

AB^{2} = 169x^{2} – 144x

AB^{2} = 25x^{2}

AB = 5x

So, Sin θ = Perpendicular/Hypotenuse = AB/AC = 5x/13x = 5/13

Cos θ = Base/Hypotenuse = BC/AC = 12x/13x = 12/13

Tan θ = Perpendicular/Base = AB/BC = 5x/12x = 5/12

Cosec θ = Hypotenuse/Perpendicular = AC/AB = 13x/5x = 13/5

Cot θ = Base/Perpendicular = BC/AB = 12x/5x = 12/5

** **

**(6) If ****∠****A**** and ****∠****B are acute angles such that cos A = cos B, then show that ****∠****A = ****∠****B****.**

**Solution: **

Let, ∆ABC has ∠A and ∠B are acute angle

Given that,

Cos A = Cos B

We can write from ∆ABC

Cos A = AC/AB & Cos B = BC/AB

AB Cos A = Cos B

AC/AB = BC/AB

Or, AC = BC

Or, ∠B = ∠A

(An angles opposite to equal side of the triangle are equal)

** **

**(7) If cot ****θ = 7/8, evaluate: **

**(i) (1+sin θ) (1-sinθ)/(1+cosθ) (1-cosθ)**

**(ii) Cot ^{2}θ **

**Solution: **

Let, ∆ABC is a right angled triangle ∠B = 90°

Given, cot θ = 7/8

BC/AB = 7/8

Let, BC = 7x

AB = 8k

Applying Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (8k)^{2} + (7k)^{2 }(Putting the value)

AC^{2} = 64x^{2} + 49x^{2}

AC^{2 }= 113x^{2}

AC = √113x

So, Sin θ AB/AC = Perpendicular/Hypotenuse = 8K/√113x = 8k/√113

Cos θ = BC/AC = Base/Hypotenuse = 7k/√113x = 7/√113

(i) (1+sinθ) (1-sinθ)/(1+cosθ) (1-cosθ)

= 1 – sin^{2} θ/1 – cos^{2} θ

= 1 – (8/√113)^{2}/1 – 7/√113)^{2}

= 1 – (64/113)/1 – (49/113)

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

(ii) cot^{2} θ = (7/8)^{2} = 49/64

** **

**(8) If 3 cot A = 4, check whether 1-tan ^{2}A/1+tan^{2}A = cos^{2}A – sin^{2}A or not. **

**Solution: **

Let ∆ABC is a right angled triangle ∠B = 90°

Given,

Cot A = 4/3

AB/BC = 4/3

Let AB = 4x, BC = 3x

Applying Pythagoras theorem

AC^{2 }= AB^{2} + BC^{2}

AC^{2} = (4x)^{2} + (3x)^{2} (Putting the value)

= AC^{2} = 16x^{2} + 9x^{2}

AC^{2} = 25x^{2}

AC = 5x

Sin A = Perpendicular/Hypotenuse = BC/AC = 3k/5x = 3/5

Cos A = Base/hypotenuse = AB/AC = 4x/5x = 4/5

Tan A = Perpendicular/Base = BC/AB = 3x/4x = 3/4

__R.H.S __

Cos^{2}A – sin^{2}A = (4/5)^{2} – (3/5)^{2} = 16/25 – 9/25

= 16-9/25 = 7/25

__L.H.S __

1-tan^{2}A/1+tan^{2}A = 1-(3/4)^{2}/1+(3/4)^{2} = (16-9/16)/(16+9/16)

= 7/25

Since R.H.S = L.H.S = 7/25

Hence (1-tan^{2}A)/(1+tan^{2}A) = cos^{2}A – sin^{2}A (Proved)

** **

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

**(i) SinA cosC + cosA sinC **

**(ii) cosA cosC – sinA sinC **

**Solution: **

Let us assume ∆ABC is a right angled triangle. ∠B = 90°

tan A = 1/√3

Or, Perpendicular/Base = 1/√3

Or, BC/AB = 1/√3

Let, BC = K, AB = √3x

From the Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (√3x)^{2} + (x)^{2} (Putting the value)

AC^{2} = 3x^{2} + x^{2}

AC^{2} = 4x^{2}

AC = 2x

Now Sin A = BC/AC = 1/2

Cos A = AB/AC = √3/2

Cos C = BC/AC = 1/2

Sin C = AB/AC = √3/2

(i) Sin A Cos C + cos A sin C

= ½ × 1/2 + √3/2 . √3/2

= 1/4 + 3/4 = 1+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

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

**Solution: **

Given ∆PQR, right angled at a,

PQ = 5cm

PR + QR = 25cm

Let, QR = x

PR = 25 – QR

PR = 25 – x

Applying Pythagoras theorem

PR^{2} = PQ^{2} + QR^{2}

(25 – K)^{2} = 5^{2} + x^{2}

625 – 50K + x^{2} = 5^{2} + x^{2 }

50x = 625 – 25

x = 600/50 = 12

∴ x = 12 = QR

So, PR = 25 – QR = 25 – 12 = 13

(i) Sin P = Perpendicular/Hypotenuse = QR/PR = 12/13

(ii) Cos P = Base/Hypotenuse = Pa/PR = 5/13

(iii) Tan P = Perpendicular/Base = QR/P = 12/5

** **

**(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**** **

**Solution:**

(i) The value of tan A always less than 1. = It is wrong statement.

Let, In ∆ABC, ∠B = 90°

AB = 6, BC = 8, AC = 10

∴ Tan A = 8/6 which is greater than 1.

According Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

(10)^{2} = (8)^{2} + (6)^{2 }

100 = 64 + 36

100 = 100

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

**Solution:-**

**-> It is a true statement. **

**Sec A = 12/5 **

**Hypotenuse/Base = 12/5 **

**AC/AB = 12/5**

Let, AC = 12k, AB = 5k

By Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

(12k)^{2} = (5k)^{2} + BC^{2}

BC^{2} = 144x^{2} – 25x^{2}

BC^{2} = 119k^{2}

∴ BC = √119 k

Such a triangle is possible.

(iii) Cos A is the abbreviation used for the cosecant of angle A –> It is a false statement.

Abbreviation used for cosecant of angle A is cosec A.

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

–> False

=> Cot A is not the product of cot and A. ∠A is the angle of cotangent.

(v) Sin θ = 4/3 for some angle θ

–> False

–> Sinθ = Perpendicular/Hypotenuse

–> Hypotenuse is the longest side.

∴ Sinθ is always less than 1.

It can never be 4/3.

** **

__Exercise – 8.2__

__Exercise – 8.2__

**(1) Evaluate the following:**

**(i) sin 60° cos 30° + sin 30° cos 60° **

**Solution: **

sin 60° cos 30° + sin 30° cos 60°

= √3/2 × √3/2 + 1/2 × 1/2

= 3/4 + 1/4

= 3+1/4 = 4/4 = 1 (Ans)

[Putting the value of sin 60° = √3/2 = co 30° Sin 30° = ½ = cos 60°]

**(ii) 2 tan ^{2} 45° + cos^{2} 30° – sin^{2} 60° **

**Solution: **

2 tan^{2} 45° + cos^{2} 30° – sin^{2} 60°

Putting the value

= 2 (1)^{2} + (√3/2)^{2} – (√3/2)^{2}

= 2 + 3/4 – 3/4

= 2 (Ans)

[We know that tan45° = 1 Sin 60° = √3/2 = cos 30°]

** **

**(iii) Cos 45°/Sec 30° + cosec 30° **

**Solution: **

Cos 45°/Sec 30° + cosec 30°

We know that

Cos 45° = 1/√2, Sec 30° = 2/√3, Cosec 30° = 2

Putting the value, we get

= (1/√2)/(2√3 + 2)

= (1√2)/(2+2√3/√3)

= √3/√2(2+2√3)

= √3/(2√2 + 2√6) × (2√2 – 2√6)/(2√2 – 2√6)

= 2 (√6 – √18)/(2√2)^{2} – (2√6)

= 2(√6 – √13)/8-24

= 2 (√6 – √18)/-16

= √6 – √18/8

= √18 – √6/8

= √2×3×3 – √6/8

= 3√2 – √6/8

** **

**(iv) Sin 30****° + tan 45° – cosec 60°/sec 30° + cos 60° + cot 45° **

**Solution: **

Sin 30° + tan 45° – cosec 60°/sec 30° + cos 60° + cot 45°

We know that

Sin 30° = 1/2, tan 45° = 1, cosec 60° = 2/√3

Sec 30° = 2/√3, cos 60° = ½, cot 45° = 1

Putting this value,

** **

**(v) 5 cos ^{2} 60° + 4 sec^{2} 30° – tan^{2} 45°/sin^{2} 30° + cos^{2} 30° **

**Solution: **

5 cos^{2} 60° + 4 sec^{2} 30° – tan^{2} 45°/sin^{2} 30° + cos^{2} 30°

We know that,

Cos 60° = 1/2, sec 30° = 2/√3, tan 45° = 1

Sin 30° = 1/2, cos 30° = √3/2

Putting this value,

** **

**(2) Choose the correct option and justify your choice:**

**(i) 2tan30°/1+tan ^{2}30° = **

**(A) sin 60° **

**(B) cos 60° **

**(C) tan 60° **

**(D) sin 30° **

**Solution:**

2tan30°/1+tan^{2}30°

We know that

tan 30° = 1/√3 and tan^{2} 30° = 1/3

Putting this value

= 2×(1/√3)/(1+1/3)

= (2/√3)/(3+1/3)

= 2/√3 × ¾

= 3/√3×2 × √3/√3

= 3√3/3×2

= √3/2 = sin 30°

Hence D option is correct answer.

[We know that sin 30° = √3/2]

** **

**(ii) 1-tan ^{2} 45**

**°/1+tan**

^{2 }45° =**(A) tan 90° **

**(B) 1 **

**(C) sin 45° **

**(D) 0 **

**Solution: **

1-tan^{2} 45°/1+tan^{2 }45°

We know that

tan 45° = 1 Hence, tan^{2} 45 = 1

Putting this value

= 1-1/1+1

= 0/2 = 0 (Ans)

Hence, (D) option is correct answer.

** **

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

**(A) 0° **

**(B) 30° **

**(C) 45° **

**(D) 60° **

**Solution: **

Sin 2A = 2 sin A

__R.H.S __

Sin 2A

Putting A = 0°

Sin θ = 0

Putting A = 30°

Sin 2×30° = sin 60°

= √3/2

Putting A = 45°

Sin 2×45° = Sin 90°

= 1

__L.H.S __

2 sin A

Putting A = 0°

2 Sin θ = 0°

Putting A = 30°

2 Sin 30° = 2×1/2 = 1

Putting A = 45°

2 sin 45° = 2 × 1/√2

= 2/√2 × √2/√2

= √2

Hence when A = 0°, the sin 2A = 2 sinA is Satisfy

Hence option (A) is correct answer.

**(iv) 2 tan30°/1-tan ^{2}30° **

**(A) cos 60° **

**(B) sin 60° **

**(C) tan 60° **

**(D) sin 30° **

**Solution: **

2 tan30°/1-tan^{2}30°

We know that

Tan 30° = 1/√3 and tan^{2} 30 = (1/√3)^{2} = 1/3

Putting this value

= (2×1/√3)/(1-1/3)

= (2/√3)/(3-1/3)

= (2√3)/(2/3)

= 2/√3 × 3/2 = 3/√3 × √3/√3

= 3√3/3

= √3 Tan 60° [We know that tan60° = √3]

Hence (c) option is correct answer.

** **

**(3) If tan (A + B) = ****√3 and tan (A – B) = 1/√3; 0****° <A + B ****≤ ****90****°; A > B, find A and B. **

**Solution: **

Given tan (A + B) = √3

Or, tan (A+B) = tan60° [We know that tan60° = √3 & 0° < A + B ≤ 90°]

Or, A+B = 60° —— eq (1)

Also given,

tan (A-B) = 1/√3

Or, tan (A – B) = tan30° [We know that tan 30° = 1/√3]

(A – B) = 30° —- eq(2)

From eq (1) & (2) we can write

A = 45°

Putting this value in eq (1)

B + 45° = 60°

B = 15°

Solving eq (1) & (2), we get

A = 45° & B = 15°

** **

**(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 = 0°.**

**Solution: **

Sin (A+B) = sinA + sinB

It is a false statement.

In a triangle, A = B ≠ 0

If we put A = 45°, and B = 45°

__R.H.S __

Sin (A+B)

= Sin (45° + 45°)

= sin 90°

= 1

__L.H.S__

SinA + sin B

= Sin 45° + sin 45°

= 1/√2 + 1/√2

= 2/√2 × √2/√2

= R√2/2 = √2

Hence R.H.S ≠ L.H.S

**In case you are missed :- NCERT Solution for Surface Areas And Volumes**

(ii) The value of sin θ increases as θ increases. The statement is correct (True)

We know that

If θ = 30°

If θ = 60°

If θ = 90°

Sin 30° = 1/2

Sin 60° = √3/2

Sin 90° = 1

Increasing the θ value, the sin θ also increases.

(iii) The value of cos θ increases as θ increase. The statement is false

We know that

θ = 0°

θ = 90°

cos 0° = 1

cos 90° = 0

When θ increases, the value of cosθ decreases.

(iv) Sinθ = cos θ for all values of θ it is false statement.

Because sin 90° ≠ cos 90°

(v) CotA is not defined for A = 0°

It is a true statement.

Cot A = CosA/SinA

Putting A = 0°

= cos0°/sin0°

= 1/0

= Not defined

** **

__EXERCISE 8.3__

__EXERCISE 8.3__

**(1) Evaluate:**

**(i) sin18°/cos72° **

**(ii) Tan26°/cot64° **

**(iii) cos48°-sin42° **

**(iv) cosec31° – sec59°**

**Solution:**

(i) We know that

cosA = sin (90° – A)

so, cos72° = sin (90° – 72°)

= sin 18°

Hence, sin18°/cos72° = sin18°/sin18° = 1

(ii) We know that

CotA = tan (90°- A)

So, cot 64° = tan (90° – 64°) = tan 26°

Hence tan26°/cot64° = tan26°/tan26° = 1

**(iii) We know that**

SinA = cos (90° – A)

SinA 42° = cos (90° – 42°) = cos 48°

Therefore, cos48° – sin42°

= cos48° – cos48° = 0

**(iv) We know that**

SecA = cosec (90° – A)

So, Sec59° = cosec (90° – 59°) = cosec 31°

Hence,

cosec 31° – sec 59°

= cosec 31° – cosec 31°

= 0

**(2) Show that:**

**(i) tan 48° tan 23° tan 42° tan 67° = 1**

**(ii) cos 38° cos 52° – sin 38° sin 52° = 0**

**Solution:**

**(i) We know that**

TanA = 1/cotA

cotA = tanA (90 – A)

(i) tan 48° tan 23° tan 42° tan 67°

= 1/cot48° × 1/cot23° × tan42° tan67° [Using this cot48 = tan (90° – 48°) = tan42°

= 1/tan42° × tan42° × 1/tan67 × tan67× [cot23° = tan (90° – 23)] = tan67

= 1 (Proved)

**(ii) We know that**

sinA = cos (90° – A)

So, sin38° = cos (90° – 38°) = cos 52°

Similarity sin52° = cos (90° – 52°) = cos 38°

Cos 38° cos52° – sin38° sin52°

= cos 38° cos 52 – cos 38° cos 52° [Putting the value]

= 0

** **

**(3) If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. **

**Solution: **

Given that

tan2A = cot (A – 18°) —– (1)

We know that

tanθ = cot (90° – θ)

tan2A = cot (90° – 2A]

We can write (1) as

cot (90° – 2A) = cot (A – 18°)

Therefore, we can write

90° – 2A = A – 18° (2A is acute angle)

3A = 90° + 18°

3A = 108°

A = 36°

** **

**(4) If tan A = cot B, prove that A + B = 90°**

**Solution:**

Given that tan A = cot B —– (1)

We know that

tan B = cot (90° – A)

We can write (1) as

cot (90° – A) = cot B

(90° – A) & B are both acute angles hence

90° – A = B

∴ A + B = 90° (Proved)

** **

**(5) If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.**

**Solution: **

Given that sec4A = cosec (A – 20°) —– (1)

We know that,

sec A = cosec (90° – A)

sec 4A = cosec (90° – 4A)

We can write (1) as

cosec (A – 20°) = cosec (90° – 4A)

Since (A – 20°) & (90° – 4A) are acute angle.

Hence, A – 20° = 90° – 4A

5A = 110°

A = 22°

** **

**(6) 6. If A, B and C are interior angles of a triangle ABC, then show that**

**Sin (B+C/2) = cos A/2 **

**Solution: **

Sum of angle in triangle ABC is A + B + C = 180°

Or, B+C = 180 – A

Or, B+C/2 = 90° – A/2 [Divide both side by 2]

Or, Sin B+C/2 = sin (90° – A/2] [Multiplying both side by sin]

Or, Sin B+C/2 = cos A/2 (Proved)

** **

**(7) Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. **

**Solution: **

Sin67° + cos75°

= cos (90° – 67°) + sin (90° – 75°)

= cos23° + sin15° [Using sinA = cos (90 – A) cosA = sin (90 – A)

__EXERCISE 8.4__

__EXERCISE 8.4__

**(1) Express the trigonometric ratios sinA, secA and tanA in terms of cotA.**

**Solution: **

sinA = 1/cosecA = 1/√1+cotA [As cosec^{2}A – cot^{2}A = 1 or cosecA = √1+cotA]

Sec^{2}A – tan^{2}A = 1

Or, sec^{2}A – 1+tan^{2}A

Or, Sec^{2}A = 1 + 1/cot^{2}A

Or, Sec^{2}A = cot^{2}A + 1/cotA

Or, Sec A = √cot^{2}A+1/cotA

TanA = 1/cotA

** **

**(2) Write all the other trigonometric ratios of ****∠****A in terms of sec A.**

**Solution:**

Sin^{2}A + cos^{2}A = 1

Or, sin^{2}A = 1 – cos^{2}A

Or, sin^{2}A = 1 – 1/sec^{2}A

Or, sin^{2}A = sec^{2}A-1/sec^{2}A

Or, sinA = √sec^{2}A – 1/secA

∴ cosec A = 1/sinA = secA/√sec^{2}A – 1

cosA = 1/secA

Sec^{2}A – tan^{2}A = 1

Or, tan^{2}A = 1 – sec^{2}A

Or, Tan^{2}A = sec^{2}A – 1

Or, tanA = √Sec^{2}A – 1

TanA = 1/cotA = 1/√Sec^{2}A – 1

**(3) Evaluate: **

**(i) Sin ^{2} 63° + sin^{2} 27°/cos^{2} 17° + cos^{2} 73°**

**(ii) sin25° cos65° + cos25° sin65°**

**Solution:**

Sin^{2} 63° + sin^{2} 27°/cos^{2} 17° + cos^{2} 73° —– (i)

We know that

SinA = cos (90° – A)

So sin27° = cos (90° – 27°) = cos 63°

Similarly cos17° = sin (90° – 17°) = sin 73

Putting this value in equation (i)

sin^{2}63° + cos^{2}63°/sin^{2}73° + cos^{2}73° [We know that sin^{2} θ + cos^{2}θ = 1]

= 1/1 = 1

(ii) sin25° cos 65° + cos 25° sin 65° —– (1)

We know that,

sinA = cos (90° – A)

So, sin25° = cos (90° – 25°) = cos 65°

Similarly cos 25° = sin (90° – 25) = sin 65°

Putting this value in equation (2)

Cos65°.cos65 + sin65°.sin65°

= cos^{2}65° + sin^{2}65 [We know that sin^{2} θ + cos^{2} θ = 1]

= 1

**(4) Choose the correct option. Justify your choice.**

**(i) 9sec ^{2}A – 9 tan^{2}A = **

**(A) 1**

**(B) 9 **

**(C) 8**

**(D) 0 **

**Solution: **

9sec^{2}A – 9 tan^{2}A

= 9 (sec^{2}A – tan^{2}A) [We know that sec^{2}A – tan^{2}A = 1]

= 9×1

= 9

** **

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

**(A) 0 **

**(B) 1 **

**(C) 2**

**(D) -1 **

**Solution: **

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

= (1 + sinθ/cotθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ)

= (cosθ + sinθ + 1/cosθ) (sinθ + cosθ – 1/sinθ)

= 1/sinθ.cosθ [{(sinθ + cosθ) +1} {(sinθ + cosθ)} – 1}]

= 1/sinθ.cosθ [(sinθ + cosθ)^{2} – 1^{2})] [(We know that (a+b) (a-b) = a^{2}-b^{2}]

= 1/sinθ.cosθ [Sin^{2}θ + cos^{2}θ + 2sinθ cosθ – 1]

= 1/sinθ.cosθ [1 + 2sinθ.cosθ – 1]

= 2sinθ.cosθ/sinθ.cosθ [sin^{2}θ + cos^{2}θ = 1]

= 2

** **

**(iii) (sec A + tan A) (1 – sin A) =**

**(A) Sec A **

**(B) Sin A **

**(C) Cosec A **

**(D) Cos A **

**Solution: **

(SecA + TanA) (1-sinA)

= (1/cosA + sinA/cosA (1-sinA)

= (1+sinA)/cosA (1-sinA)

= [(1)^{2} – (sinA)^{2}]/cosA [We know that (A-b) (a+b) = a^{2} – b^{2}]

= 1 – sin^{2}A/cosA

= cos^{2}A/cosA [(Cos^{2}A = 1-sin^{2}A]

= cosA

** **

**(iv) 1+tan ^{2} A/1 + cot^{2} A = **

**(A) Sec ^{2}A **

**(B) -1 **

**(C) cot ^{2} A **

**(D) tan ^{2} A **

**Solution: **

1 + tan^{2}A/1+cot^{2}A

** **

**(5) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.**

**(i) (cosec****θ – cotθ) ^{2} = 1 – cosθ/1+cosθ **

**Solution:**

L.H.S = (cosecθ – cotθ)^{2}

= (1/sinθ – cosθ/sinθ)^{2}

= (1-cosθ/sinθ)^{2}

= (1-cosθ)^{2}/sin^{2}θ

= (1-cosθ)^{2}/(1-cos^{2}θ [We know that sin^{2}θ + cos^{2}θ = 1 sin^{2}θ = 1 – cos^{2}θ]

= (1 – cosθ)^{2}/(1 – cosθ) (1 + cosθ)

= – 1 – cosθ/1+cosθ =- R.H.S (proved)

** **

**(ii) cosA/1+sinA + 1+sinA/cosA = 2sec A **

**Solution: **

L.H.S = cosA/1+sinA + 1+sinA/cosA

= cos^{2}A + (1 + sinA)^{2}/cosA (1 + sinA)

= cos^{2}A + 1 + 2sinA + sin^{2}A/cosA (1 + sinA)

= 1+1+2 sinA/cosA (1 + sinA) [sin^{2}A + cos^{2}A = 1]

= 2/cosA

= 2sec A = R.H.S = (proved)

**(iii) tanθ/1-cotθ + cotθ/1-tanθ = 1 + secθ cosecθ **

**Solution: **

L.H.S = tanθ/1-cotθ + cotθ/1-tanθ

= cosecθ.secθ + 1

= 1 + secθ.cosecθ

= R.H.S (Proved)

**(iv) 1 + secA/secA = sin ^{2}A/1-cosA **

**Solution: **

= (1 + cosA)

= (1 + cosA) × (1-cosA)/(1-cosA)

= 1 – cos^{2}A/1 – cosA

= sin^{2}A/1 – cos A = R.H.S = (Proved)

(As sin^{2}A = 1 – cos^{2}A)

**(v) cosA – sinA + 1/cosA + sinA – 1 = cosecA + cot A using the identity cosec ^{2} A = 1 + cot^{2}A. **

**Solution: **

L.H.S = cosA – sinA + 1/cosA + sinA – 1

**(vi) ****√1+sinA/1-sinA = secA + tanA **

**Solution: **

L.H.S = √1+sinA/1-sinA

= 1+sinA/cosA

= 1/cosA + sinA/cosA

= secA + tanA = (R.H.S) (Proved)

** **

**(vii) sin****θ-2sin ^{3}θ/2cos^{3}θ – cosθ = tanθ **

**Solution: **

L.H.S = sinθ-2sin^{3}θ/2cos^{3}θ – cosθ

**(viii) (SinA + cosecA) ^{2} + (cosA + secA)^{2} = 7 + tan^{2}A + cot^{2}A **

**Solution: **

L.H.S = (SinA + cosecA) + (cosA + secA)^{2}

= (sin^{2}A + 2sinA.cosecA + cosec^{2}A) + (cos^{2}A + 2cosA secA + sec^{2}A)

= (sin^{2}A + 2sinA.1/sinA + cosec^{2}A + cos^{2}A + 2.cosA. 1/cosA + sec^{2}A)

= (sin^{2}A + cos^{2}A) + 2 + 2 + cosec^{2}A + sec^{2}A

= 1 + 4 + cosec^{2}A + sec^{2}A

= 5 + 1 + cot^{2}A + 1 + tan^{2}A [As cosec^{2} – cot^{2}A = 1 sec^{2}A – tan^{2}A = 1]

= 7 + cot^{2}A + tan^{2}A

= R.H.S (Proved)

** **

**(ix) (cosecA – sinA) (secA – cosA) = 1/tanA + cotA **

**Solution: **

L.H.S = (cosecA – sinA) (secA – cosA)

= (1/sinA – sinA) (1/cosA – cosA)

= (1 – sin^{2}A/sinA) × (1 – cos^{2}A/cosA) [As sin^{2}A + cos^{2}A = 1]

= cos^{2}A/sinA × sin^{2}A/cosA

= cosA.sinA

R.H.S = 1/tanA + cotA

** **

**(x) (1+tan ^{2}A/1+cot^{2}A) = (1-tanA/1-cotA)^{2} = tan^{2}A **

**Solution: **

(1+tan^{2}A/1+cot^{2}A) [As cosec^{2}A – cot^{2}A = 1 sec^{2}A – tan^{2}A = 1]

= tan^{2}A

1+tan^{2}A/1+cot^{2}A = (1-tanA/1-cotA)^{2} = tan^{2 }(Proved)

**In case you are missed :- NCERT Solution for Triangles**