Class 10 - Maths - Introduction to Trigonometry

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

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.

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.

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.

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.

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

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

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.

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

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.

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

Question 11:

(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.

(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)

(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

(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.

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:

(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.

(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

(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

(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.

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.

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.

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

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.

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.

(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

(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:

1. 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

(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

(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