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Chap 8 : Introduction to Trigonometry

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CHAPTER 8

Introduction to Trigonometry

1. OBJECTIVE QUESTIONS

1. If x = p sec q and y = q tan q, then

(a) x2 - y2 = p2q2

(c)

x2q2 -

y2p2

=

1 p2q2

(b) x2q2 - y2p2 = pq (d) x2q2 - y2p2 = p2q2

Ans : (d) x2q2 - y2p2 = p2q2

We know,

sec2q - tan2q = 1

and

sec q

=

x p

tan q

=

y q

x2q2 - y2p2 = p2q2

2.

If b tan q = a ,

the value

of

a sin q - b cos q a sin q + b cos q

is

(a)

a-b a2 + b2

(b)

a+b a2 + b2

(c)

a2 + b2 a2 - b2

Ans

:

(d)

a2 - b2 a2 + b2

(d)

a2 - b2 a2 + b2

tan q

=

a b

a sin q - b cos q a sin q + b cos q

=

a a

tan tan

q q

+

b b

=

a2 a2

+

b2 b2

3. The value of tan 1ctan 2c tan 3c... tan 89c is

(a) 0

(b) 1

(c) 3

(d) None of these

Ans : (b) 1

Given, tan 1c tan 2c tan 3c... tan 89c = tan (90c - 89c) tan (90c - 88c)

tan^90c - 87ch... tan 87c tan 88c tan 89c = cot 89c cot 88c cot 87c... tan 87c

tan 88c tan 89c = ^cot 89c tan 89ch^cot 88c tan 88ch

^cot 87c tan 87ch...^cot 44c tan 44chtan 45c = 1 # 1 # 1...1 # 1 = 1

4. (cos4A - sin4A) is equal to

(a) 1 - 2 cos2A (c) sin2A - cos2A

(b) 2 sin2A - 1 (d) 2 cos2A - 1

Ans : (d) 2 cos2A - 1 (cos4A - sin4A) = (cos2A)2 - (sin2A)2 = (cos2A - sin2A) (cos2A + sin2A)

= (cos2A - sin2A) (1) = cos2A - (1 - cos2A) = 2 cos2A - 1

5. If sec 5A = cosec^A + 30ch, where 5A is an acute

angle, then the value of A is

(a) 15c

(b) 5c

(c) 20c

(d) 10c

Ans : (d) 10c We have, sec 5A = cosec^A + 30ch

sec 5A = sec690c - ^A - 30ch@ 6sec^90c - qh = cosec q@

sec 5A = sec^60c - Ah 5A = 60c - A 6A = 60c A = 10c

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6. If x sin3q + y cos3q = sin q cos q and x sin q = y cos q , than x2 + y2 is equal to

(a) 0

(b) 1/2

(c) 1

(d) 3/2

Ans : (c) 1

We have, x sin3q + y cos3q = sin q cos q

^x sin qhsin2q + ^y cos qhcos2q = sin q cos q

x sin q^sin2qh + ^x sin qhcos2q = sin q cos q

x sin q^sin2q + cos2qh = sin q cos q

x sin q = sin q cos q & x = cos q

Now,

x sin q = y cos q

cos q sin q = y cos q

y = sin q

Hence,

x2 + y2 = cos2q + sin2q = 1

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Chap 8 : Introduction to Trigonometry

7. If tan 2A = cot^A - 18ch, where 2A is an acute angle,

then the value of A is

(a) 12c

(b) 18c

(c) 36c

(d) 48c

Ans : (c) 36c

Given,

tan 2A = cot^A - 18ch

cot^90c - 2Ah = cot^A - 18ch

90c - 2A = A - 18c

[since, ^90c - 2Ah and ^A - 18ch

both are acute angles]

90c + 18c = A + 2A

3A = 108c

A

=

108c 3

= 36c

8. If tan q + sin q = m and tan q - sin q = n , then m2 - n2 is equal to

(a) mn (c) 4 mn

(b)

m n

(d) None of these

Ans : (c) 4 mn

Given, tan q + sin q = m and tan q - sin q = n

m2 - n2 = ^tan q + sin qh2 - ^tan q - sin qh2

= 4 tan q sin q

= 4 tan2q sin2q

= 4

sin2

q

sin2 q cos2 q

=4

sin2 q cos2 q

-

sin2 q

= 4 tan2q - sin2q

= 4 ^tan q + sin qh^tan q - sin qh

= 4 mn

9.

If is

0

<

q

<

p 4

,

then

the

simplest

form

of

1 - 2 sin q cos q

(a) sin q - cos q

(b) cos q - sin q

(c) cos q + sin q

(d) sin q cos q

Ans : (b) cos q - sin q

1 - 2 sin q cos q = sin2q + cos2q - 2 sin q cos q

= ^cos q - sin qh2

8a2 + b2 - 2ab = ^a - bh2 and 1 = sin2q + cos2qB

For 0c < q < 45c

= cos q - sin q

0

p/6

p/4

cos q

1

3 /2

1/ 2

sin q

0

1/2

1/ 2

Here, we see that cos q > sin q , when 0 < q < p/4, that's why we take ^cos q - sin qh2 instead of taking ^sin q - cos qh2.

10. If ^sec A + tan Ah^sec B + tan Bh^sec C + tan C h= ^sec A - tan Ah ^sec B - tan Bh^sec C - tan C h = x then the value/

.in

values of x is/are (a) ! 1

(c) ! 2

(b) 0 (d) 1

Ans : (a) !1

We have,^sec A + tan Ah^sec B + tan Bh^sec C + tan C h = ^sec A - tan Ah ^sec B - tan Bh^sec C - tan C h

On multiplying both sides by ^sec A - tan Ah^sec B - tan Bh^sec C - tan C h, we get ^sec A + tan Ah^sec B + tan Bh^sec C + tan C h

#^sec A - tan Ah^sec B - tan Bh^sec C - tan C h = ^sec A - tan Ah2 ^sec B - tan Bh2 ^sec C - tan C h2

^sec2A - tan2Ah^sec2B - tan2Bh^sec2C - tan2C h = ^sec A - tan Ah2 ^sec B - tan Bh2 ^sec C - tan C h2

6^a + bh^a - bh = a2 - b2@

1 # 1 # 1 = 7^sec A - tan Ah^sec B - tan Bh^sec C - tan C hA2 6sec2q - tan2q = 1@

^sec A - tan Ah^sec B - tan Bh^sec C - tan C h =! 1

11. If sin q + sin2q = 1, then find the value of cos12q + 3 cos10q + 3 cos8q + cos6q + 2 cos4q + 2 cos2q - 2.

(a) 0

(b) 1

(c) cos q

(d) sin q

Ans : (b) 1

We have,

sin q + sin2q = 1

sin q = 1 - sin2q

sin q = cos2q

6sin2q + cos2q = 1@

cos12q + 3 cos10q + 3 cos8q + cos6q +2 cos4q + 2 cos2q - 2

= ^cos12q + 3 cos10q + 3 cos8q + cos6qh

+ 2^cos4q + cos2q - 1h

= ^cos4qh3 + 3 cos6q^cos4q + cos2qh

+^cos2qh3 + 2^cos4q + cos2q - 1h

= ^cos4q + cos2qh3 + 2^cos4q + cos2q - 1h 8^a + bh3 = a3 + b3 + 3ab^a + bhB

= ^sin2q + cos2qh3 + 2^sin2q + cos2q - 1h

6cos2q = sin q & cos4q = sin2q@

= ^1h3 + 2^1 - 1h = 1

6sin2q + cos2q = 1@

12. If 0c < x < 90c and 2 sin x + 15 cos2x = 7, then find

the value of tan x .

(a) 4/5

(b) 3/5

(c) 3/4

(d) 4/3

Ans : (d) 4/3

Given,

2 sin x + 15 cos2x = 7

2 sin x + 15^1 - sin2x h = 7 6sin2x + cos2x = 1@

2 sin x + 15 - 15 sin2x = 7

15 sin2x - 2 sin x - 8 = 0

Let,

y = sin x , then

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Chap 8 : Introduction to Trigonometry

15y2 - 2y - 8 = 0 15y2 - 12y + 10y - 8 = 0

[by splitting the middle term] 3y^5y - 4h + 2^5y - 4h = 0

^5y - 4h^3y + 2h = 0

5y - 4 = 0

and

3y + 2 = 0 & y = 4/5

and

y = - 2/3

sin x

=

4 5

and

sin x

=

-

2 3

[put y = sin x ]

But, Now,

0c < x < 90c sin x = 4/5 [for 0c < x < 90c, sin x is not negative] cos x = 1 - sin2x

=

1

-

b

4 5

2

l

=

1

-

16 25

=

9 25

=

3 5

tan x

=

sin x cos x

=

4/5 3/5

=

4 3

13. If f^x h = cos2x + sec2x , then f^x h

(a) $ 1 (c) $ 2

(b) # 1 (d) # 2

Ans : (c) $ 2 Given, f^x h = cos2x + sec2x

= cos2x + sec2x - 2 + 2

[adding and subtracting 2] = cos2x + sec2x - 2 cos x $ sec x + 2

6cos x $ sec x = 1@ = ^cos x - sec x h2 + 2

8a2 + b2 - 2ab = ^a - bh2B We know that, square of any expression is always greater than equal to zero.

f^x h $ 2

Hence proved.

14. If ABC is a right angled triangle, then find the

relation between

tan

b

A

-

B 2

-

C

land

-

tan

b

A

+

B 2

-

C

l

(a) equal

(b) unequal

(c) sum of these equal to 1 (d) None of the above

Ans : (b) unequal

Given, ABC is a right angled triangle. Since, the sum of the angles of a triangle is 180c.

A + B + C = 180c

Now,

tan

b

A

-

B 2

-

C

l

=

tan ;A

-

^180c 2

-

AhE

[from Eq. (1)]

= tanb 2A -2180cl

cbse.online

= tan^A - 90ch

= tan6-^90c - Ah@

=- tan^90c - Ah 6tan^-qh =- tan q@

= - cot A

6tan^90c - Ah = cot A@ ...(2)

and

-

tan

b

A

+

B 2

-

C

l

=

-

tan

;^180c

2

C

hC

E

[from Eq. (1)]

=- tan^90c - C h

= cot C

...(3)

From. Eq. (2) and Eq. (3), we get

tanb A

-

B 2

-

C

l

!-

tanb A

+

B 2

-

C

l

15. If sin q + sin2q + sin3q = 1, then cos6q - 4 cos4q + 8 cos2q

is equal to

(a) 1

(b) 2

(c) 3

(d) 4

Ans : (d) 4 Given, sin q + sin2q + sin3q = 1

sin q + sin3q = 1 - sin2q sin q^1 + sin2qh = cos2q sin q^1 + 1 - cos2qh = cos2q

6cos2q + sin2q = 1@ sin q^2 - cos2qh = cos2q On squaring both sides, we get,

sin2q^2 - cos2qh2 = cos4q ^1 - cos2qh^4 + cos4q - 4 cos2qh = cos4q 4 + cos4q - 4 cos2q - 4 cos2q - cos6q + 4 cos4q

= cos4q 4 = cos6q - 4 cos4q - cos4q + 4 cos2q + 4 cos2q + cos4q 4 = cos6q - 4 cos4q + 8 cos2q

16. If m = a cos3q + 3a cos q sin2q and n = a sin3q + 3a cos2q sin q , then ^m + nh2/3 + ^m - nh2/3 is equal to

(a) 2a2/3 (c) 2a3/2

(b) a2/3 (d) a3/2

Ans : (a) 2a2/3

Given,

m = a cos3q + 3a cos q sin2q ...(1)

and

n = a sin3q + 3a cos2q sin q ...(2)

On adding Eqs. (1) and (2), we get

m + n = a cos3q + 3a cos q sin2q + a sin3q + 3a cos2q sin q

= a6cos3q + sin3q + 3 cos q sin q^sin q + cos qh@

= a^cos q + sin qh3

(3)

8^a + bh3 = a3 + b3 + 3ab^a + bhB

On subtracting Eq. (2) from Eq. (1), we get

m - n = a cos3q + 3a cos q sin2q - a sin3q - 3a cos2q sin q

= a8cos3q - sin3q - 3 cos q sin q^cos q - sin qhB

= a6cos q - sin q@3

...(4)

8^a - bh3 = a3 - b3 - 3ab^a - bhB

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Chap 8 : Introduction to Trigonometry

Now, ^m + nh2/3 + ^m - nh2/3 = 8a^cos q + sin qh3B2/3 +8a^cos q - sin qh3B2/3

[from Eq. (3) and (4)]

= a2/3 ^cos q + sin qh2 + a2/3 ^cos q - sin qh2

= a2/3 8^cos q + sin qh2 + ^cos q - sin qh2B

= a2/3 6cos2q + sin2q + 2 cos q sin q + cos2q

+ sin2q - 2 sin q cos q@

8^a + bh2 = a2 + b2 + 2ab, ^a - bh2 = a2 + b2 - 2abB

= a2/3 61 + 1@

6cos2q + sin2q = 1@

= 2a2/3

17.

If a b

tan q =

=

1

a sin f - a cos f

and

tan

f

=

1

b sin q - b cos

q

,

then

(a)

sin q 1 - cos q

(b)

sin q 1 - cos f

(c)

sin f sin q

(d)

sin q sin f

Ans : (d) We have,

sin q sin f

tan q

=

a sin f 1 - a cos f

cot q

=

a

1 sin

f

-

cot

f

cot q + cot f

=

1 a sin f

tan f

=

b sin q 1 - b cos q

...(1)

cot f

=

b

1 sin

q

- cot q

cot f + cot q

=

b

1 sin

q

...(2)

From (1) and (2), we have

1 a sin f

=

b

1 sin

q

a b

=

sin q sin f

18. If a sec q + b tan q + c = 0 and p sec q + q tan q + r = 0, then ^br - qch2 - ^pc - ar h2 is equal to

(a) ^ap - bqh2 (c) ^ap - bqh

(b) ^aq - bph2 (d) ^aq - bph

Ans : (b) ^aq - bph2

We have, a sec q + b tan q + c = 0

and

p sec q + q tan q + r = 0

Solving these two equations for sec q and tan q by the

cross-multiplication method, we get

sec q br - qc

=

tan cp -

q ar

=

aq

1 -

bp

sec q

=

br aq

-

cq bp

and

tan q

=

cp aq

-

ar bp

Now,

sec2q - tan2q = 1

d

br aq

-

cq bp

2

n

-

e

cp aq

-

ar bp

2

o

= 1

.in ^br - cqh2 - ^cp - ar h2 = ^aq - bph2

2. FILL IN THE BLANK

1. sin 60c cos 30c + sin 30c cos 60c = .......... Ans : 1

2. sin2q + sin2(90c - q) = .......... Ans : 1 [Hint : sin2(90c - q)] = cos2q]

3. 2 tan245c + 3 cos230c - sin260c = ..........

Ans

:

7 2

4. Triangle in which we study trigonometric ratios is called ..........

Ans : Right Triangle

5.

cos 45c sec 30c + cosec 30c

=

..........

Ans : 3 (

3 - 1) 4

6.

sin 18c cos 72c

=

..........

Ans : 1

7. cos 48c - sin 42c = .......... Ans : 0

8. Cosine of 90c is .......... Ans : Zero

9. If 15 cot A = 8, sec A = .......... Ans : 17/8

10. The value of sin A or cos A never exceeds .......... Ans : 1

11. sine of ^90 - qh is .......... Ans : cos q

12. sin2A + cos2A = .......... Ans : 1

13. It tan A = 4/3 then sin A .......... Ans : 4/5

14. In a right triangle ABC , right angled at B , if tan A = 1, sin A cos A = ..........

Ans

:

1 2

15. Reciprocal of sin q is .......... Ans : cosec q

16. InTABC , right-angled at B , AB = 24 cm, BC = 7 cm . sin A = .......... Ans : 7/25

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Chap 8 : Introduction to Trigonometry

17. Maximum value for sine of any angle is .......... Ans : 1

18. In TPQR , right-angled at Q , PR + QR = 25 cm and PQ = 5 cm. The value of tan P is .......... Ans : 12/5

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15. The value of the expression ^sin 80c - cos 80ch is negative Ans : False

16. tan 48c tan 23c tan 42c tan 67c ! 1 Ans : False

19. Sum of .......... of sine and cosine of angle is one. Ans : Square

3 . T RU E/FALSE

1. The value of sin q increases as q increases. Ans : True

2. ^1 - cos2qhsec2q = tan q Ans : True

3.

sec A

=

12 5

for

some

value

of

angel

A.

Ans : True

4. sin (A + B) = sin A + sin B . Ans : False

5. The value of cos q increases as q increases. Ans : False

6.

sin q =

5 3

for

some

angle

q .

Ans : False

7. The value of tan A is always less than 1. Ans : False

8. The value of the expression ^cos223c - sin267ch is positive. Ans : False

9. cot A is not defined for A = 0c. Ans : True

10. sin^90c - Ah = cos A Ans : True

11. If +B and +Q are acute angles such that sin B = sin Q , then +B ! +Q . Ans : False

12. ^tan q + 2h^2 tan q + 1h = 5 tan q + sec2q Ans : False

13.

tan tan

65c 25c

=

1

Ans : False

14. sin q = cos q for all values of q. Ans : False

17. Trigonometry deals with measurement of components of triangles. Ans : True

18. sin2q # cos2q = 1 Ans : False

19. The value of tan A is always less than 1. Ans : False

20.

tan cot

47c 43c

=

1

Ans : True

21. sec A = 12/5 for some value of angle A. Ans : True

22. If cos A + cos2A = 1, then sin2A + sin4A = 1. Ans : True

23. cos A is the abbreviation used for the cosecant of angle A.

Ans : False

24. cot A is the product of cot and A. Ans : False

25. The value of sin q + cos q is always greater than 1. Ans : False

26.

sin

q

=

4 3

for some angle q .

Ans : False

4 . M ATCH I N G QU EST I ON S

DIRECTION : Each question contains statements given in two Columns which have to be matched. Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II.

1. In TABC , +B = 90c, AB = 3 cm and BC = 4 cm then match the column.

Column-I

Column-II

(A) sin C

(p) 3/5

(B) cos C

(q) 4/5

(C) tan A

(r) 5/3

(D) sec A

(s) 4/3

Ans : (A) - p, (B) - q, (C) - s, (D) - r

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