(free sample)...171 iit foundation & olympiad explorer 8. trigonometry-ii mathematics class - x...

CLASS - X

IIT Foundation

& O

lympia

d E

xplorer - M

ath

ematics C

lass - X

FOUNDATION OLYMPIAD&

Integrated Syllabus

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UNIQUE ATTRACTIONSl Concept Maps

l Cross word Puzzles

l Graded Exercise

n Basic Practice

n Further Practice

n Brain Works

l Multiple Answer Questions

l Paragraph Questions

l Concept Drill

l Simple, clear and systematic presentation

l Concept maps provided for every chapter

l Set of objective and subjective questions at the end of each chapter

l Previous contest questions at the end of each chapter

l Designed to fulfill the preparation needs for international/national talent exams, olympiads and all competitive exams

India’s FIRST scientifically designed portal for Olympiad preparation� Olympiad & Talent Exams preparation packages Analysis Reports Previous question papers � � Free Demo Packages Free Android Mobile App� �

YOUR

COACH

A unique opportunity to take about 50 tests per subject.

School Edition

` 250

ISBN 978-93-82058-44-1

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MATHEMATICSCLASS - X

FOUNDATION & OLYMPIAD

School Edition

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Published by:

Brain Mapping Academy#16–11–16/1/B, First Floor,Farhat Hospital Road,Saleem Nagar, Malakpet,Hyderabad–500 036Andhra Pradesh, India.✆ 040–65165169, 66135169E–mail: [email protected]: www.bmatalent.com

© Brain Mapping Academy

ALL RIGHTS RESERVEDNo part of this book may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher.

Publication Team

Chief Mentor: Srinivas Kalluri

Author: Y.S. Srinivasu

Design & Typing: P. S. Chakravarthy & M.S.M. Lakshmi

ISBN: 978-93-82058-44-1

Disclaimer

Every care has been taken by the compilers andpublishers to give correct, complete and updated information. In case there is any omission, printing mistake or anyother error which might have crept in inadvertently,neither the compiler / publisher nor any of thedistributors take any legal responsibility.

In case of any dispute, all matters are subject to the exclusive jurisdiction of the courts in Hyderabad only.

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Preface

Speed and accuracy play an important role in climbing the competitive ladder and reaching

it. Students have to integrate the habit of calculating quickly as well as that of functioning

efficiently in order to thrive in the learning culture. They need to think on their feet by

understanding the basic requirements, identifing appropriate information sources and using

that to their advantage.

The preparation required for the tough competitive examinations is fundamentally different

from that of qualifying ones like the board examinations. A student can emerge successful in

a qualifying examination by merely scoring the minimum percentage of marks, whereas in a

competitive examination, he has to score the highest and perform better than the others.

This book provides all types of questions that a student would be required to tackle at the

foundation level. The questions in the exercises are graded as Basic Practice, Further Practice,

Brainworks, Multiple Answer Questions, Concept drill and Paragraph Questions. Simple

questions involving a direct application of the concepts are given in Basic Practice. More

challenging questions on direct application are given in Brain Works. Questions involving

higher order thinking or an open-ended approach to problems are given in Concept drill.

These questions encourage students to think analytically, to be creative and to come up with

solutions of their own. Constant practice and familiarity with these questions will not only

make him/her conceptually sound, but also give the student the confidence to face any

entrance examination with ease.

Valuable suggestions as well as criticism from the teacher and student community are most

welcome and will be incorporated in the ensuing edition.

Publisher

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1. Number System ......................................... 7

2. Polynomials - III ......................................... 34

3. Quadratic Equations - II ............................. 50

Crossword - I ............................................... 69

4. Progressions .............................................. 70

5. Mensuration - III ......................................... 92

6. Co-ordinate Geometry - III ........................ 118

Crossword - II ............................................. 145

7. Plane Geometry - III ................................... 146

8. Trigonometry - II ........................................ 170

9. Probability - II ............................................ 194

Crossword - III ............................................ 213

10. Functions ................................................... 214

11. Limits ......................................................... 239

12. Binomial Theorem ..................................... 251

Crossword - IV ............................................ 264

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IIT Foundation & Olympiad Explorer 8. Trigonometry-II

Mathematics170 Class - X

SYNOPSIS

Trigonometric Identities

Angle

An angle is the amount of rotation of a revolving line with respect to a fixed line.

Note: If the rotation is in clockwise sense, the angle measured is negative and it ispositive if the rotation is in anti�clockwise sense.

O

Q

+ve angle

X

−ve angle

O

Q

X

Different Units for Measuring Angles

I. Sexagesimal system (or) British system

In sexagesimal system, a right angle is divided into 90 equal parts called degrees. Further,each degree is divided into sixty equal parts called minutes and each minute is dividedinto sixty equal parts called seconds.

Thus, 1 right angle = 90 degrees (900)

10 = 60 minutes (60|)

1| = 60 seconds (60||)

II. Centesimal system or french system

1 right angle is divided into 100 equal parts. Each part is called a grade.

Thus, 1 right angle = 100 grades (100g)

1 grade = 100 minutes (100|)

1 minute = 100 seconds (100||)

Force and PressureForce and Pressure1Mathematical Induction

Chapter

TRIGONOMETRY - II

Chapter

8

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Mathematics Class - X

III. Radian Measure

An angle made by an arc of length equal to radius of a given circle at its centre is calledone radian.

Relation between degree and radian

If D is the degree measure of an angle and R is its measure in radians then

D

90=

2R

ð

∴ 1 Radian =180°

π degrees = 570 17|45|| (approximately)

and 1 degree = ð

180Radian = 0.01746 Radian (approximately)]

Trigonometric Ratio’s

I. In a right angled triangle

ON = x, NP = y and

OP = r, PON = PNO = 90°∠ θ ∠,

opposite side NP ysin = = =

hypotenuse OP rθ

Hypotenuse OP rcosec = = =

Opposite side NP yθ

Adjacent side ON xcos = = =

Hypotenuse OP rθ

Hypotenuse OP rsec = = =

Adjacent side ON xθ

Opposite side NP ytan = = =

Adjacent side ON xθ Adjacent side ON x

cot = = =Oppositeside NP y

θ

II. sin θ . cosec θ = 1 sin = coseccosec sin

⇒ θ ⇒ θ =θ θ

1 1

III. cos sec = 1 cos secsec cos

θ θ ⇒ θ = ⇒ θ =θ θ

1 1.

Degrees 300 450 600 900 1800 2700 3600

Radians 6

π4

π3

π2

ππ

3

2

π 2 π

O

Hyp

otenuse

P

r

y

Opposite

side

Adjacent sidex

θX

Y

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IV. tan θ . cot θ = 1 ⇒ tan θ = 1 1

cot =cot tan

⇒ θθ θ

V. tan θ = sin

cos

θθ ⇒ cot θ =

cos

sin

θθ

For All Values of θI. sin2 θ +cos2 θ = 1 ⇒ 1 � sin2 θ = cos2 θ ⇒ 1 � cos2 θ = sin2 θ

II. sec2 θ � tan2 θ = 1 ⇒ 1+ tan2 θ = sec2 θ ⇒ sec2 θ � 1 = tan2 θ

III. cosec2 θ � cot2 θ = 1 ⇒ 1+ cot2 θ = cosec2 θ ⇒ cosec2 θ � 1 = cot2 θ

Signs of Trigonometric Functions

Table below shows the proper sign (+) the trigonometric functions of angles in each ofthe four quadrants of a complete cycle.

Quadrant – I Quadrant–II Quadrant–III Quadrant–IV

All silver tea cups

all +ve sin +ve tan +ve cos +ve

Trigonometric Ratios of Standard Angles

Angle 00 300=6

π450=

4

π600=

3

π900=

2

π

sin 012 2

1 3

21

cos 13

2

1

2

1

20

tan 03

11 3 ∞

cot ∞ 3 13

10

sec 12

3 2 2 ∞

cosec ∞ 2 22

31

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Trigonometric Ratios of Allied Angles

Allied angles : Two angles are said to be allied when their sum or difference is a multipleof 90°.

Thus θ and � θ ; θ and 900� θ ; θ and 900 + θ ; θ and 1800� θ ; θ and 1800 + θ ; θ and 2700 + θ ;

θ and 3600 � θ ; θ and 3600 + θ are called Allied angles, it is assumed that θ is measured

in degrees.

I. sin(� θ ) = �sin θ , cos(� θ )= cos θ , tan(� θ ) = �tan θcot(� θ ) = �cot θ , sec(� θ ) = sec θ , cosec(� θ ) = �cosec θ

e.g., sin(�600) = �sin600 = 3

2

−; cos(�600) = cos 600 =

1

2

II. Complementary Angles

Two angles are said to be complementary when their sum is 900.

Thus θ and 900 � θ are complementary angles.

e.g., cos300 = cos (900 � 600) = sin600 = 3

2

III. Supplementary Angles

Two angles are said to be supplementary when their sum is 1800.

Thus, θ and 1800 � θ are supplementary angles.

e.g., sin 1200 = sin(1800 � 600) = sin 600 = 3

2

Note 1: If the allied angle is 90 ± θ , 270 ± θ , then trigonometric ratios changed as

sin ↔ cos, tan ↔ cot, sec ↔ cosec

Note 2: If the allied angle is � θ , 180 ± θ , 360 ± θ , then trigonometric ratio is not

changed.

Note 3: If the allied angle is in the form of (n. 360 ± θ ) where�n� is an integer, then

trigonometric ratio is not changed.

e.g., sin7500 = sin (2.3600 + 300) = sin300 = 1

2

Compound Angle

A compound angle is that which is made up of algebraic sum of two or more angles.

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Addition and Subtraction of two Angles

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

(ii) sin (A � B) = sin A . cos B � cos A . sin B

(iii) cos (A + B) = cos A . cos B � sin A . sin B

(iv) cos (A � B) = cos A . cos B + sin A . sin B

(v) tan (A + B) = tanA + tanB

1 tanA × tanB−

(vi) tan (A � B) = tanA tanB

1 + tanA × tanB

−

(vii) cot (A + B) = cotA × cotB 1

cotA + cotB

−

(viii) cot (A � B) = cotA × cotB + 1

cotB cotA−

(ix) sin (A + B) sin (A � B) = sin2A � sin2B = cos2B � cos2A

(x) cos (A + B) cos (A � B) = cos2A � sin2B = cos2B � sin2A

(xi) tan (A + B) tan (A � B) = 2 2

2 2

tan A tan B

1 - tan A × tan B

−

(xii) cot (A + B) cot (A � B) = 2 2

2 2

cot A × cot B 1

cot B cot A

−−

Some Important Results

(i) tan1 + tanA cosA + sinA

+ A = =4 1 tanA cosA sinA

π − −

(ii) tan1 tanA cosA sinA

A = =4 1 + tanA cosA + sinA

π − − −

Trigonometric Ratios of Multiple and Submultiple Angles

If A is an angle, then the angles 2A, 3A, 4A etc., are called multiple angles of A. And the

angles A A A

, ,2 3 4

etc., are called submultiple angles of A.

1. sin 2A = 2sin A . cos A = 2

2tanA

1 + tan A

2. cos 2A = cos2A � sin2A = 1 � 2sin2A = 2cos2A � 1 = 2

2

1 tan A

1 + tan A

−

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3. tan 2A = 2

2tanA

1 tan A−

4. cot 2A = 2cot A 1

2cotA

−

5. sin 3A = 3 sinA � 4 sin3A

6. cos 3A = 4cos3A � 3cosA

7. tan 3A = 3

2

3tanA tan A

1 3tan A

−−

8. cot 3A = 3 3

2 2

cot A 3cotA 3cotA cot A=

3cot A 1 1 3cot A

− −− −

Heights and Distances

Angle of Elevation

XOM∠ i.e., the angle in which the line joining the object and the eye makes with the

horizontal through the eye is called the angle of elevation of M as seen from O.

O

M

X

Angle of Depression

XOP∠ i.e., the angle in which the line joining the object and the eye makes with the

horizontal through the eye is called the angle of depression of P as seen from O.

O

Q

X

(or)

The angle between the horizontal line drawn through the observer�s eye and the linejoining the eye to any object is called, the angle of elevation of than object when it is athigher level than the eye.

The angle of depression of the object when it is at a lower level than the eye.

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SOLVED EXAMPLES

Example 1:

If cos sin 2 cosθ + θ = θ , show that cos sin 2 sinθ − θ = θ .

Solution:

cos sin 2 cosθ + θ = θ

( )sin 2 1 cos⇒ θ = − θ

1cos sin

2 1⇒ θ = θ

−

( )sin 2 1 sin cos sin 2sin⇒ θ = + θ ⇒ θ − θ = θ

Example 2:

(i) Prove that 2 2A A 1

sin sin sin A8 2 8 2 2

π π + − − = .

Solution:

We have,

L.H.S.2 2A A

sin sin8 2 8 2

π π = + − −

A A A A

sin sin8 2 8 2 8 2 8 2

π π π π = + + − + − −

sin sin A4

π=

1

sin A R.H.S.2

= =

(ii) Find the value of cos2 45° � sin2 15°.

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(ii) If x y

sin cos 1a b

θ + θ = and x y

cos sin 1a b

θ − θ = prove that 2 2

2 2

x y2

a b+ =

Solution:

Example 5:

(i) If A + B = 45°, prove that (cot A � 1) (cot B � 1) = 2 and hence deduce that o1

cot 22 2 1.2

= +

Solution:

A + B = 45° ⇒ cot (A + B) = cot 450 cot A cot B 1

1cot A cot B

−⇒ =+

⇒ cot A cot B � 1= cot A + cot B ⇒ cot A cot B � cot A � cot B = 1

⇒ cot A . cot B . cot A � cot B + 1 = 1 + 1 ⇒ cot A (cot B � 1) � 1 (cot B � 1) = 2

⇒ (cot A � 1) (cot B �1) = 2

Let A = B = 01

222

. Then A + B = 450 and hence (cot A � 1) (cot B � 1) = 2

0 0 001 1 1 1cot22 1 cot22 1 2 cot22 1 2 cot22 2 1

2 2 2 2

⇒ − − = ⇒ − = ⇒ = +

(ii) If tan (A � B) = 7

24 and tan A =

4

3 where A and B are acute, show that A + B =

2

π

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Example 6:

(i) Prove that cot cosec 1 1 cos

cosec cotcot cosec 1 sin

θ + θ − + θ= θ + θ =θ − θ + θ

Solution:

LHS =

2 2cot cosec 1 (cot cosec ) (cosec cot )

cot cosec 1 cot cosec 1

θ + θ − θ + θ − θ − θ=

θ − θ + θ − θ +

(cot cosec ) (cosec cot ) (cosec cot )

cot cosec 1

θ + θ − θ + θ θ − θ=

θ − θ +

(cot cosec ) (1 cosec cot )

(cot cosec 1)

θ + θ − θ + θ=

θ − θ +

cos 1 1 coscot cosec R.H.S.

sin sin sin

θ + θ= θ + θ = + = =θ θ θ

(ii) Prove that 1 1 1 1

sec A tan A cos A cos A sec A tan A− = −

− + .

Solution:

Example 7:

(i) If ( ) ( )sin cosθ + α = θ + α then express tan θ in terms of α .

Solution:

( ) ( ) ( ) tan tansin cos tan 1 1

1 tan tan

θ + αθ + α = θ + α ⇒ θ + α = ⇒ =− θ α

tan tan 1 tan tan tan tan tan 1 tan⇒ θ + α = − θ α ⇒ θ + θ α = − α

( )tan 1 tan 1 tan⇒ θ + α = − α

1 tan

tan1 tan

− αθ =+ α

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(ii) Find the value of ( ) ( )tan / 4 .tan / 4 .π + θ π − θ

Solution:

Example 8:

A man on the deck of a ship is 16m above water level. He observes that the angle ofelevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculatethe distance of the cliff from the ship and the height of the cliff.

Solution:

Let the man be at M, 16 m above water level WB. AB = h m is the cliff.

x

N

h-16

h

A

BW

M45°

30°16 m

Let WB = x m be the distance of the ship from the cliff. MN is the horizontal levelthrough M.

AMN 45∠ = ° and NMB 30∠ = °

∴ MBW 30∠ = °

Also MN = WB = xm

Now, MW

tan30WB

= ° ⇒16 1

x 3=

⇒ x 16 3 16 1.732 27.712= = × =

andAN

tan 45MN

= ° ⇒h 16 h 16

1 1x 16 3

− −= ⇒ =

⇒ h 16 16 3− = ⇒ h 16 3 16 16( 3 1)= + = +

⇒ h 16(1.732 1) 16 2.732 43.712.= + = × =

Hence, the distance of the cliff from the ship = 27.712 mand the height of the cliff 43.712 m.

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Example 10:

The angle of elevation of a cliff from a fixed point is θ . After going up a distance of

k metres towards the top of the cliff at an angle of φ , it is found that the angle of

elevation is α . Show that the height of the cliff is

k(cos sin cot )metres.

cot cot

φ − φ αθ − α

Solution:

MP = h be the height of the top of the cliff.

MAP∠ = θ is the angle of elevation of the top P are observed from the fixed point A.

PCD∠ = α where AC = k and CAB∠ = φDraw CD MP⊥Here, CD = BM

and BC = MD

From right angled ABC∆ , we have

BCsin

AC= φ and

ABcos

AC= φ

⇒BC

sink

= φ andAB

cosk

= φ

⇒ BC k sin= φ and AB = k cosφ ...... (1)

From right angled AMP,∆MP

tanAM

= θ ⇒h

tanAM

= θ

⇒h

AMtan

=θ ⇒ AM hcot= θ ...... (2)

From right angled CDP,∆DP

tanCD

= α ⇒MP MD MP BC

tanBM AM AB

− −α = =−

⇒h ksin

tanh cot k cos

− φα =θ − φ (from (1) and (2))

⇒1 h ksin

cot hcot k cos

− φ=α θ − φ

⇒ hcot k cos cot (h ksin )θ − φ = α × − φ

⇒ h(cot cot ) k(cos cot sin )θ − α = φ − α φ

⇒ k(cos cot sin )h

(cot cot )

φ − α φ=θ − α

⇒ k(cos sin cot )

h(cot cot )

φ − φ α=θ − α

M

(Foot of

the cliff)

(Top of the

cliff)

Aφ

θD

α

B

C

P

h

k

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CONCEPT MAP

ysin =

rθ

rcosec =

yθ

x

cos =r

θ rsec =

xθ

y

tan =x

θx

cot =y

θ

Trigonometry: The word trigonometry means�three angles measurement�.

Measurement of angles

Trigonometric

Ratio�s:

Relation between degree and radian:

∴ 1 Radian = 180

π degrees

= 57° 17|45|| (approximately) and

1 degree = 180

π

Radian = 0.01746 Radian (approximately)

Centesimal system or french system:

Thus, 1 right angle = 100 grades (100g)

1 grade = 100 minutes (100|)

1 minute = 100 seconds (100||)

Sexagesimal system (or) British system:

Thus, 1 right angle = 90 degrees (90°)

1° = 60 minutes (60|); 1| = 60 seconds (60||)

Reciprocal relations:

sin θ . cosec θ = 1

cos .sec 1θ θ =

tan θ . cot θ = 1

Quotient relations:

tan θ = sin

cos

θθ cot θ =

cos

sin

θθ

Identities:

sin2 θ + cos2 θ = 1

sec2 θ � tan2 θ = 1

cosec2 θ � cot2 θ = 1

Addition and subtraction of two angles:

(i) sin (A ± B) = sin A . cos B ± cos A . sin B

(ii) cos (A ± B) = cos A . cos B m sin A . sin B

(iii) tan (A ± B) = tan A tanB

1 tan A .tanB

±m

(iv) cot (A ± B) = cot A . cotB 1

cot A cotB±m

(v) sin (A + B) sin (A � B)= sin2A � sin2B = cos2B � cos2A

(vi) cos (A + B) cos (A � B)= cos2A � sin2B = cos2B � sin2A

(vii) tan (A + B) tan (A � B)

=

2 2

2 2

tan A tan B

1 tan A . tan B

−−

1. sin 2A = 2sin A . cos A

= 2

2 tan A

1 tan A+

2. cos 2A = cos2A � sin2A = 1�2sin2A = 2cos2A � 1

= 2

2

1 tan A

1 tan A

−+

3. tan 2A = 2

2 tan A

1 tan A−4. sin 3A

= 3 sinA � 4 sin3A5. cos 3A

= 4cos3A � 3cosA

Trigonometric ratios of

multiple and submultiple angles:

sin A = 2 sin A/2 . cos A/ 2

2

2 tan A / 2

1 tan A / 2=

+cos A = cos2A/2 � sin2A/2= 1�2sin2A/2 = 2cos2A /2� 1

=

2

2

1 tan A / 2

1 tan A / 2

−

+

tan A = 2

2 tan A / 2

1 tan A / 2−

tan 3A = 3

2

3 tan A tan A

1 3 tan A

−−

Hyp

oten

use

O

Y

R

Y

Opposite side

X

Adjacent side

Xθ

r

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BASIC PRACTICE

1. In a circle of diameter 40 cm the length of a cord is 20 cm. Find the length of minor arccorresponding to the cord.

2. Find the angle between the minute hand of a clock and the hour hand when the timeis 7 : 20 a.m.

Solution:

3. If ax by

cos sin+

θ θ = a2 � b2 and 2 2

ax sin by cos

cos sin

θ θ−

θ θ = 0 prove that (ax)2/3 + (by)2/3 = (a2 � b2)2/3.

4. Given that (1 + cos α ) (1 + cos β ) (1 + cos γ ) = (1 � cos α ) (1 � cos β ) (1 � cos γ ),

Show that one of the values of each member of his equality is sin α sin β sin γ .

5. If sec 2θ = and 3

22

π < θ < π , find the value of 1 tan cosec

1 cot cosec

+ θ + θ+ θ − θ .

6. In any quadrilateral ABCD, prove that

(i) sin (A + B) + sin (C + D) = 0 (ii) cos (A + B) = cos (C + D)

7. If 4 12 3

cos A ,cosB , A,B 25 13 2

π= = < < π , find the values of the following.

(i) cos (A + B) (ii) sin (A � B)

8. Prove that sin2 A = cos2 (A � B) + cos2 B � 2 cos (A � B) cos A cos B.

Solution:

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FURTHER PRACTICE

1. For an acute angle θ , sin θ + cos θ takes the greatest value when θ is

(A) 30o (B) 45o (C) 60o (D) 90o

2. If sin θ + cos θ = a and sec θ + cosec θ = b, find the value of b (a2 � 1).

(A) 2a (B) 3a (C) 0 (D) 2ab

3. If 7 cosec θ � 3 cot θ = 7, find the value of 7 cot θ � 3 cosec θ .

(A) 5 (B) 3 (C)7

3 (D)3

7

4. If a sec θ + b tan θ = 1 and a2 sec2 θ � b2tan2 θ = 5, find a2b2 + 4a2.

(A) 9b2 (B) 2

9

a (C)−2

b(D) 9

5. If θ lies in the first quadrant and 5 tan θ = 4, find 5sin 3cos

sin 2cos

θ − θθ + θ

(A)5

14(B)

3

14(C)

1

14(D) 0

6. If sin θ + sin2 θ = 1, what is the value of cos2 θ + cos4 θ ?

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

7. Find tan 75° � cot 75°.

(A) 2 3 (B) 3 2 (C) 3 (D) 1

8. Find the value of sin 120° cos 150° � cos 240° sin 330°.

(A) 1 (B) � 1 (C)2

3(D)

3 1

4

+− 9. If A, B, C, D are the angles of a cyclic qauadrilateral, find cos A + cos B + cos C + cos D.

(A) 4 (B) 1 (C) 0 (D) � 1

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10. If tan θ =4

3

−, find the value of sin θ .

(A)4

5

− but not

4

5(B)

4

5

− or

4

5(C)

4

5 but not

4

5

−(D)

2

5

11. If 1

tan7

θ = and θ is an acute angle, find 2 2

2 2

cosec sec

cosec sec

θ − θθ + θ

.

(A)3

4(B)

1

2(C) 2 (D)

5

4

Solution:

12. A 25 m ladder is placed against a vertical wall of a building. The foot of the ladder is 7mfrom the base of the building. If the top of the ladder slips 4m, when the foot of the ladderwill slide?

(A) 5m (B) 8 m (C) 9m (D) 15m

13. A tree 6m tall casts a 4m long shadow. At the same time, a flag pole casts a shadow50m long. How long is the flag pole?

(A) 75 m (B) 100 m (C) 150 m (D) 50 m

14. Find the angle of elevation of the sum when the length of the shadow of a pole is 3times the height of the pole.

(A) 30° (B) 45° (C) 60° (D) 75°

15. The distance between the tops of two trees 20 m and 28 m high is 17m. What is thehorizontal distance between the two tree?

(A) 9 m (B) 11 m (C) 15 m (D) 31 m

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16. A tree breaks due to storm and the broken part bends to that the top of the tree firsttouches the ground, making an angle of 30 with the horizontal. The distance from thefoot of the tree to the point where the top touches the ground is 10m. Find the heightof the tree.

(A) +10( 3 1)m (B) 10 3m (C) −10( 3 1)m (D)10

m3

17. One side of a parallelogram is 12 cm and its area is 60 cm2. If the angle between theadjacent side is 30°, find its other side.

(A) 10 cm (B) 8 cm (C) 6 cm (D) 4 cm

18. A vertical pole breaks due to storm and the broken part bends, so that the top of thepole touches the ground, making an angle of 30° with ground. The distance from thefoot of the pole to the point where the top touches the ground is 10m. Find the heightof the pole.

(A) 10 +( 3 1)m (B) 10 3m (C) ( )−10 3 1 m (D)10

m3

Solution:

BRAIN WORKS

1. If 0 / 4< θ < π , show that ( )2 2 1 cos 4 2cos+ + θ = θ .

2. If cosec sin m and sec cos nθ − θ = θ − θ = , prove that (m2n)2/3 + (mn2)2/3 = 1

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MULTIPLE ANSWER QUESTIONS

1.

22 (x y)

sin4xy

+θ = is possible only, when

(A) x > 0, y > 0 and x y≠ (B) x > 0, y > 0 and x = y

(C) x < 0, y < 0 and x = y (D) x > 0, y < 0 and x y≠

2. (m + 2) sin θ + (2m � 1) cos θ = 2m + 1 is true if

(A)3

tan4

θ = (B)4

tan3

θ = (C) 2

2mtan

m 1θ =

+ (D) 2

2mtan

m 1θ =

−

3. If 3 5

cos and cos ,5 13

α = β = which of the following is true?

(A)33

cos( )65

α + β = (B)56

sin( )65

α + β =

(C)2 1

sin2 65

α − β = (D)

63cos( )

65α − β =

4. Find 3 cos76 cot16

cot76 cot16

+ ° °° + ° .

(A) tan16° (B) cot76° (C) tan 46° (D) cot44°

5. If cot tan x and sec cos yθ + θ = θ − θ = , which is the true statement?

(A)1

sin cosx

θ θ = (B) sin tan yθ θ =

(C) 2 2 / 3 2 2 / 3(x y) (xy ) 1− = (D) 2 1 / 3 2 1 / 3(x y) (xy ) 1= =

6. Which of the following have the same value?

(A) sin2 θ + cos2 θ (B) sin2 θ cos2 θ (C) cosec2 θ � cot2 θ (D) sec2 θ � tan2 θ

7. Which of the following ratios are equal?

(A) sin 30° (B) cos 30° (C) sin 60° (D) cos 90°

8. Given tan θ = 1, which of the following is equal to tan θ ?

(A) sin 0° (B) sin 90° (C) cot 45° (D) cos 0°

9. Which of the following are equivalent to sec θ ?

(A)1

cos θ (B) cosθ (C) 2tan 1θ − (D) 21 tan+ θ

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191



10. Find the trigonometric ratios equivalent to cosec 45o.

(A) 2 (B) sec 45° (C)1

sin 45° (D) 21 cot 45+ °

11. Find the expressions that are not equivalent to 1.

(A) sin3 φ + cos3 φ (B) sin2 θ + cos2 φ (C) cosec2 φ � cot2 φ (D)2

2

1 cot A

cosec A

+

12. Given 3

cos ,2

θ = which of the following are not the possible values of sin 2θ ?

(A)3

2(B)

1

2− (C) 1 (D) 0

13. Which of the following values are possible?

(A) sin θ = 0 (B) sec θ = 1

2(C) tan θ = 1 (D) cosec θ = 3

14. Which of the following statements are not true?

(A) For θ = 45°, cot θ and tan θ are equal.

(B) For θ = 90°, cos θ and cot θ are not defined.

(C) For θ = 60°, sec θ is 4 cos θ .

(D) For θ = 0°, sin θ and tan θ are not defined.

15. Identify the equivalent expressions.

(A) cos (A � B) (B) sin (A + B)

(C) cos A cos B + sin A sin B (D) sin A cos B � cos A sin B

PARAGRAPH QUESTIONS

� If D, G and C are respectively the measures of an angle in degrees, grades and radians,

then D G C

180 200= =

π.

� By using given equations involving � θ � and trigonometrical identities, we shall obtain

an equation not involving � θ �. This is called method of eliminating � θ �.

(i) The angles of a triangle are in AP and the number of degrees in the least is to thenumber of radians in the greatest as 60 : π . Find the angles in degrees.

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3. (i) x = a sin θ � b cos θy = a cos θ + b sin θ

(A) x2 + y2 = a2 + b2 (B) x2 + y2 = a2 (C) x2y2 = a2 + b2 (D) x2 � y2 = a2 + b2

(ii) If p = sin θ + cos θ and q = sec θ + cosec θ then determine the value of q(p2 � 1).

(A) 2p � 1 (B)p

2(C)

12p

2+ (D) 2p

PREVIOUS CONTEST QUESTIONS

1. Prove that sin2A = cos2(A � B) + cos2 B � 2 cos(A � B) cosA cosB.

Solution:

2. If sec θ = 2 and 3

22

π < θ < π , find the value of 1 tan cosec

.1 cot cosec

+ θ + θ+ θ − θ

3. If sin α + cosec α = 2 determine the value of sinn α + cosecn α .

4. Prove that

2 2 2 2cos8 2cos .+ + + θ = θ

5. Evaluate 2 2

2 2

tan A sec A.

cos B cot B−

6. If tan ( π cos θ ) = cot( π sin θ ), prove that cos1

.4 2 2

π θ − = ±

7. If a cos θ � b sin θ = c, prove that a sin θ + b cos θ = 2 2 2a b c+ − .

8. If sin θ + cosθ = a and secθ + cosecθ = b, show that b(a2 � 1) = 2a.

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CLASS - X

IIT Foundation

& O

lympia

d E

xplorer - M

ath

ematics C

lass - X

FOUNDATION OLYMPIAD&

Integrated Syllabus

www.bmatalent.com

UNIQUE ATTRACTIONSl Concept Maps

l Cross word Puzzles

l Graded Exercise

n Basic Practice

n Further Practice

n Brain Works

l Multiple Answer Questions

l Paragraph Questions

l Concept Drill

l Simple, clear and systematic presentation

l Concept maps provided for every chapter

l Set of objective and subjective questions at the end of each chapter

l Previous contest questions at the end of each chapter

l Designed to fulfill the preparation needs for international/national talent exams, olympiads and all competitive exams

India’s FIRST scientifically designed portal for Olympiad preparation� Olympiad & Talent Exams preparation packages Analysis Reports Previous question papers � � Free Demo Packages Free Android Mobile App� �

YOUR

COACH

A unique opportunity to take about 50 tests per subject.

School Edition

` 250

ISBN 978-93-82058-44-1

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