FOUNDATION & OLYMPIAD

Transcription

1 Concept maps provided for every chapter Set of objective and subjective questions at the end of each chapter Previous contest questions at the end of each chapter Designed to fufi the preparation needs for internationa/nationa taent exams, oympiads and a competitive exams UNIQUE ATTRACTIONS Concept Maps Cross word Puzzes Graded Exercise n Basic Practice n Further Practice n Brain Works Mutipe Answer Questions Paragraph Questions Concept Dri w w. (F bm re a e ta Sa e n m t. p co e) m Simpe, cear and systematic presentation IIT Foundation & Oympiad Exporer - Cass - X Integrated Syabus FOUNDATION & OLYMPIAD CLASS - X Schoo Edition ` 50 w ISBN YOUR COACH India s FIRST scientificay designed porta for Oympiad preparation Oympiad & Taent Exams preparation packages Anaysis Reports Previous question papers Free Demo Packages Free Android Mobie App A unique opportunity to take about 50 tests per subject.

2 FOUNDATION & OLYMPIAD MATHEMATICS CLASS - X Schoo Edition

3 Pubished by: Brain Mapping Academy #6 6//B, First Foor, Farhat Hospita Road, Saeem Nagar, Maakpet, Hyderabad Andhra Pradesh, India , E mai: info@bmataent.com Website: Brain Mapping Academy ALL RIGHTS RESERVED No part of this book may be reproduced, stored in a retrieva system or transmitted in any form or by any means, eectronic, mechanica, photocopying, recording or otherwise without the prior written permission of the pubisher. Pubication Team Chief Mentor: Srinivas Kauri Author: Y.S. Srinivasu Design & Typing: P. S. Chakravarthy & M.S.M. Lakshmi ISBN: Discaimer Every care has been taken by the compiers and pubishers to give correct, compete and updated information. In case there is any omission, printing mistake or any other error which might have crept in inadvertenty, neither the compier / pubisher nor any of the distributors take any ega responsibiity. In case of any dispute, a matters are subject to the excusive jurisdiction of the courts in Hyderabad ony.

4 Preface Speed and accuracy pay an important roe in cimbing the competitive adder and reaching it. Students have to integrate the habit of cacuating quicky as we as that of functioning efficienty in order to thrive in the earning cuture. 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 fundamentay different from that of quaifying ones ike the board examinations. A student can emerge successfu in a quaifying examination by merey 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 a types of questions that a student woud be required to tacke at the foundation eve. The questions in the exercises are graded as Basic Practice, Further Practice, Brainworks, Mutipe Answer Questions, Concept dri and Paragraph Questions. Simpe questions invoving a direct appication of the concepts are given in Basic Practice. More chaenging questions on direct appication are given in Brain Works. Questions invoving higher order thinking or an open-ended approach to probems are given in Concept dri. These questions encourage students to think anayticay, to be creative and to come up with soutions of their own. Constant practice and famiiarity with these questions wi not ony make him/her conceptuay sound, but aso give the student the confidence to face any entrance examination with ease. Vauabe suggestions as we as criticism from the teacher and student community are most wecome and wi be incorporated in the ensuing edition. Pubisher

5 . Number System Poynomias - III Quadratic Equations - II Crossword - I Progressions Mensuration - III Co-ordinate Geometry - III... 8 Crossword - II Pane Geometry - III Trigonometry - II Probabiity - II Crossword - III Functions Limits Binomia Theorem... 5 Crossword - IV... 64

6 IIT Foundation & Oympiad Exporer SYNOPSIS Trigonometric Identities Ange An ange is the amount of rotation of a revoving ine with respect to a fixed ine. Note: If the rotation is in cockwise sense, the ange measured is negative and it is positive if the rotation is in anti cockwise sense. Different Units for Measuring Anges O Q +ve ange I. Sexagesima system (or) British system II. Chapter 8 X O ve ange In sexagesima system, a right ange is divided into 90 equa parts caed degrees. Further, each degree is divided into sixty equa parts caed minutes and each minute is divided into sixty equa parts caed seconds. Thus, right ange = 90 degrees (90 0 ) 0 = 60 minutes (60 ) = 60 seconds (60 ) Centesima system or french system right ange is divided into 00 equa parts. Each part is caed a grade. Thus, right ange = 00 grades (00 g ) grade = 00 minutes (00 ) minute = 00 seconds (00 ) Mathematica TRIGONOMETRY - Force and Pressure Induction II Q X 70 Cass - X

7 IIT Foundation & Oympiad Exporer III. Radian Measure An ange made by an arc of ength equa to radius of a given circe at its centre is caed one radian. Reation between degree and radian If D is the degree measure of an ange and R is its measure in radians then D 90 = R ð Radian = 80 degrees = (approximatey) ð and degree = Radian = Radian (approximatey)] 80 Trigonometric Ratio s I. In a right anged triange ON = x, NP = y and Degrees Radians 6 OP = r, PON = θ, PNO = 90 opposite side NP y sin θ = = = hypotenuse OP r Adjacent side ON x cos θ = = = Hypotenuse OP r Opposite side NP y tan θ = = = Adjacent side ON x θ O x Adjacent side Hypotenuse OP r cosec θ = = = Opposite side NP y Hypotenuse OP r sec θ = = = Adjacent side ON x Adjacent side ON x cot θ = = = Opposite side NP y II. sin θ. cosec θ = sin θ = cosec θ = cosec θ sinθ III. cos θ. sec θ = cos θ = secθ = secθ cosθ 4 r Hypotenuse P y Opposite side Y X Cass - X 7

8 IIT Foundation & Oympiad Exporer IV. tan θ. cot θ = tan θ = cot θ = cotθ tanθ V. tan θ = sin θ cosθ cot θ = cosθ sinθ For A Vaues of θ I. sin θ +cos θ = sin θ = cos θ cos θ = sin θ II. III. sec θ tan θ = + tan θ = sec θ sec θ = tan θ cosec θ cot θ = + cot θ = cosec θ cosec θ = cot θ Signs of Trigonometric Functions Tabe beow shows the proper sign (+) the trigonometric functions of anges in each of the four quadrants of a compete cyce. Quadrant I Quadrant II Quadrant III Quadrant IV A siver tea cups a +ve sin +ve tan +ve cos +ve Trigonometric Ratios of Standard Anges Ange = 6 sin 0 cos tan = = 90 0 = cot 0 0 sec cosec 7 Cass - X

9 IIT Foundation & Oympiad Exporer Trigonometric Ratios of Aied Anges Aied anges : Two anges are said to be aied when their sum or difference is a mutipe of 90. Thus θ and θ ; θ and 90 0 θ ; θ and θ ; θ and 80 0 θ ; θ and θ ; θ and θ ; θ and 60 0 θ ; θ and θ are caed Aied anges, it is assumed that θ is measured in degrees. I. sin( θ ) = sin θ, cos( θ )= cos θ, tan( θ ) = tan θ cot( θ ) = cot θ, sec( θ ) = sec θ, cosec( θ ) = cosec θ II. III. e.g., sin( 60 0 ) = sin60 0 = Compementary Anges ; cos( 60 0 ) = cos 60 0 = Two anges are said to be compementary when their sum is Thus θ and 90 0 θ are compementary anges. e.g., cos0 0 = cos ( ) = sin60 0 = Suppementary Anges Two anges are said to be suppementary when their sum is Thus, θ and 80 0 θ are suppementary anges. e.g., sin 0 0 = sin( ) = sin 60 0 = Note : Note : Note : If the aied ange is 90 ± θ, 70 ± θ, then trigonometric ratios changed as sin cos, tan cot, sec cosec If the aied ange is θ, 80 ± θ, 60 ± θ, then trigonometric ratio is not changed. If the aied ange is in the form of (n. 60 ± θ ) where n is an integer, then trigonometric ratio is not changed. e.g., sin750 0 = sin ( ) = sin0 0 = Compound Ange A compound ange is that which is made up of agebraic sum of two or more anges. Cass - X 7

10 IIT Foundation & Oympiad Exporer Addition and Subtraction of two Anges (i) (ii) (iii) (iv) sin (A + B) = sin A. cos B + cos A. sin B sin (A B) = sin A. cos B cos A. sin B cos (A + B) = cos A. cos B sin A. sin B cos (A B) = cos A. cos B + sin A. sin B (v) tan (A + B) = (vi) tan (A B) = (vii) cot (A + B) = (viii) (ix) (x) tana + tanb tana tanb tana tanb +tana tanb cota cotb cota + cotb cot (A B) = cota cotb+ cotb cota sin (A + B) sin (A B) = sin A sin B = cos B cos A cos (A + B) cos (A B) = cos A sin B = cos B sin A (xi) tan (A + B) tan (A B) = (xii) cot (A + B) cot (A B) = Some Important Resuts (i) (ii) +tana cosa+sina tan +A = = 4 tana cosa sina tana cosa sina tan A = = 4 +tana cosa +sina tan A tan B -tan A tan B cot A cot B cot B cot A Trigonometric Ratios of Mutipe and Submutipe Anges If A is an ange, then the anges A, A, 4A etc., are caed mutipe anges of A. And the anges A, A, A etc., are caed submutipe anges of A. 4 tana. sin A = sin A. cos A = +tan A. cos A = cos A sin A = sin A = cos A = tan A +tan A 74 Cass - X

11 IIT Foundation & Oympiad Exporer tana. tan A = tan A 4. cot A = cot A cota 5. sin A = sina 4 sin A 6. cos A = 4cos A cosa 7. tan A = 8. cot A = tana Heights and Distances Ange of Eevation tan A tan A cot A cota cota cot A = cot A cot A XOM i.e., the ange in which the ine joining the object and the eye makes with the horizonta through the eye is caed the ange of eevation of M as seen from O. Ange of Depression O XOP i.e., the ange in which the ine joining the object and the eye makes with the horizonta through the eye is caed the ange of depression of P as seen from O. O M Q X X (or) The ange between the horizonta ine drawn through the observer s eye and the ine joining the eye to any object is caed, the ange of eevation of than object when it is at higher eve than the eye. The ange of depression of the object when it is at a ower eve than the eye. Cass - X 75

12 IIT Foundation & Oympiad Exporer Soution: SOLVED EXAMPLES Exampe : If cos θ+ sin θ= cosθ, show that cosθ sinθ= sin θ. cosθ+ sin θ= cosθ ( ) sin θ = cosθ cosθ = sin θ ( ) sin θ = + sin θ cos θ sin θ = sin θ Exampe : A A (i) Prove that + = 8 8 Soution: We have, A A L.H.S. = sin + sin 8 8 sin sin sin A A A A A sin sin = = sin sin A 4 = sin A = R.H.S. (ii) Find the vaue of cos 45 sin 5. Soution:. 76 Cass - X

13 IIT Foundation & Oympiad Exporer (ii) If x sin θ+ y cosθ= and x cosθ y sin θ= prove that a b a b x a y + = b Soution: Exampe 5: (i) If A + B = 45, prove that (cot A ) (cot B ) = and hence deduce that cot. = + Soution: A + B = 45 cot (A + B) = cot 45 0 cot A cot B = cot A + cot B cot A cot B = cot A + cot B cot A cot B cot A cot B = cot A. cot B. cot A cot B + = + cot A (cot B ) (cot B ) = (cot A ) (cot B ) = Let A = B = 0. Then A + B = 45 0 and hence (cot A ) (cot B ) = cot cot = cot = cot = + (ii) If tan (A B) = 7 4 and tan A = 4 where A and B are acute, show that A + B = Soution: o 78 Cass - X

14 IIT Foundation & Oympiad Exporer Exampe 6: (i) Prove that Soution: (ii) LHS = Prove that cot θ+ cosec θ cos = cosecθ+ cot θ = + θ cot θ cosecθ+ sin θ cot cosec (cot cosec ) (cosec cot ) θ+ θ θ+ θ θ θ = cot θ cosec θ+ cot θ cosec θ+ (cot θ+ cosec θ) (cosec θ+ cot θ) (cosec θ cot θ) = cot θ cosec θ+ (cotθ+ cosec θ) ( cosec θ+ cot θ) = (cotθ cosec θ+ ) cos θ + cos θ = cot θ+ cosec θ = + = = R.H.S. sinθ sin θ sin θ Soution: Exampe 7: (i) If sin( ) cos( ) =. sec A tan A cos A cos A sec A + tan A θ+α = θ+α then express tan θ in terms of α. Soution: tan θ+ tan α sin( θ+α ) = cos( θ+α) tan( θ+α ) = = tanθtanα tanθ+ tan α = tanθtanα tan θ+ tanθtanα = tan α tanθ ( + tanα ) = tanα tanα tan θ= + tan α Cass - X 79

15 IIT Foundation & Oympiad Exporer (ii) Find the vaue of tan ( / 4 +θ).tan( / 4 θ). Soution: Soution: Exampe 8: A man on the deck of a ship is 6m above water eve. He observes that the ange of eevation of the top of a ciff is 45 and the ange of depression of the base is 0. Cacuate the distance of the ciff from the ship and the height of the ciff. Let the man be at M, 6 m above water eve WB. AB = h m is the ciff. M 6 m 45 0 W x B Let WB = x m be the distance of the ship from the ciff. MN is the horizonta eve through M. AMN = 45 and NMB = 0 A N h-6 MBW = 0 Aso MN = WB = xm Now, MW tan0 WB = 6 = x x = 6 = 6.7 = 7.7 and AN tan 45 MN = h 6 h 6 = = x 6 h 6 = 6 h = = 6( + ) h = 6(.7 + ) = 6.7 = 4.7. Hence, the distance of the ciff from the ship = 7.7 m and the height of the ciff 4.7 m. h 80 Cass - X

16 IIT Foundation & Oympiad Exporer Soution: Exampe 0: The ange of eevation of a ciff from a fixed point is θ. After going up a distance of k metres towards the top of the ciff at an ange of φ, it is found that the ange of eevation is α. Show that the height of the ciff is k(cosφ sinφcot α) metres. cotθ cotα MP = h be the height of the top of the ciff. MAP = θ is the ange of eevation of the top P are observed from the fixed point A. PCD = α where AC = k and CAB = φ (Top of the P ciff) Draw CD MP Here, CD = BM and BC = MD From right anged ABC, we have h BC sin AC = φ AB and cos AC = φ BC α sin k = φ AB C and cos k = φ k D θ φ A M BC = k sin φ and AB = kcosφ... () B (Foot of From right anged AMP, the ciff) MP tan AM = θ h tan AM = θ h AM = tan θ AM = h cot θ... () From right anged CDP, DP tan CD = α MP MD MP BC tan α= = BM AM AB h ksinφ tan α= h cotθ kcosφ (from () and ()) h ksinφ = cot α hcot θ k cosφ h cotθ kcosφ= cot α (h ksin φ) h(cot θ cot α ) = k(cosφ cotαsin φ) k(cosφ cotαsin φ) h = (cotθ cot α) k(cosφ sinφcot α) h = (cotθ cot α) 8 Cass - X

17 IIT Foundation & Oympiad Exporer CONCEPT MAP Measurement of anges Sexagesima system (or) British system: Thus, right ange = 90 degrees (90 ) = 60 minutes (60 ); = 60 seconds (60 ) Centesima system or french system: Thus, right ange = 00 grades (00 g ) grade = 00 minutes (00 ) minute = 00 seconds (00 ) Reation between degree and radian: Radian = 80 degrees = (approximatey) and degree = 80 Radian = Radian (approximatey) Addition and subtraction of two anges: (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) = Trigonometry: The word trigonometry means three anges measurement. tan A ± tan B m tan A.tanB cot A. cot B m (iv) cot (A ± B) = cot A ± cotb (v) sin (A + B) sin (A B) = sin A sin B = cos B cos A (vi) cos (A + B) cos (A B) = cos A sin B = cos B sin A (vii) tan (A + B) tan (A B) tan A tan B = tan A. tan B Trigonometric Ratio s: y sin θ = r x cos θ = r y tan θ = x Reciproca reations: sin θ. cosec θ = cos θ.sec θ= tan θ. cot θ = Quotient reations: tan θ = sin θ cos θ. sin A = sin A. cos A tana = + tan A. cos A = cos A sin A = sin A = cos A tan A = + tan A tana. tan A = tan A 4. sin A = sina 4 sin A 5. cos A = 4cos A cosa r cosec θ = y r sec θ = x x cot θ = y Trigonometric ratios of mutipe and submutipe anges: Y O rr Hypotenuse Identities: sin θ + cos θ = sec θ tan θ = cosec θ cot θ = cot θ = cos θ sin θ sin A = sin A/. cos A/ tana/ = + tan A/ cos A = cos A/ sin A/ = sin A/ = cos A / tan A / = + tan A / tana/ tan A = tan A/ tana tan A tan A = tan A Y X Adjacent side Opposite side θ X Cass - X 8

18 IIT Foundation & Oympiad Exporer BASIC PRACTICE. In a circe of diameter 40 cm the ength of a cord is 0 cm. Find the ength of minor arc corresponding to the cord.. Find the ange between the minute hand of a cock and the hour hand when the time is 7 : 0 a.m. Soution: ax by. If + cos θ sin θ = a b ax sin θ by cos θ and = 0 prove that (ax )/ + (by) / = (a b ) /. cos θ sin θ 4. Given that ( + cos α ) ( + cos β ) ( + cos γ ) = ( cos α ) ( cos β ) ( cos γ ), Show that one of the vaues of each member of his equaity is sin α sin β sin γ. 5. If sec θ= and + tanθ+ cosecθ <θ<, find the vaue of + cotθ cos ecθ. 6. In any quadriatera ABCD, prove that (i) sin (A + B) + sin (C + D) = 0 (ii) cos (A + B) = cos (C + D) 4 7. If cos A =,cos B =, < A,B <, find the vaues of the foowing. 5 (i) cos (A + B) (ii) sin (A B) 8. Prove that sin A = cos (A B) + cos B cos (A B) cos A cos B. Soution: 84 Cass - X

19 IIT Foundation & Oympiad Exporer FURTHER PRACTICE. For an acute ange θ, sin θ + cos θ takes the greatest vaue when θ is (A) 0 o (B) 45 o (C) 60 o (D) 90 o. If sin θ + cos θ = a and sec θ + cosec θ = b, find the vaue of b (a ). (A) a (B) a (C) 0 (D) ab. If 7 cosec θ cot θ = 7, find the vaue of 7 cot θ cosec θ. (A) 5 (B) (C) 4. If a sec θ + b tan θ = and a sec θ b tan θ = 5, find a b + 4a. 9 (A) 9b (B) a (C) 5. If θ ies in the first quadrant and 5 tan θ = 4, find (A) 5 4 (B) 4 (C) 7 b 4 6. If sin θ + sin θ =, what is the vaue of cos θ + cos 4 θ? 5sinθ cosθ sinθ+ cosθ (D) 7 (D) 9 (D) 0 (A) 0 (B) (C) (D) 7. Find tan 75 cot 75. (A) (B) (C) (D) 8. Find the vaue of sin 0 cos 50 cos 40 sin 0. + (A) (B) (C) (D) 4 9. If A, B, C, D are the anges of a cycic qauadriatera, find cos A + cos B + cos C + cos D. (A) 4 (B) (C) 0 (D) Soution: 86 Cass - X

20 IIT Foundation & Oympiad Exporer 0. If tan θ = 4, find the vaue of sin θ. (A) 4 5 but not 4 5 (B) 4 5 or 4 5 cos ec. If tan θ= and θ is an acute ange, find 7 cos ec (A) (B) (C) 4 Soution: (C) but not 5 θ sec θ θ+ sec θ.. A 5 m adder is paced against a vertica wa of a buiding. The foot of the adder is 7m from the base of the buiding. If the top of the adder sips 4m, when the foot of the adder wi side? (A) 5m (B) 8 m (C) 9m (D) 5m. A tree 6m ta casts a 4m ong shadow. At the same time, a fag poe casts a shadow 50m ong. How ong is the fag poe? (A) 75 m (B) 00 m (C) 50 m (D) 50 m 4. Find the ange of eevation of the sum when the ength of the shadow of a poe is times the height of the poe. (A) 0 (B) 45 (C) 60 (D) The distance between the tops of two trees 0 m and 8 m high is 7m. What is the horizonta distance between the two tree? (A) 9 m (B) m (C) 5 m (D) m Soution: (D) (D) Cass - X 87

21 IIT Foundation & Oympiad Exporer 6. A tree breaks due to storm and the broken part bends to that the top of the tree first touches the ground, making an ange of 0 with the horizonta. The distance from the foot of the tree to the point where the top touches the ground is 0m. Find the height of the tree. 0 (A) 0( + )m (B) 0 m (C) 0( )m (D) m 7. One side of a paraeogram is cm and its area is 60 cm. If the ange between the adjacent side is 0, find its other side. (A) 0 cm (B) 8 cm (C) 6 cm (D) 4 cm 8. A vertica poe breaks due to storm and the broken part bends, so that the top of the poe touches the ground, making an ange of 0 with ground. The distance from the foot of the poe to the point where the top touches the ground is 0m. Find the height of the poe. 0 (A) 0 ( + )m (B) 0 m (C) 0( ) m (D) m Soution: BRAIN WORKS. If 0 /4 + + cos 4θ = cosθ. <θ<, show that ( ). If cosecθ sin θ= m and sec θ cosθ= n, prove that (m n) / + (mn ) / = Soution: 88 Cass - X

22 IIT Foundation & Oympiad Exporer MULTIPLE ANSWER QUESTIONS (x + y). sin θ= is possibe ony, when 4xy (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. (m + ) sin θ + (m ) cos θ = m + is true if 4 m m (A) tan θ= (B) tan θ= (C) tanθ= (D) tanθ= 4 m + m. If (A) (C) 4. Find 5 cosα= and cos β=, which of the foowing is true? 5 cos( α+β ) = 65 sin α β = 65 + cos76 cot6 cot 76 + cot6. (B) (D) 56 sin( α+β ) = 65 6 cos( α β ) = 65 (A) tan6 (B) cot76 (C) tan 46 (D) cot If cot θ+ tan θ= x and sec θ cos θ= y, which is the true statement? (A) (C) sin θcosθ = (B) sinθ tan θ = y x / / (x y) (xy ) = (D) 6. Which of the foowing have the same vaue? / / (x y) (xy ) = = (A) sin θ + cos θ (B) sin θ cos θ (C) cosec θ cot θ (D) sec θ tan θ 7. Which of the foowing ratios are equa? (A) sin 0 (B) cos 0 (C) sin 60 (D) cos Given tan θ =, which of the foowing is equa to tan θ? (A) sin 0 (B) sin 90 (C) cot 45 (D) cos 0 9. Which of the foowing are equivaent to sec θ? (A) cos θ (B) cos θ (C) tan θ (D) + tan θ 90 Cass - X

23 IIT Foundation & Oympiad Exporer 0. Find the trigonometric ratios equivaent to cosec 45 o. (A) (B) sec 45 (C). Find the expressions that are not equivaent to.. Given sin 45 (A) sin φ + cos φ (B) sin θ + cos φ (C) cosec φ cot φ (D) (A) (D) + cot 45 + cot A cosec A cos θ=, which of the foowing are not the possibe vaues of sin θ? (B). Which of the foowing vaues are possibe? (A) sin θ = 0 (B) sec θ = 4. Which of the foowing statements are not true? (A) For θ = 45, cot θ and tan θ are equa. (C) (D) 0 (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. 5. Identify the equivaent expressions. (C) tan θ = (D) cosec θ = (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 respectivey the measures of an ange in degrees, grades and radians, D G C then = = By using given equations invoving θ and trigonometrica identities, we sha obtain an equation not invoving θ. This is caed method of eiminating θ. (i) The anges of a triange are in AP and the number of degrees in the east is to the number of radians in the greatest as 60 :. Find the anges in degrees. Cass - X 9

24 IIT Foundation & Oympiad Exporer. (i) x = a sin θ b cos θ y = a cos θ + b sin θ (A) x + y = a + b (B) x + y = a (C) x y = a + b (D) x y = a + b (ii) If p = sin θ + cos θ and q = sec θ + cosec θ then determine the vaue of q(p ). (A) p (B) p PREVIOUS CONTEST QUESTIONS (C) p + (D) p. Prove that sin A = cos (A B) + cos B cos(a B) cosa cosb. Soution:. If sec θ = and + tanθ+ cosec θ <θ<, find the vaue of. + cot θ cosec θ. If sin α + cosec α = determine the vaue of sin n α + cosec n α. 4. Prove that cos8θ = cos θ. 5. Evauate tan A sec A. cos B cot B 6. If tan ( cos θ ) = cot( sin θ ), prove that cos θ =± If a cos θ b sin θ = c, prove that a sin θ + b cos θ = a + b c. 8. If sin θ + cos θ = a and sec θ + cosec θ = b, show that b(a ) = a. Cass - X 9

25 Concept maps provided for every chapter Set of objective and subjective questions at the end of each chapter Previous contest questions at the end of each chapter w w. (F bm re a e ta Sa e n m t. p co e) m Simpe, cear and systematic presentation Designed to fufi the preparation needs for internationa/nationa taent exams, oympiads and a competitive exams UNIQUE ATTRACTIONS Concept Maps Cross word Puzzes Graded Exercise n Basic Practice n Further Practice n Brain Works Mutipe Answer Questions Paragraph Questions Concept Dri ` 50 w ISBN YOUR COACH India s FIRST scientificay designed porta for Oympiad preparation Oympiad & Taent Exams preparation packages Anaysis Reports Previous question papers Free Demo Packages Free Android Mobie App A unique opportunity to take about 50 tests per subject. IIT Foundation & Oympiad Exporer - Cass - X Integrated Syabus FOUNDATION & OLYMPIAD CLASS - X Schoo Edition