HomeMath Class 11Trigonometry MCQ test 2 Trigonometry MCQ test 2 Trigonometry 11 CBSE - MCQ test 2 Question 1.Which of following is equal to cos²x+cos²y-2cosx.cosy.cos(x+y): sin(x+y) sin²(x+y) sin³(x+y) None of these Click Here For Solution Solution: cos² x + cos² y - 2cos x cos y cos (x + y) => cos² x + cos² y - 2cos x cos y [cos x cos y - sin x sin y] => cos² x + cos² y - 2cos² x cos² y + 2sin x sin y cos x cos y => cos² y - cos² x cos² y + cos² x - cos²y cos² x + 2sin x sin y cos x cos y => (1 - cos²x)cos²y + (1 - cos²y)cos²x + 2sin x sin y cos x cos y => sin²x cos²y + cos²x sin²y + 2sin x sin y cos x cos y => [ sin x cos y + cos x sin y ]² => [sin (x + y)]² = sin² (x + y) Question 2. Determine the value of (asinX+bcosX)² if acosX-bsinX=c : a²+b²+c² a²+b²-c² a²-b²-c² a²-b²+c² Click Here For Solution Given, acosX+bcosY=C Squaring both sides, (acosX-bsinX)²=c² => (a²cos²X+b²sin²X-2acosX.bsinX)=c² => a²(1-sin²X)+b²(1-cos²X)+2acosX.bsinX=c² => a²-a²sin²X+b²-b²cos²X+2acosX.bsinX=c² => a²+b²-c²=a²sin²X+b²cos²X-2acosX.bsinX => a²+b²-c²=(asinX+bcosX) Question 3. If in △ABC, cosA+2cosB+cosC=2,What is relation between a,b and c. a-c=2b a+c=2b a+b=2c 2a+c=b Click Here For Solution A,B,C are the angles of a △ABC. Therefore, A+B+C=180° Given, cosA+2cosB+cosC=2 cosA+cosC=2(1−cosB) 2cos(A+C)/2.cos(A-C)/2=2.2sin²B/2 2.sinB/2.cos(A-C)/2=2.2sin²B/2 [ since,cos(A+C)/2=cos(180°-B)/2= sinB/2 ] cos(A-C)/2=2sinB/2 Multiplying both sides by 2cosB/2 cos(A-C)/2*2cosB/2 =2sinB/2 *2cosB/2 [ since,sin(A+C)/2=sin(180°-B)/2= cosB/2 ] [cos(A-C)/2]*[2sin(A+C)/2] =2sinB/2 *2cosB/2 2*[cos(A-C)/2]*[sin(A+C)/2] =2sinB 2*[sin(A+C)/2]*[cos(A-C)/2] =2sinB sinA+sinC=2sinB a+c=2b Question 4. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? 12π 10π 8π 6π Click Here For Solution 360 revolutions in 1 minute. 360 revolutions in 60 second. 6 revolutions in 1 second. 6 *2π radian in 1 second. 12π radian in 1 second. Question 5. What will be the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7). 10°36' 12°36' 14°36' 11°36' Click Here For Solution Angle subtended at the centre of a circle=22/100 radian Angle subtended at the centre of a circle=22/100*180/π degree Angle subtended at the centre of a circle=(22/100)*(180*7/22) degree Angle subtended at the centre of a circle=(180*7/100)=12.6°=12°36' Question:6 If tanA=(cos9°-sin9°)/(cos9°+sin9°) what will be vaule of A? 36° 54° 45° None Click Here For Solution tanA=(cos9°-sin9°)/(cos9°+sin9°) Dividing RHS part by cos9° in numerator and denumarator tanA=(1-tan9°)/(1+tan9°) tanA=(tan45°-tan9°)/(1+tan45°tan9°) tanA=tan(45°-9°)=tan36° Question:7 In a triangle ABC, cosec A (sin B cos C + cos B sin C) is equal to : a/c 1 c/a None of these Click Here For Solution cosec A (sin B cos C + cos B sin C) cosec A *sin (B + C) cosec A *sin (180°-A) cosec A *sin A 1 Question:8 The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is π/3 2π/3 π/2 None Click Here For Solution a+b+c=6(sinA+sinB+sinC)/3 a+b+c=2(sinA+sinB+sinC) (a+b+c)/2=(sinA+sinB+sinC) (sinA+sinB+sinC)/(a+b+c)=1/2----Eqn 1 We know that: sinA/a=sinB/b=sinC/c sinA/a=sinB/b=sinC/c=(sinA+sinB+sinC)/(a+b+c) sinA/a=sinB/b=sinC/c=1/2 sinB/b=1/2 sinB=b/2=2/2=1=sin(π/2) B=π/2 Question:9 If tan A – tan B = x and cot B – cot A = y, then What will be the value of cot (A – B): x + y 1/x + 1/y 1/x + y None of these Click Here For Solution tan A – tan B = x ----Eqn1 cot B – cot A = y ----Eqn2 Eqn1 tan A – tan B = x ( 1/cotA ) – ( 1/cotB ) = x ( cotB - cotA )/cotA*cotB = x putting (cot B – cot A )= y y/cotA*cotB = x tanA*tanB = x/y ----Eqn3 cot (A – B)=1/tan (A – B) (1+tanA*tanB)/(tan A – tan B) From Eqn1 and Eqn3 we get, [1+(x/y)]/x 1/x + 1/y (y+x)/xy 1/x + 1/y Question:10 If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (√5 – 1):4 3 : (√5 – 1) 4 : (√5 – 1) None Click Here For Solution Calculation of angles of triangle: Calculation of value of sin18°: Let θ = 18°….(i) We can write 90° as: 90° = 36° + 54° 90 = 2 × 18° + 3 × 18° Substituting (i) in the above equation, 90° = 2θ + 3θ 3θ = 90° – 2θ Taking “sin” on both sides, sin 3θ = sin(90° – 2θ) Using the identity sin 3A = 3 sin A – 4 sin³A, 3 sin θ – 4 sin³θ = cos 2θ Now, using the identity cos 2θ = 1 – 2 sin²θ, 3 sin θ – 4 sin³θ = 1 – 2 sin²θ = 0 4 sin³θ – 2 sin²θ – 3 sin θ + 1 = 0 Let sin θ = x. 4x³ – 2x² – 3x + 1 = 0 (x – 1)(4x² + 2x – 1) = 0 x – 1 = 0, 4x² + 2x – 1 = 0 x = 1, 4x² + 2x – 1 = 0 Here x can not be equal to 1 as x=sin18° which must be less tha 1. 4x² + 2x – 1 = 0 Using quadratic formula, x = [-2 ± √{4 – 4(-4)}]/2(4) = [-2 ± √(4 + 16)]/8 = [-2 ± √20]/8 = [-2 ± 2√5]/8 = 2[-1 ± √5]/8 = (-1 ± √5)/4 The value of sin θ is positive as x=sin18°. x=sin18°=(-1 + √5)/4 Let greatest side =c in front of angle 90° and the smallest side = a infront of angle 18°. sin90°/c=sin18°/a sin90°/sin18°=c/a 1/[(-1 + √5)/4]=c/a 4/(-1 + √5)=c/a Question:11 If cosecA-cotA=1/5 then what will be sinA: 48/7 10/23 5/13 None Click Here For Solution cosecA-cotA=1/5 ------Eqn 1 so, cosecA+cotA=5 ------Eqn 2 Adding Eqn 1 & Eqn 2,We get 2cosecA=5+1/5=26/5 cosecA=13/5 sinA=5/13 Question:12 If secA-tanA=5 then what will be cosA: 5/3 10/3 3/5 None of these Click Here For Solution secA-tanA=3 ------Eqn 1 secA+tanA=1/3 ------Eqn 2 Adding Eqn 1 & Eqn 2,We get 2secA=3+1/3=10/3 secA=5/3 cosA=3/5 Question:13 If sinx + cosecx =2, then which of following is equal to (sin³x + cosec³x) : 10 2 2³ None Click Here For Solution sinx+cosecx=2 sinx+1/sinx=2 sin²x+1=2sinx sin²x+1-2sinx=0 (sinx-1)²=0 (sinx-1)=0 sinx=1 so,cosecx =1 (sin³x + cosec³x)=1³+1³=2 Question:14 If sinx + sin²x = 1 then the value ofcos^8 x + 2cos^6 x+ cos^4 x √3/2 √3 1 1:3 Click Here For Solution Given, sinx + sin²x = 1 sinx = 1- sin²x =cos²x We have to find value of : cos^8 x + 2cos^6 x+ cos^4 x (cos^2 x)^4 + 2(cos^2)^3 x+ (cos^2 x)2 (cos^2 x)^4 + 2(cos^2)^3 x+ (cos^2 x)2 (sinx)^4 + 2(sinx)^3 x+ (sinx)2 (sin²x)² + 2(sin²x)(sinx)+ (sinx)2 (sinx + sin²x )²=1 Question:15 Which of following is equal to sin420°.cos390°+cos(-300°)sin(-330°) 11/√3 √3 1 None Click Here For Solution sin420°.cos390°+cos(-300°)sin(330°) sin(360°+60°).cos(360°+30°)-cos(360°-60°)sin(360°-30°) sin(60°).cos(30°)+cos(60°)sin(30°) sin(60°+30°)=1 Question:16 If tan x=1/3–√ What will equal to cos2x? 1/2 -1/2 √3/2 None of these Click Here For Solution We know, cos2x = cos²x – sin²x => (cos²x – sin²x)/(cos²x+sin²x) {1 = sin²x + cos²x} => (1-tan²x)/(1+tan²x) => (1-(1/3–√)2)/(1+(1/3–√)2) =>(1-1/3)/(1+1/3) = (2/3)/(4/3) = 1/2. Question:17 Which of following is correct? cos 3x = 3cosx – 4cos³x cos 3x = 4cosx – 3cos³x cos 3x = 3cos³x – 4cosx cos 3x = 4cos³x – 3cosx Click Here For Solution cos 3x = cos (2x +x) => cos 2x cos x – sin 2x sin x => (2cos² x – 1) cos x – 2sin x cos x sin x => (2cos² x – 1) cos x – 2cos x (1 – cos² x) => 2cos³ x – cos x – 2cos x + 2 cos³ x => 4cos³ x – 3cos x. Question:18 Which of following is equal to 5 tan² A – 5 sec² A + 1 -2 1 3 -4 Click Here For Solution 5 tan² A – 5 sec² A + 1 -5(sec² A-tan² A)+1 -5*1+1 -4 Question:19 If sinX/sinY=1/2 and cosX/cosY=3/2 where x,y∈(0,π/2), then the value of tan(x+y) is equal to: √12 0 √13 √15 Click Here For Solution Question:20 lf in a ΔABC,∠C=90° , then What will be the maximum value of sinAsinB : 1 1/√2 2 1/2 Click Here For Solution Submit Newer Older
