SSC CGL TRIGONOMETRY MCQ ENGLISH

Correct Answer : (C)

Explanation : Given:A=π6 and B=π3⇒A+B=π6+π2=π2As we know,sin⁡(π2−θ)=cos⁡θcos⁡(π2−θ)=sin⁡θtan⁡(π2−θ)=cot⁡θcot⁡(π2−θ)=tan⁡θStatement:Isin⁡A+sin⁡B=sin⁡(π2−A)+sin⁡(π2−B)⇒sin⁡A+sin⁡B=cos⁡A+cos⁡BStatement: IItan⁡A+tan⁡B=tan⁡(π2−B)+tan⁡(π2−B)⇒tan⁡A+tan⁡B=cot⁡A+cot⁡BWe can see that, both statement are correct.

Correct Answer : (C)

Explanation : Let, f(x)=sin⁡x.cos⁡x=12×(2sin⁡x×cos⁡x)=12sin⁡2xAs we know, the maximum value of sin⁡x is 1.Therefore the maximum value of sin⁡2x is 1.Hence maximum value of f(x) is 12×1=12

Correct Answer : (C)

Explanation : Given:sin⁡(A+B)=32⇒sin⁡(A+B)=sin⁡60∘⇒A+B=60∘…….(1)Also, cos⁡B=32⇒cos⁡B=cos⁡30∘⇒B=30∘Substituting the value of B in equation (1), we get⇒A=60∘−30∘=30∘∴tan⁡(2A−B)=tan⁡(60∘−30∘)=tan⁡30∘=13

Correct Answer : (B)

Explanation : Given:cos260∘+4sec230∘−tan245∘sin230∘+cos230∘⇒122+4×232-1122+322⇒14+43×4-114+34⇒14+163-1⇒3+6412-1⇒6712-1⇒5512∴ The value is 5512 .

Correct Answer : (B)

Explanation : Given,(1+cot⁡θ−cosecθ)(1+tan⁡θ+sec⁡θ)To find value of (1+cot⁡θ−cosecθ)(1+tan⁡θ+sec⁡θ)=(1+cos⁡θsin⁡θ−1sin⁡θ)(1+sin⁡θcos⁡θ+1cos⁡θ) [∵cot⁡θ=cos⁡θsin⁡θ,cosecθ=1sin⁡θ,sec⁡θ=1cos⁡θ]=(sin⁡θ+cos⁡θ−1)(sin⁡θ+cos⁡θ+1)sin⁡θcos⁡θ=(sin⁡θ+cos⁡θ)2−1sin⁡θcos⁡θ=sin2⁡θ+cos2⁡θ+2sin⁡θcos⁡θ−1sin⁡θcos⁡θ [∵sin2⁡θ+cos2⁡θ=1]=1+2sin⁡θcos⁡θ−1sin⁡θcos⁡θ=2sin⁡θcos⁡θsin⁡θcos⁡θ=2

Correct Answer : (D)

Explanation : Given:tan (A + B) = 3⇒ tan (A + B) = tan 60°⇒ A + B = 60° —-(1)tan (A – B) = 0⇒ tan (A – B) = tan 0°⇒ A – B = 0° —-(2)Adding equations (1) and (2), we get:(A – B) + (A + B) = 0° + 60°⇒ 2A = 60°⇒ A = 30°Puting the value of A in equation (1), we get⇒ B = 60° – 30° = 30°∴ The value of A and B are 30° and 30°.

Correct Answer : (C)

Explanation : We know that,⇒tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy),We have,tan−1⁡(17)+tan−1⁡(113)=tan−1⁡(17+1131−17×113)=tan−1⁡(2090)=tan−1⁡(29)

Correct Answer : (C)

Explanation : We know that,tan⁡2A=2tan⁡A1−tan2⁡A=2×121−(12)2=11−14=134=43Now,tan⁡(2A+B)=tan⁡2A+tan⁡B1−tan⁡2Atan⁡B∴ tan⁡2A+tan⁡B1−tan⁡2Atan⁡B=43+131−43×13=5/35/9=53×95=3

Correct Answer : (A)

Explanation : Given:sin⁡3x+cos⁡3x+4sin3⁡x−3sin⁡x+3cos⁡x−4cos3⁡xWe know that,sin⁡3x=3sin⁡x−4sin3⁡xcos⁡3x=4cos3⁡x−3cos⁡xUsing above formula we get,=3sin⁡x−4sin3⁡x+4cos3⁡x−3cos⁡x+4sin3⁡x−3sin⁡x+3cos⁡x−4cos3⁡x=0

Correct Answer : (C)

Explanation : Given: cosec θ = 222………Let, x = 222………On squaring both sides we get,x2 = 222………⇒ x2 = 2x∵x=222………⇒ x2−2x=0⇒ x(x – 2) = 0⇒ x = 2 or x = 0If x = 2:cosec θ = 2⇒ θ = 30°∴ tan θ = tan 30° = 13

Correct Answer : (C)

Explanation : Given that, sin⁡x+sin⁡y=cos⁡y−cos⁡x2sin⁡(x+y2)⋅cos⁡(x−y2)=−2sin⁡(x+y2)⋅sin⁡(y−x2)=2sin⁡(x+y2)⋅sin⁡(x−y2)⇒sin⁡(x+y2)⋅cos⁡(x−y2)=sin⁡(x+y2)⋅sin⁡(x−y2)dividing by cos⁡(x−y2)⇒sin⁡(x+y2)=sin⁡(x+y2)tan⁡(x−y2)⇒tan⁡(x−y2)=1

Correct Answer : (D)

Explanation : Given:(3sin⁡θ+5cos⁡θ)=4Squaring both sides,⇒(3sin⁡θ+5cos⁡θ)2=42⇒9sin2⁡θ+25cos2⁡θ+30sin⁡θcos⁡θ=16 [∵(a+b)2=a2+b2+2ab]⇒30sin⁡θcos⁡θ=16−9sin2⁡θ−25cos2⁡θ …….(1)Now,(3cos⁡θ−5sin⁡θ)2=9cos2⁡θ+25sin2⁡θ−30sin⁡θcos⁡θ=9cos2⁡θ+25sin2⁡θ−16+9sin2⁡θ+25cos2⁡θ [ Substituting from eq. (1) ]=34(sin2⁡θ+cos2⁡θ)−16=34−16[∵sin2⁡θ+cos2⁡θ=1]= 18

Correct Answer : (A)

Explanation : Given: 2 cos x – 1 = 0⇒ cos x = 12As we know that if cos θ = cos α then θ = 2nπ ± α, α ∈ [0, π], n ∈ Z.∴The general solution of the given equation is: x = 2nπ ± π4 , where n ∈ Z.

Correct Answer : (C)

Explanation : Considering statement 1,Suppose cos⁡θ+sec⁡θ=1.5cos⁡θ+1cos⁡θ=32⇒2(cos2⁡θ+1)=3cos⁡θ⇒2cos2⁡θ−3cos⁡θ+3=0For a quadratic equation of form ax2+bx+c,x=x=−b±b2−4ac2acos⁡θ=[3±(9−24)]4⇒cos⁡θ=(3±−15)4 , which is not possible as the value of cos⁡θ is always a real number.Statement 1 is correctConsidering statement 2,Suppose sec2⁡θ+cosec2θ<41cos2⁡θ+1sin2⁡θ<4⇒sin2⁡θ+cos2⁡θ<4sin2⁡θcos2⁡θ⇒1

Correct Answer : (C)

Explanation : Given: x=tan−1⁡(15)∴tan⁡x=15As we know that, cos⁡2θ=1−tan2⁡θ1+tan2⁡θ∴cos⁡2x=1−tan2⁡x1+tan2⁡x=1−(15)21+(15)2=(25−125)(25+125)=2426=1213

Correct Answer : (C)

Explanation : Considering statement 1 ,[cos⁡θ(1−sin⁡θ)]+[cos⁡θ(1+sin⁡θ)]=4⇒[cos⁡θ(1+sin⁡θ)+cos⁡θ(1−sin⁡θ)(1−sin2⁡θ)]=4⇒2cos⁡θcos2⁡θ=4[∵sin2⁡θ+cos2⁡θ=1]⇒cos⁡θ=12⇒θ=60∘[∵0<θ<90∘]⇒ Statement 1 is correctConsidering statement 2 ,3tan⁡θ+cot⁡θ=5cosecθ⇒3(sin⁡θcos⁡θ)+(cos⁡θsin⁡θ)=5sin⁡θ⇒3sin2⁡θ+cos2⁡θ=5cos⁡θ⇒3−3cos2⁡θ+cos2⁡θ=5cos⁡θ[∵sin2⁡θ+cos2⁡θ=1]⇒2cos2⁡θ+5cos⁡θ−3=0⇒(2cos⁡θ−1)(cos⁡θ+3)=0⇒cos⁡θ=12,−3⇒θ=60∘[∵0<θ<90∘]⇒ Statement 2 is correct∴ Both statements 1 and 2 are correct

Correct Answer : (A)

Explanation : Minimum value of cos⁡x is -1.So, minimum value of 3cos⁡(A+π3)=3(−1)=−3

Correct Answer : (D)

Explanation : Given:4(cosec257−tan2⁡33)−cos⁡90+y×tan2⁡66×tan2⁡24=y24(cosec2(90−33)−tan2⁡33)−cos⁡90+y×tan2⁡66×tan2⁡(90−66)=y2⇒4(sec2⁡57−tan2⁡33)−cos⁡90+y×tan2⁡66×cot2⁡66=y2⇒4×1−0+y×tan2⁡66×1tan2⁡66=y2⇒4+y=y2⇒y−y2=−4⇒y2=−4⇒y=−8∴ The value of y is −8 .

Correct Answer : (C)

Explanation : We know that,sec⁡(y)=1312= Hypotenuse Base⇒( Height )2=( Hypotenuse )2−( Base )2⇒( Height )2=(13)2−(12)2⇒( Height )2=169−144=25Height =5tan⁡(y)= Height Base=512tan⁡(3y)=3tan⁡(y)−tan3⁡(y)1−3tan2⁡(y)=3×512−(512)31−3×(512)2=54−(1251728)1−3×(25144)⇒2160−12517281−2548 =203517282348⇒9768039744 =2035828

Correct Answer : (C)

Explanation : Given:[cos2⁡(45∘+θ)+cos2⁡(45∘−θ)][tan⁡(60∘+θ)tan⁡(30∘−θ)]As we know,cos⁡A=sin⁡(90∘−A)tan⁡A=cot⁡(90∘−A)tan⁡Acot⁡A=1By substituting it we get,=[cos2⁡(45∘+θ)+sin2⁡(90∘−45∘+θ)][tan⁡(60∘+θ)cot⁡(90∘−30∘+θ)]So,=[cos2⁡(45∘+θ)+sin2⁡(45∘+θ)][tan⁡(60∘+θ)cot⁡(60∘+θ)]As we know cos2⁡A+sin2⁡A=1 and tan⁡Acot⁡A=1By substituting it,=11=1

Correct Answer : (A)

Explanation : Given: Sin A = 513To find: tan ASin A = 513 = PerpendicularHypotenusePerpendicular = 5 and Hypotenuse = 13Hypotenuse2 = Perpendicular2 + Base2⇒ 132 = 52 + Base2169 = 25 + Base2Base2 = 169 – 25 = 144⇒ Base = 12So, tan A =Perpendicular Base = 512

Correct Answer : (D)

Explanation : Given: cos⁡2B=3sin2⁡A …(1)and 3sin⁡2 A=2sin⁡2B …(2)Dividing equation (1) by (2),cos⁡2B2sin⁡2B=3sin2⁡A3sin⁡2A⇒12cos⁡2 Bsin⁡2 B=sin2⁡A2sin⁡A.cos⁡A⇒cos⁡2 Bsin⁡2 B=sin⁡Acos⁡A⇒cos⁡2B.cos⁡A−sin⁡A.sin⁡2B=0⇒cos⁡(A+2 B)=0∴A+2B=π2

Correct Answer : (B)

Explanation : When the value of θ ranges as 0 < θ < 90°, the value of cos θ ranges as 0 > cos θ > 1Similarly,When the value of φ ranges as 0 < φ < 90°, the value of cos φ ranges as 0 > cos φ > 1Now given,⇒ cos θ < cos φ⇒ θ > φ

Correct Answer : (C)

Explanation : Given: tan⁡θ=8We know that, 1+tan2⁡θ=sec2⁡θsec2⁡θ=1+(8)2=1+8=9∴sec⁡θ=9=3

Correct Answer : (C)

Explanation : Given:1+cot2⁡θ=1+cos2⁡θsin2⁡θ∴[cot⁡θ=cos⁡θsin⁡θ]=sin2⁡θ+cos2⁡θsin2⁡θ=1sin2⁡θ=cosec2θ.

Correct Answer : (C)

Explanation : sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡BComplex variable:In mathematics it’s a variable that can take on the value of a complex number.In basic algebra the variable ′x′ and ′y′ stand for value of real numbers.The algebra of complex number uses the complex variable ‘z’ to represent a number of the form x+iy .x : Real party : Imaginary partGiven function is log⁡sin⁡(x+iy)log⁡[sin⁡xcos⁡iy+sin⁡iycos⁡x]….(1)sin⁡(x+iy)=sin⁡xcos⁡(iy)+cos⁡xsin⁡iyWe know that{cosh⁡x=cos⁡(ix) and sinh⁡x=−isin⁡(ix),tan⁡hx=−itan⁡(ix)sin⁡(x+iy)=sin⁡xcosh⁡x+cos⁡x(isinh⁡x)=sin⁡xcosh⁡x+i(cos⁡xsinh⁡x)∴log⁡[sin⁡(x+iy)]=log⁡[sin⁡xcosh⁡y+icos⁡xsinh⁡y]∵cos⁡(ix)=cosh⁡x;sin⁡(x)=isinh⁡x∴log⁡[sin⁡(x+iy)]=12log⁡[sin2⁡xcos2⁡hy+cos2⁡xsinh2⁡y]+itan−1⁡[cos⁡xsinh⁡ysin⁡xcosh⁡y]log⁡(sin⁡(x+iy))=12log⁡[sin2⁡xcos2⁡hy+cos2⁡xsinh2⁡y]+itan−1⁡[cot⁡xtan⁡hy]Imaginary part is:tan−1⁡[cot⁡xtan⁡hy]

Correct Answer : (B)

Explanation : Given:sin⁡θ+cosecθ=2We know that:cosecθ=1sin⁡θsin⁡90∘=1cos⁡45∘=12cot⁡45∘=1(a−b)2=a2−2ab+b2∵sin⁡θ+cosecθ=2⇒sin⁡θ+(1sin⁡θ)=2⇒sin2⁡θ+1=2sin⁡θ⇒sin2⁡θ−2sin⁡θ+1=0⇒(sin⁡θ−1)2=0⇒sin⁡θ−1=0⇒sin⁡θ=1∴θ=90∘θ2=45∘∴2[cos2⁡(θ2)+cot2⁡(θ2)]=2×(cos2⁡45∘+cot2⁡45∘)=2×[12+1]=2×32=3∴ The value of 2[cos2⁡(θ2)+cot2⁡(θ2)] is 3.

Correct Answer : (D)

Explanation : Given,α=β=15∘sin⁡(α+1∘)+cos⁡(β+1∘)=sin⁡(15∘+1∘)+cos⁡(15∘+1∘)=sin⁡15∘cos⁡1∘+cos⁡15∘sin⁡1∘+cos⁡15∘cos⁡1∘−sin⁡15∘sin⁡1∘=cos⁡1∘(sin⁡15∘+cos⁡15∘)+sin⁡1∘(cos⁡15∘−sin⁡15∘) …(1)sin⁡15∘+cos⁡15∘=sin⁡15∘+sin⁡75∘=2sin⁡(75∘+15∘2)cos⁡(75∘−15∘2)=2sin⁡(45∘)cos⁡30∘=2×12×32=32Andcos⁡15∘−sin⁡15∘=cos⁡(90∘−75∘)−sin⁡15∘=sin⁡75∘−sin⁡15∘=2cos⁡(75∘+15∘2)sin⁡(75∘−15∘2)=2cos⁡(45∘)sin⁡30∘=2×12×12=12So, (1) will becomecos⁡1∘(32)+sin⁡1∘(12)12(3cos⁡1∘+sin⁡1∘)

Correct Answer : (B)

Explanation : Given:cos⁡θ=15∵sin⁡θ=(1−cos2⁡θ)⇒sin⁡θ=(1−15)=(45)=25⇒tan⁡θ=sin⁡θcos⁡θ=(25)(15)=2⇒2tan⁡θ(1−tan2⁡θ)=(2×2)(1−4)=4(−3)=(−43)

Correct Answer : (A)

Explanation : Here, cos(2010°)⇒ cos (6 × 360° – 150°)= cos(150°)(∵ cos (2nπ ± θ) = cos θ)= cos(90° + 60°)= – sin(60°)(∵ cos (90 + θ) = -sin θ)= -32