0
793

# SSC CHSL Trigonometry Questions:

Download Top-20 Trigonometry questions for SSC CHSL exam 2020. Most important Trigonometry questions based on asked questions in previous exam papers for SSC CHSL.

Question 1:Â What is the value of $(1 + cotA)^{2} + (1-cotA)^{2}$ ?

a)Â $2cosec^{2}A$

b)Â $2sec^{2}A$

c)Â $1-2cosec^{2}A$

d)Â $1-2sec^{2}A$

Question 2:Â If cot 135Â° = x, then the value of x is

a)Â -1/âˆš3

b)Â -âˆš3

c)Â -1

d)Â -1/2

Question 3:Â If âˆš$(1 – cos^2A)/cosA$ = x, then the value of x is

a)Â cotA

b)Â cosecA

c)Â tanA

d)Â secA

Question 4:Â What is the value of âˆš[(1 – sinA)/(1 + sinA)]?

a)Â cosA/(1 – sinA)

b)Â cosecA/(1 + sinA)

c)Â cosA/(1 + sinA)

d)Â cosecA/(1 – sinA)

Question 5:Â If cosecA/(cosecA – 1) + cosecA/(cosecA + 1) = x, then x is

a)Â $2cosec^2A$

b)Â $2cosecA$

c)Â $2secA$

d)Â $2sec^2A$

Question 6:Â If cot 30Â° – cos 45Â° = x, then x is

a)Â âˆš3+2

b)Â (âˆš6-1)/âˆš2

c)Â (âˆš3-2âˆš2)/âˆš6

d)Â (1+âˆš2)/2

Question 7:Â If tan2A = x, then x is

a)Â $2tanA/(1 – tan^2A)$

b)Â $(1 – tan^2A)/2tanA$

c)Â $2tanA/(1 + tan^2A)$

d)Â $(tan^2A – 1)/2tanA$

Question 8:Â What is the value of sin 300Â°?

a)Â -1/2

b)Â 2

c)Â 2/âˆš3

d)Â -âˆš3/2

Question 9:Â If âˆš$(1 – cos^2A)$ = x, then the value of x is

a)Â cosecA

b)Â sinA

c)Â tanA

d)Â secA

Question 10:Â If cosA/(1 – sinA) = x, then the value of x is

a)Â secA – tanA

b)Â cosecA + tanA

c)Â secA + tanA

d)Â cosecA – tanA

Question 11:Â  Â­ If $\sqrt{(sec^2A-1)}=x$, then the value of x is

a)Â cotA

b)Â tanA

c)Â cosecA

d)Â cosA

Question 12:Â What is the value of $cos^2A/(1 + sinA)$?

a)Â 1 + sinA

b)Â 1 -Â­ sinA

c)Â 1 Â­- secA

d)Â 1 + secA

Question 13:Â If $(cotA – cosecA)^2 = x$, then the value of x is

a)Â (1 – cosA)/(1 + cosA)

b)Â (1 – sinA)/(1 + sinA)

c)Â (1 – secA)/(1 + secA)

d)Â (1 – cosecA)/(1 + cosecA)

Question 14:Â If cotA = x, then the value of x is

a)Â âˆš$(sec^2A + 1)$

b)Â âˆš$(cosec^2A + 1)$

c)Â âˆš$(sec^2A – 1)$

d)Â âˆš$(cosec^2A – 1)$

Question 15:Â If cos 240Â° = x, then value of x is

a)Â -1/âˆš2

b)Â Â­âˆš3/2

c)Â 1/2

d)Â -1/2

Question 16:Â If 1/âˆš(1+$tan^{2}$A) = x, then value of x is

a)Â cosA

b)Â sinA

c)Â cosecA

d)Â secA

Question 17:Â In which of the following quadrilaterals only one pair of opposite angles is supplementary?

a)Â Isosceles Trapezium

b)Â Parallelogram

d)Â Rectangle

Question 18:Â What is the value of âˆš[(1 + sinA)/(1 Â­- sinA)]?

a)Â secA Â­- tanA

b)Â cosecA + tanA

c)Â secA + tanA

d)Â cosecA Â­- tanA

Question 19:Â If (1 + sinA)/cosA + cosA/(1 + sinA) = x, then the value of x is

a)Â 2cosecA

b)Â 2$cosec^{2}$A

c)Â 2$sec^{2}$A

d)Â 2secA

Question 20:Â If tan 330Â° = x, then the value of x is

a)Â $\frac{-1}{\sqrt{3}}$

b)Â ${-\sqrt{3}}$

c)Â $\frac{-1}{2}$

d)Â $\frac{-1}{\sqrt{2}}$

ExpressionÂ :Â $(1 + cotA)^{2} + (1-cotA)^{2}$

= $(1+cot^2A+2cotA)+(1+cot^2A-2cotA)$

= $2+2cot^2A=2(1+cot^2A)$

$\because$ $(cosec^2A-cot^2A=1)$

= $2cosec^2A$

=> Ans – (A)

ExpressionÂ :Â cot 135Â° = x

= $cot(180-45)$

= $-cot(45)$

= $-1$

=> Ans – (C)

ExpressionÂ : $\frac{\sqrt{1-cos^2A}}{cosA}$

$\because (sin^2A+cos^2A=1)$

= $\frac{\sqrt{sin^2A}}{cosA}$

= $\frac{sinA}{cosA} = tanA$

=> Ans – (C)

ExpressionÂ :Â $\sqrt{[\frac{(1 – sinA)}{(1 + sinA)}}]$

Multiplying both numerator and denominator by $(\sqrt{1+sinA})$

=Â $\sqrt{[\frac{(1 – sinA)}{(1 + sinA)}}]$ $\times \sqrt{\frac{(1+sinA)}{(1+sinA)}}$

= $\sqrt{\frac{1-sin^2A}{(1+sinA)^2}} = \sqrt{\frac{cos^2A}{(1+sinA)^2}}$

= $\frac{cosA}{1+sinA}$

=> Ans – (C)

ExpressionÂ : $\frac{cosecA}{cosecA-1}+\frac{cosecA}{cosecA+1}$

= $[(\frac{1}{sinA})\div(\frac{1}{sinA}-1)]+[(\frac{1}{sinA})\div(\frac{1}{sinA}+1)]$

= $[(\frac{1}{sinA})\div(\frac{1-sinA}{sinA})]+[(\frac{1}{sinA})\div(\frac{1+sinA}{sinA})]$

= $[(\frac{1}{sinA}) \times (\frac{sinA}{1-sinA})]+[(\frac{1}{sinA}) \times (\frac{sinA}{1+sinA})]$

= $(\frac{1}{1-sinA})+(\frac{1}{1+sinA})$

= $\frac{(1+sinA)+(1-sinA)}{(1+sinA)(1-sinA)} = \frac{2}{1-sin^2A}$

= $\frac{2}{cos^2A} = 2sec^2A$

=> Ans – (D)

ExpressionÂ :Â cot 30Â° – cos 45Â° = x

= $\sqrt{3} – \frac{1}{\sqrt{2}}$

= $\frac{\sqrt{6}-1}{\sqrt{2}}$

=> Ans – (B)

ExpressionÂ : $tan(2A)$

$\because tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$

Replacing B by A

=> $tan(A+A)=\frac{tanA+tanA}{1-tanAtanA}$

=> $tan(2A)=\frac{2tanA}{1-tan^2A}$

=> Ans – (A)

ExpressionÂ :Â sin 300Â°

= $sin(360-60)$

= $-sin(60)$

= $\frac{-\sqrt{3}}{2}$

=> Ans – (D)

ExpressionÂ : $\sqrt{1-cos^2A}$

$\because (sin^2A+cos^2A=1)$

= $\sqrt{sin^2A}=sinA$

=> Ans – (B)

ExpressionÂ :Â cosA/(1 – sinA) = x

Multiplying both numerator and denominator by $(1+sinA)$

= $\frac{cosA}{1-sinA} \times \frac{(1+sinA)}{(1+sinA)}$

= $\frac{cosA(1+sinA)}{1-sin^2A} = \frac{cosA(1+sinA)}{cos^2A}$

= $\frac{1+sinA}{cosA} = (\frac{1}{cosA}+\frac{sinA}{cosA})$

= $secA+tanA$

=> Ans – (C)

ExpressionÂ :Â $\sqrt{(sec^2A-1)}=x$

$\because (sec^2A-tan^2A=1)$

= $\sqrt{tan^2A} = tanA$

=> Ans – (B)

ExpressionÂ :Â $cos^2A/(1 + sinA)$

Multiplying both numerator and denominator by $(1-sinA)$

= $\frac{cos^2A}{1+sinA} \times \frac{(1-sinA)}{(1-sinA)}$

= $\frac{cos^2A(1-sinA)}{1-sin^2A} = \frac{cos^2A(1-sinA)}{cos^2A}$

= $1-sinA$

=> Ans – (B)

ExpressionÂ :Â $(cotA – cosecA)^2 = x$

= $(\frac{cosA}{sinA}-\frac{1}{sinA})^2$

= $(\frac{cosA-1}{sinA})^2$

= $\frac{(1-cosA)^2}{sin^2A} = \frac{(1-cosA)^2}{1-cos^2A}$

= $\frac{(1-cosA)^2}{(1-cosA)(1+cosA)}$

= $\frac{1-cosA}{1+cosA}$

=> Ans – (A)

We know that, $cosec^2A-cot^2A=1$

=> $cot^2A=cosec^2A-1$

=> $cotA=\sqrt{cosec^2A-1}$

=> Ans – (D)

ExpressionÂ :Â cos 240Â° = x

= $cos(180+60)$

= $-cos(60)$

= $\frac{-1}{2}$

=> Ans – (D)

ExpressionÂ :Â 1/âˆš(1+$tan^{2}$A) = x

$\because (sec^2A-tan^2A=1)$

= $\frac{1}{\sqrt{sec^2A}}$

= $\frac{1}{secA} = cosA$

=> Ans – (A)

In rectangle, all angles are 90Â° and in a parallelogram, adjacent angles are supplementary while in a cyclic quadrilateralÂ only one pair of opposite angles is supplementary.

=> Ans – (C)

ExpressionÂ :Â $\sqrt{[\frac{(1 + sinA)}{(1 – sinA)}}]$

Multiplying both numerator and denominator by $(\sqrt{1+sinA})$

=Â $\sqrt{[\frac{(1 + sinA)}{(1 – sinA)}}]$ $\times \sqrt{\frac{(1+sinA)}{(1+sinA)}}$

= $\sqrt{\frac{(1+sinA)^2}{1-sin^2A}} = \sqrt{\frac{(1+sinA)^2}{cos^2A}}$

= $\frac{1+sinA}{cosA} = (\frac{1}{cosA}+\frac{sinA}{cosA})$

= $secA+tanA$

=> Ans – (C)

ExpressionÂ : $\frac{(1+sinA)}{cosA}+\frac{cosA}{(1+sinA)}$

=Â $\frac{(1+sinA)^2+(cosA)^2}{cosA(1+sinA)}$

= $\frac{(1+2sinA+sin^2A)+cos^2A}{cosA(1+sinA)}$

= $\frac{1+2sinA+1}{cosA(1+sinA)}$ Â  Â $[\because sin^2A+cos^2A=1]$

= $\frac{2+2sinA}{cosA(1+sinA)} = \frac{2(1+sinA)}{cosA(1+sinA)}$

= $\frac{2}{cosA} = 2secA$

=> Ans – (D)

= $tan(360-30)$
= $-tan(30)$
= $\frac{-1}{\sqrt{3}}$