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.

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

a) $2cosec^{2}A$

b) $2sec^{2}A$

c) $1-2cosec^{2}A$

d) $1-2sec^{2}A$

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

a) -1/√3

b) -√3

c) -1

d) -1/2

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

a) cotA

b) cosecA

c) tanA

d) secA

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

a) cosA/(1 – sinA)

b) cosecA/(1 + sinA)

c) cosA/(1 + sinA)

d) cosecA/(1 – sinA)

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

a) $2cosec^2A$

b) $2cosecA$

c) $2secA$

d) $2sec^2A$

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

a) √3+2

b) (√6-1)/√2

c) (√3-2√2)/√6

d) (1+√2)/2

Question 7: If tan2A = x, then x is

a) $2tanA/(1 – tan^2A)$

b) $(1 – tan^2A)/2tanA$

c) $2tanA/(1 + tan^2A)$

d) $(tan^2A – 1)/2tanA$

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

a) -1/2

b) 2

c) 2/√3

d) -√3/2

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

a) cosecA

b) sinA

c) tanA

d) secA

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

a) secA – tanA

b) cosecA + tanA

c) secA + tanA

d) cosecA – tanA

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

a) cotA

b) tanA

c) cosecA

d) cosA

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

a) 1 + sinA

b) 1 -­ sinA

c) 1 ­- secA

d) 1 + secA

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

a) (1 – cosA)/(1 + cosA)

b) (1 – sinA)/(1 + sinA)

c) (1 – secA)/(1 + secA)

d) (1 – cosecA)/(1 + cosecA)

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

a) √$(sec^2A + 1)$

b) √$(cosec^2A + 1)$

c) √$(sec^2A – 1)$

d) √$(cosec^2A – 1)$

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

a) -1/√2

b) ­√3/2

c) 1/2

d) -1/2

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

a) cosA

b) sinA

c) cosecA

d) secA

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

a) Isosceles Trapezium

b) Parallelogram

c) Cyclic quadrilateral

d) Rectangle

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

a) secA ­- tanA

b) cosecA + tanA

c) secA + tanA

d) cosecA ­- tanA

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

a) 2cosecA

b) 2$cosec^{2}$A

c) 2$sec^{2}$A

d) 2secA

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

a) $\frac{-1}{\sqrt{3}}$

b) ${-\sqrt{3}}$

c) $\frac{-1}{2}$

d) $\frac{-1}{\sqrt{2}}$

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Answers & Solutions:

1) Answer (A)

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

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

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

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

= $2cosec^2A$

=> Ans – (A)

2) Answer (C)

Expression : cot 135° = x

= $cot(180-45)$

= $-cot(45)$

= $-1$

=> Ans – (C)

3) Answer (C)

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

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

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

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

=> Ans – (C)

4) Answer (C)

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

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

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

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

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

=> Ans – (C)

5) Answer (D)

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

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

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

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

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

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

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

=> Ans – (D)

6) Answer (B)

Expression : cot 30° – cos 45° = x

= $\sqrt{3} – \frac{1}{\sqrt{2}}$

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

=> Ans – (B)

7) Answer (A)

Expression : $tan(2A)$

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

Replacing B by A

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

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

=> Ans – (A)

8) Answer (D)

Expression : sin 300°

= $sin(360-60)$

= $-sin(60)$

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

=> Ans – (D)

9) Answer (B)

Expression : $\sqrt{1-cos^2A}$

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

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

=> Ans – (B)

10) Answer (C)

Expression : 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)

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11) Answer (B)

Expression : $\sqrt{(sec^2A-1)}=x$

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

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

=> Ans – (B)

12) Answer (B)

Expression : $cos^2A/(1 + sinA)$

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)

13) Answer (A)

Expression : $(cotA – cosecA)^2 = x$

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

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

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

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

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

=> Ans – (A)

14) Answer (D)

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

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

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

=> Ans – (D)

15) Answer (D)

Expression : cos 240° = x

= $cos(180+60)$

= $-cos(60)$

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

=> Ans – (D)

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16) Answer (A)

Expression : 1/√(1+$tan^{2}$A) = x

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

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

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

=> Ans – (A)

17) Answer (C)

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

=> Ans – (C)

18) Answer (C)

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

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

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

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

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

= $secA+tanA$

=> Ans – (C)

19) Answer (D)

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

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

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

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

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

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

=> Ans – (D)

20) Answer (A)

Expression : tan 330° = x

= $tan(360-30)$

= $-tan(30)$

= $\frac{-1}{\sqrt{3}}$

=> Ans – (A)

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