# Trigonometry Questions for SSC CHSL Exam PDF

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## Trigonometry Questions for SSC CHSL Exam PDF:

SSC CHSL Trignometry Questions download PDF based on previous year question paper of SSC CHSL exam. 25 Very important Trignometry questions for SSC CHSL Exam.

Question 1:Â If $\sin x+\frac{1}{\sin x}=2$ then $\cos ^5 x+\cot ^{5} x=?$

a)Â 0
b)Â -1
c)Â 1
d)Â 2

Question 2:Â If $cos^2x + (1 + sinx)^2 = 3$, then find the value of cosec x + sin(60+x), where 0Âº < x < 90Âº?

a)Â 3
b)Â 2
c)Â $2\frac{1}{2}$
d)Â $3\frac{1}{2}$

Question 3:Â If $x$ and $y$ are acute angles such that their sum is less than $90^{\circ}$ . If $\cos(2x-20^{\circ})=\sin(2y+20^{\circ})$ , then the value of $\tan(x+y)$ is

a)Â $0$
b)Â $1$
c)Â $\frac{1}{\sqrt{3}}$
d)Â $\sqrt{3}$

Question 4:Â If x is an acute angle such that
Tan (4x – 50Â°) = Cot ( 50Â° – x), then the value of x would be?

a)Â 60
b)Â 45
c)Â 50
d)Â 30

Question 5:Â If cosec 9x= sec 9x (0 <x<10), what is the value of x?

a)Â 9Â°
b)Â 3Â°
c)Â 5Â°
d)Â 6Â°

Question 6:Â If $\sin x+1=\frac{2}{\sin x}$ then $\cos ^3 x+ \mathrm{cosec} ^{-3} x=?$

a)Â 0
b)Â -1
c)Â 1
d)Â 2

Question 7:Â If $\frac{sin2x}{sinx} + cos2x + sin^2x = 0$, find the value of x, if $0 \leq x \leq 180$.

a)Â 45Âº
b)Â 90Âº
c)Â 180Âº
d)Â More than one value is possible

Question 8:Â If $\frac{sin^2 \Theta}{4} = \frac{cos^2 \Theta}{9}$. Find the value of $tan^2 \Theta – cot^2 \Theta$.

a)Â 1/6
b)Â 13/27
c)Â 1/18
d)Â None of these

Question 9:Â If $\sec^4 x-\tan^4 x=4+4sin^2 x$ then $x=?$ (0<x<90Â°)

a)Â 45Â°
b)Â 30Â°
c)Â 75Â°
d)Â 60Â°

Question 10:Â Simplify the expression $\sqrt{\frac{1+cosx}{1-cosx}}$.

a)Â sec x + tan x
b)Â sin x
c)Â cosec x + cot x
d)Â cos x

Question 11:Â If $sinx = 1 – cos x$, find the value of cos 2x.

a)Â 0
b)Â 1
c)Â $\frac{1}{2}$
d)Â $\frac{\sqrt{3}}{2}$

Question 12:Â If it is known that
$\tan\theta + \cot\theta = 2$, then find the value of $\tan^3\theta + \cot^3\theta$

a)Â $\sqrt{5}$
b)Â 2
c)Â $1 + \sqrt{3}$
d)Â $2 + \sqrt{3}$

Question 13:Â If $\sin x+\frac{1}{\sin x}=\frac{5}{2}$ then $\tan {x}+\cot {x}=?$

a)Â 0
b)Â $\frac{2}{\sqrt3}$
c)Â $\frac{4}{\sqrt3}$
d)Â $\frac{\sqrt3}{2}$

Question 14:Â If $\cos x+\frac{1}{\cos x}=2$ then $\sin ^{5} x+\sec ^{-5} x=?$

a)Â -1
b)Â 0
c)Â 1
d)Â 2

Question 15:Â If $\tan x+\frac{1}{\tan x}=2$ then $\sin ^{5} x+\cos ^{5} x=?$

a)Â $\frac{1}{4\sqrt{2}}$
b)Â $\frac{1}{2\sqrt{2}}$
c)Â $\frac{1}{\sqrt{2}}$
d)Â $\frac{1}{8\sqrt{2}}$

Question 16:Â The maximum value of sin Î¸ + cos Î¸ is

a)Â $1$
b)Â $\sqrt{2}$
c)Â $2$
d)Â $3$

Question 17:Â If $sin \Theta + cosec \Theta = 5$, Find the value of $sin^3 \Theta + cosec ^3 \Theta$.

a)Â 96
b)Â 115
c)Â 110
d)Â None of these

Question 18:Â The value of $cos^2 30^{\circ} + sin^2 60^{\circ} + tan^2 45^{\circ} + sec^2 60^{\circ} + cos0^{\circ}$ is

a)Â $4\frac{1}{2}$
b)Â $5\frac{1}{2}$
c)Â $6\frac{1}{2}$
d)Â $7\frac{1}{2}$

Question 19:Â If $cos x + cos^{2} x = 1,$ then $sin^{8} x + 2 sin^{6} x + sin^{4}$ x is equal to

a)Â 0
b)Â 3
c)Â 2
d)Â 1

Question 20:Â If $x= p$ $cosec Î¸$ and $y= q$ $cot Î¸$, then the value of $\frac{x^2}{p^2}-\frac{y^2}{q^2}$ is

a)Â $sin^{2} Î¸$
b)Â $tanÎ¸$
c)Â $1$
d)Â $0$

Question 21:Â Find the value of tan 4Â° tan 43Â° tan 47 tan 86Â°

a)Â $\frac{2}{3}$
b)Â $1$
c)Â $\frac{1}{2}$
d)Â $2$

Question 22:Â If $x cosÎ¸ – sinÎ¸ = 1,$ then $x^{2} – (1 +x^{2}) sinÎ¸$ equals

a)Â 2
b)Â 1
c)Â – 1
d)Â 0

Question 23:Â If $sin Î¸ + sin^{2} Î¸ = 1$ then $cos^2 Î¸ + cos^4 Î¸$ is equal to

a)Â None
b)Â 1
c)Â $Sin Î¸ /cos^{2} Î¸$
d)Â $cos^{2}Î¸ / Sin Î¸$

Question 24:Â The value of tan1Â°tan2Â°tan3Â° â€¦â€¦â€¦â€¦â€¦tan89Â° is

a)Â 1
b)Â -1
c)Â 0
d)Â None of the options

Question 25:Â If 0Â° â‰¤ A â‰¤ 90Â°, the simplified form of the given expression sin A cos A (tan A – cot A) is

a)Â 1
b)Â 1 – 2 $sin^2$ A
c)Â 2 $sin^2$ A – 1
d)Â 1 – cos A

$\sin x+\frac{1}{\sin x}=2$
=>$\sin^2 x+1=2\sin x$
=>$(\sin x-1)^2=0$
=>$\sin x = 1$
=>$\cos x = 0$
The given expression would be,
$\cos ^5 x+\cot ^{5} x$
$\cos ^5 x+\frac{\cos^5 x}{\sin^5 x}$
$=0$

$cos^2x + (1 + sinx)^2 = 3$
=> $cos^2x + 1 + sin^2x + 2sinx = 3$ (Since $cos^2x + sin^2x = 1$)
=> $2 sin x = 1$ or sin x = Â½ which means x=30 degrees [Solution in the first quadrant]
=> cosec x = 1/sin x = 2 and sin (60+x) = sin 90 = 1
=> The value of the expression is 2+1 = 3

Given that , $x+y<90^{\circ} \dots (1)$
$\implies\cos(2x-20^{\circ})=\cos(90^{\circ}-(2y+20^{\circ}))$
Using the given property in $(1)$
$\implies 2x-20^{\circ}=70^{\circ}-2y$
$\implies 2x+2y=90^{\circ}$
$\implies x+y=45^{\circ}$
$\implies \tan(x+y)=1$
Hence , the correct option is B

We know that Tan ( 90 – x) = Cot x and Cot (90 – x) = Tanx
So 4x – 50 = 90 – y
Then 50 – x will be y
=> y = 50 – x
x + y = 50
Moreover,
90 – y = 4x – 50
=> 4x + y = 140
Subtracting the two equations, we get
3x = 90
=> x = 30
Hence the correct answer is option D.

If cosec 9x= sec 9x,
Then $\sin 9x=\cos 9x$
In the first quadrant cos and sin are equal when $9x =45$Â°
Hence, x = 5Â°

$\sin x+1=\frac{2}{\sin x}$
=>$\sin^2 x+\sin x=2$
=>$(\sin x-1)(\sin x+2)=0$
As $\sin x$ can only take values between -1 and 1,
=>$\sin x = 1$
=>$\cos x = 0$
The given expression would be,
$\cos ^3 x+ \mathrm{cosec}^{-3} x$
$\cos ^3 x + \sin^3 x$
$=1$

$\frac{sin2x}{sinx} + cos2x + sin^2x = 0$
We know that $sin2x = 2 sinx cosx$ and $cos2x=cos^2x-sin^2x$
=> $2cosx + cos^2x = 0$
=> cos x =0 or cos x =2
Now, cos x =2 is not possible.
=> cos x = 0
In the range, $0 \leq x \leq 180$, cos x =0 at only 90Âº
Thus, B is the correct answer.

$\frac{sin^2 \Theta}{4} = \frac{cos^2 \Theta}{9}$
==> $\frac{sin^2 \Theta}{cos^2 \Theta}$ = $\frac{4}{9}$
==> $tan^2 \Theta = \frac{4}{9}$
==> $cot^2 \Theta = \frac{9}{4}$
==> $tan^2 \Theta – cot^2 \Theta$ = 4/9 – 9/4 = -65/36
Since there is no such option, the correct option to choose is D.

$\sec^4 x-\tan^4 x=4+4sin^2 x$
$(sec^2 x-\tan^2 x)( sec^2 x+\tan^2 x)=4+4sin^2 x$
$(1)( sec^2 x+\tan^2 x)=4+4sin^2 x$ (as $sec^2 x-\tan^2 x = 1$)
$\frac{(1+ sin^2 x)}{cos^2 x} = 4(1+sin^2 x)$
$cos^2 x = \frac{1}{4}$
$cos x = \frac{1}{2} or \frac{-1}{2}$
Here, $cosx$ cannot be negative as (0<x<90Â°)

Hence, $cos x = \frac{1}{2}$ and $x = 60$Â°

$\sqrt{\frac{1+cosx}{1-cosx}}$
Multiplying the numerator and denominator by 1+cosx, we get
$\sqrt{\frac{(1+cosx)^2}{(1-cosx)(1+cosx)}}$
=> $\sqrt{\frac{(1+cosx)^2}{(1- cos^2x}}$
=> $\frac{1+cosx}{sinx}$ = cosecx + cotx$11)Â AnswerÂ (B)$sinx = 1 – cos x$=>$sinx + cos x = 1 $Squaring both sides we get,$sin^2x + cos^2x + 2 sinx cos x = 1$=>$1 + sin2x = 1$=> sin 2x = 0 or x=0 => cos 2x = 1 12)Â AnswerÂ (B)$\tan\theta + \cot\theta = 2$Cubing both sides we get$\tan^3\theta + \cot^3\theta + 3\tan\theta*\cot\theta(\tan\theta+\cot\theta) = 8$=>$\tan^3\theta + \cot^3\theta + 3*2 = 8$=>$\tan^3\theta + \cot^3\theta = 2$13)Â AnswerÂ (C)$\sin x+\frac{1}{\sin x}=\frac{5}{2}$=>$2\sin^2 x-5\sin x+2=0$=>$(\sin x-1/2)(\sin x-2)=0$=>$\sin x = 1/2$=>$x=30$degrees =>$\tan x = \frac{1}{\sqrt3}$=>$\cot x = \sqrt3$The given expression would be,$\tan {x}+\cot {x}=\frac{1}{\sqrt3}+\sqrt3=\frac{4}{\sqrt3}$14)Â AnswerÂ (C)$\cos x+\frac{1}{\cos x}=2$=>$\cos^2 x+1=2\cos x$=>$(\cos x-1)^2=0$=>$\cos x = 1$=>$\sin x = 0$The given expression is,$\sin ^5 x+\sec ^{-5} x= \sin ^5 x+\frac{1}{\cos^5 x}=0+1 = 1$15)Â AnswerÂ (B)$\tan x+\frac{1}{\tan x}=2$=>$\tan^2 x+1=2\tan x$=>$(\tan x-1)^2=0$=>$\tan x = 1$=>$\sin x = \frac{1}{\sqrt{2}}$=>$\cos x = \frac{1}{\sqrt{2}}$The given expression would be,$ \sin ^5 x+\cos ^{5} x \sin ^5 x+\cos^5 {x}=\frac{1}{4\sqrt{2}}+\frac{1}{4\sqrt{2}} = \frac{1}{2\sqrt{2}}$16)Â AnswerÂ (B) for asin Î¸ + bcos Î¸ + c, maximum value =$c+\sqrt{a^{2}+b^{2}}$minimum value =$c-\sqrt{a^{2}+b^{2}}$for sin Î¸ + cos Î¸ , a = 1, b = 1, c = 0 maximum value =$c+\sqrt{a^{2}+b^{2}}=0+\sqrt{1^{2}+1^{2}}=\sqrt{2}$so the answer is option B. 17)Â AnswerÂ (C)$sin \Theta + cosec \Theta = 5$==>$(sin \Theta + cosec \Theta)^3 = 5^3$==>$ sin^3 \Theta + cosec ^3 \Theta + 3*sin \Theta * cosec \Theta (sin \Theta + cosec \Theta)$= 125 ==>$sin^3 \Theta + cosec ^3 \Theta + 3*5 = 125$==>$sin^3 \Theta + cosec ^3 \Theta $= 110. 18)Â AnswerÂ (D) Substituting values of angles, we get, 3/4 + 3/4+ 1 + 4 + 1 = 7.5. Option D is the right answer. 19)Â AnswerÂ (D)$cos x + cos^2 x = 1$=>$cos x = 1 – cos^2 x$=>$cos x = sin^2 x\thereforesin^{8} x + 2 sin^{6} x + sin^{4} x$=$(sin^4 x + sin^2 x)^2$=$((cos x)^2 + sin^2 x)^2$=$(cos^2 x + sin^2 x)^2 = 1$20)Â AnswerÂ (C)$x= pcosec Î¸$=>$cosec\theta$=$\frac{x}{p}$Also,$y= qcot Î¸$=>$cot\theta$=$\frac{y}{q}\becausecosec^2\theta-cot^2\theta$= 1 =>$\frac{x^2}{p^2}$–$\frac{y^2}{q^2}$= 1 21)Â AnswerÂ (B) Expression : tan 4Â° tan 43Â° tan 47 tan 86Â°$\becausetan(90-\theta) = cot\theta$=>$tan 4^{\circ} = tan(90^{\circ}-86^{\circ}) = cot 86^{\circ}$Similarly,$tan 43^{\circ} = cot 47^{\circ}$=>$(cot 86^{\circ} \times tan 86^{\circ}) * (tan 47^{\circ} \times cot 47^{\circ})$Using,$tan\theta cot\theta$= 1 => 1*1 = 1 22)Â AnswerÂ (B) Expression :$x cosÎ¸ – sinÎ¸ = 1$=>$x = \frac{1}{cos\theta} + \frac{sin\theta}{cos\theta}$=>$x = sec\theta + tan\theta$————–Eqn(1)$\becausesec^2\theta – tan^2\theta = 1$=>$(sec\theta + tan\theta)(sec\theta – tan\theta) = 1$=>$(sec\theta – tan\theta) = \frac{1}{x}$————–Eqn(2) Adding eqns(1)&(2) =>$2sec\theta = x + \frac{1}{x} = \frac{x^2 + 1}{x}$=>$sec\theta = \frac{x^2 + 1}{2x}$Subtracting eqn(2) from (1) =>$2tan\theta = x – \frac{1}{x} = \frac{x^2 – 1}{x}$=>$tan\theta = \frac{x^2 – 1}{2x}$We know that,$sin\theta = \frac{tan\theta}{sec\theta}$=>$sin\theta = \frac{x^2 – 1}{2x} * \frac{2x}{x^2 + 1}
=> $sin\theta = \frac{x^2 – 1}{x^2 + 1}$
To find : $x^2 – (1 + x^2) sinÎ¸$
= $x^2 – (1 + x^2) * \frac{x^2 – 1}{x^2 + 1}$
= $x^2 – (x^2 – 1)$
= 1

Expression : $sin\theta + sin^2\theta = 1$
=> $sin\theta = 1 – sin^2\theta$
=> $sin\theta = cos^2\theta$
To find : $cos^2\theta + cos^4\theta$
= $cos^2\theta + (cos^2\theta)^2$
= $cos^2\theta + sin^2\theta$
= 1

Expression : tan1Â°tan2Â°tan3Â° â€¦â€¦â€¦â€¦â€¦tan88Â°tan89Â°
$\because$ $tan(90^{\circ}-\theta) = cot\theta$
=> tan 89Â° = tan(90Â°-1) = cot 1Â°
Similarly, tan 88Â° = cot 2Â° and so on till tan 46Â° = cot 44Â°
=> (tan1Â°tan2Â°tan3Â°…….tan45Â°……cot3Â°cot2Â°cot1Â°)
Using, $tan\theta cot\theta$ = 1 and $tan45^{\circ}$ = 1
=> 1*1 = 1

Expression : $sin A cos A (tan A – cot A)$
= $sin A cos A (\frac{sin A}{cos A} – \frac{cos A}{sin A})$
= $sin A cos A (\frac{sin^2 A – cos^2 A}{sin A cos A})$
= $sin^2 A – cos^2 A$
= $sin^2 A – (1 – sin^2 A)$
= $2sin^2 A – 1$