Trigonometry Questions For SSC GD PDF
SSC GD Constable Trigonometry Question paper with answers download PDF based on SSC GD exam previous papers. 40 Very important Trigonometry questions for GD Constable.
Trigonometry QUESTIONS FOR SSC GD PDF
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Question 1: If $\sec \theta +\tan \theta=2$, then the value of $\sin \theta$ is:
a) $\frac{4}{5}$
b) $\frac{\sqrt{3}}{5}$
c) $\frac{2}{5}$
d) $\frac{3}{5}$
Question 2: If $\alpha + \theta$ = $\frac{7\pi}{12}$ and $\tan \theta = \sqrt{3}$, then the value of $\tan \theta$ is:
a) $\sqrt{3}$
b) 1
c) 0
d) $\frac{1}{\sqrt{3}}$
Question 3: If $\cos\theta + \sec\theta = \sqrt{3}$, then the value of $(\cos^{3}\theta + \sec^{3}\theta)$ is:
a) 1
b) $\frac{1}{\sqrt{2}}$
c) 0
d) ${\sqrt{2}}$
Question 4: The value of the following is: $\frac{\sin\theta\ \cos ec\theta\ \tan\theta\ \cot\theta}{\sin^{2}\theta+\cos^{2}\theta}$
a) 1
b) $\tan\theta$
c) 0
d) 2
Question 5: In a right angled triangle ABC and right angled at B, AB=BC, what is the value of sinA$\star$cosC ?
a) $\frac{1}{4}$
b) $\frac{1}{2}$
c) $\frac{1}{6}$
d) $\frac{1}{5}$
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Question 6: If $\sin\theta + \frac{1}{\sin\theta} =\frac{5}{2}$ and $\theta\in Q1$ then what is the value of $\tan\theta ?$
a) $\frac{1}{\sqrt{3}}$
b) $1$
c) $\sqrt{3}$
d) $\frac{1}{\sqrt{2}}$
Question 7: The elevation of the top of a tower from a point on the ground is $\ 45^\circ$. On travelling 60 m from the point towards the tower, the alevation of the top becomes $\ 60^\circ$. The height of the tower, in metres, is
a) 30
b) 30(3- $\ \sqrt{3}$)
c) 30(3 +$\ \sqrt{3}$)
d) 30$\ \sqrt{3}$
Question 8: The expression $\sqrt{\frac{1+sin\ \theta}{1-sin\ \theta}}+\sqrt{\frac{1-sin\ \theta}{1+sin\ \theta}}$ is equal to
a) $2\ sec\ \theta$
b) $2\ tan\ \theta$
c) $\frac{2\ (1-sin\ \theta)}{cos\ \theta}$
d) $2\ sin\ \theta$
Question 9: If $13sinA = 12$ then find out the value of $\frac{3sinA + 4cosA}{3sinA – 4cosA}$? Assume that ($0 \leq A \leq 90$)
a) $\frac{7}{2}$
b) $\frac{7}{4}$
c) $\frac{7}{16}$
d) $\frac{7}{8}$
Question 10: What is the minimum and maximum value that the expression $12Sin\theta – 35cos\theta$ can take?
a) $-23, 47$
b) $-47, 23$
c) $-47, 47$
d) $-37, 37$
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Question 11: If $cos\theta_{1} = \frac{35}{37}$, $sin\theta_{2} = \frac{3}{5}$,then find out $\tan(\theta_{2} – \theta_{1})$ ? (Assume that $0 \leq \theta_{1},\theta_{2} \leq 90$)
a) $\frac{49}{71}$
b) $\frac{57}{76}$
c) $\frac{75}{176}$
d) $\frac{114}{352}$
Question 12: If $tan\theta = \frac{11}{60}$ then find out $\frac{cos\theta – sin\theta}{cos\theta + sin\theta}$ ?
a) $\frac{49}{71}$
b) $\frac{55}{65}$
c) $\frac{65}{55}$
d) $\frac{71}{64}$
Question 13: If $tan θ + cot θ = x$, then what is the value of $tan^4θ + cot^4θ$?
a) $x^2(x^2-4)+6$
b) $x^2(x^2-4)+2$
c) $x^2(x^2-6)+2$
d) $x^2(x^2-2)+2$
Question 14: If $sin^2\theta – cos^2\theta = \frac{16}{25}$, what is the value of $cos^4\theta – sin^4\theta$?
a) $-\frac{4}{25}$
b) $-\frac{16}{5}$
c) $-\frac{16}{25}$
d) $-\frac{4}{5}$
Question 15: If $cosec^2\theta + cot^2\theta = \frac{11}{5}$, what is the value of $cosec^4\theta – cot^4\theta$?
a) $\frac{11}{5}$
b) $\frac{121}{25}$
c) $\frac{5}{11}$
d) $\frac{11}{25}$
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Question 16: If $tan^2\theta + sec^2\theta = \frac{9}{8}$, what is the value of $ tan^4\theta – sec^4\theta $?
a) $-\frac{81}{64}$
b) $-\frac{9}{8}$
c) $-\frac{81}{8}$
d) $-\frac{9}{64}$
Question 17: If $cos θ + sec θ = a$, then what is the value of $cos^6θ + sec^6θ$?
a) $a^6 + 6a^4 + 6a^2 + 2$
b) $a^6 – 6a^4 + 18a^2 – 14$
c) $a^6 – 6a^4 + 18a^2 + 2$
d) $a^6 – 6a^4 + 6a^2 – 2$
Question 18: Shadow of a tower is 60 m long and the towers’s elevation angle is 60° then what is the height of the tower?
a) 120 m
b) $60\sqrt{3}$ m
c) $\frac{60}{\sqrt{3}}$ m
d) None of these
Question 19: If $tanA-cotA =0$ then find out the value of ($tan^4A+cot^4A$) ?
a) 2
b) 4
c) 0
d) None of these
Question 20: If $5tanA = 12$ then find out the value of $\frac{12sinA + 5cosA}{12sinA – 5cosA}$?
a) $\frac{129}{179}$
b) $\frac{179}{129}$
c) $\frac{119}{169}$
d) $\frac{169}{119}$
Question 21: It is given that sinA = $\frac{-\sqrt{3}}{2}$, cosB = $\frac{\sqrt{3}}{2}$ then what is the value of (A+B) ?
(Assume that 180° $\leq$ A, B $\leq$ 360°)
a) 600°
b) 630°
c) 570°
d) More than one of the above
Question 22: It is given that cosA = -$\frac{1}{\sqrt{2}}$, cosB = -$\frac{1}{2}$, then what is the value of (A+B) ?
(Assume that 180° $\leq$ A, B $\leq$ 270°)
a) 465°
b) 435°
c) 150°
d) More than one of the above
Question 23: It is given that sinA = $\frac{\sqrt{3}}{2}$, cosB = $\frac{1}{2}$ then what is the value of (A+B) ?
(Assume that 0° $\leq$ A, B $\leq$ 180°)
a) 120°
b) 90°
c) 150°
d) More than one of the above
Question 24: If $\cos\theta_{1} = \frac{\sqrt{3}}{2}$, $\cos\theta_{2} = \frac{24}{25}$ then find out the value of $\tan(\theta_{1}-\theta_{2})$ ? (Assume that $0 \leq \theta_{1},\theta_{2} \leq 90$)
a) $\frac{7 + 24\sqrt{3}}{7 – 24\sqrt{3}}$
b) $\frac{7 – 24\sqrt{3}}{7 + 24\sqrt{3}}$
c) $\frac{24 – 7\sqrt{3}}{24\sqrt{3} + 7}$
d) $\frac{24 +7\sqrt{3}}{24\sqrt{3} – 7}$
Question 25: If $\cos\theta_{1} = \frac{1}{\sqrt{2}}$, $\cos\theta_{2} = \frac{40}{41}$ then find out the value of $\cos(\theta_{1}-\theta_{2})$ ? (Assume that $0 \leq \theta_{1},\theta_{2} \leq 90$)
a) $\frac{49\sqrt{2}}{82}$
b) $\frac{49}{82\sqrt{2}}$
c) $\frac{41\sqrt{2}}{98}$
d) $\frac{41\sqrt{2}}{49}$
Question 26: If $\sin\theta_{1} = \frac{3}{5}$, $\cos\theta_{2} = \frac{7}{25}$ then find out the value of $\sin(\theta_{1}+\theta_{2})$ ? (Assume that $0 \leq \theta_{1},\theta_{2} \leq 90$)
a) $\frac{99}{125}$
b) $\frac{117}{125}$
c) $\frac{96}{125}$
d) $\frac{107}{125}$
Question 27: $\frac{\sin x}{1 + \cos x} + \frac{1 +\ cos x}{\sin x} = 4$ for 0° < $x$ < 90°
What is the value of $\cot x$
a) $\frac{1}{\sqrt{3}}$
b) $\sqrt{3}$
c) 1
d) 0
Question 28: If $\cos 4x = \sin x$ where, $0 < x < 90$, what is the value of $\sin 5x$?
a) 1
b) 0
c) 0.5
d) 0.33
Question 29: $2a + a^2\tan 2x = \tan 2x$
What is the value of $\sec^2x$?
a) $a^2$
b) $1 + a^2$
c) $a^2 – 1$
d) $2a^2 + 1$
Question 30: If $\cos\theta = \frac{3}{5}$ then find out the value of $\frac{\tan\theta – \cot\theta}{\tan\theta + \cot\theta }$ ? (Assume that $0 \leq \theta \leq 90$)
a) $\frac{16}{25}$
b) $\frac{18}{25}$
c) $\frac{7}{25}$
d) $\frac{9}{25}$
Question 31: If $\frac{\tan^2\theta+3}{\tan^2\theta-3} = 2$, what is the value of $\frac{\sec^2\theta+\cot^2\theta}{\sec^2\theta-\cot^2\theta}$?(Assume $\theta$ is in 1st quadrant)
a) $\frac{89}{91}$
b) $\frac{114}{71}$
c) $\frac{6}{5}$
d) $\frac{91}{89}$
Question 32: If $\cos\theta = \frac{\sqrt{2+\sqrt{3}}}{2}$, then what is the value of $cos\theta +sec\theta$? (Assume that $0 \leq \theta \leq 90$)
a) $\frac{6 +\sqrt{3}}{2\sqrt{2+\sqrt{3}}}$
b) $\frac{3 +\sqrt{3}}{\sqrt{2+\sqrt{3}}}$
c) $\frac{3 +\sqrt{6}}{\sqrt{2+\sqrt{6}}}$
d) $\frac{6 +\sqrt{6}}{\sqrt{3+\sqrt{6}}}$
Question 33: If $\frac{\sec^2\theta+1}{\sec^2\theta-1} = \frac{37}{35}$, what is the value of $\frac{\tan\theta+\sin\theta}{\tan\theta-\sin\theta}$ ?(Assume $\theta$ is in 1st quadrant)
a) $\frac{5}{7}$
b) $\frac{7}{5}$
c) $\frac{7}{6}$
d) $\frac{6}{7}$
Question 34: From the top of 600 m high tower, the angles of depression of the top and bottom of a flag are observed to be 30° and 45° respectively. What is the height of the flag?
a) $200(\sqrt{3} – 1)$
b) $200(\sqrt{3} – \sqrt{2})$
c) $200\sqrt{3}(\sqrt{3} – 1)$
d) None of the above
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Question 35: If $cos\theta – sec\theta = 7$, what is the value of $cos^2\theta +sec^2\theta$?
a) 51
b) 36
c) 49
d) None of the above
Question 36: If $\sin\theta + \cosec\theta = 2$ then find out the value of $\cosec^{32}\theta + \sin^{32}\theta$ ?
a) 32
b) 16
c) 64
d) None of the above
Question 37: If $\sec\theta = \frac{15}{9}$ then find out the value of $\frac{\cot\theta + \cosec\theta}{\cot\theta – \cosec\theta }$ ? (Assume that $0 \leq \theta \leq 90$)
a) 3
b) -3
c) -4
d) 4
Question 38: If sec θ = 25/24, then what is the value of cot θ? (Assume that $0 \leq \theta \leq 90$)
a) $\frac{7}{24}$
b) $\frac{25}{7}$
c) $\frac{24}{7}$
d) $\frac{7}{25}$
Question 39: If cos θ + sec θ = a, then what is the value of $cos^4θ + sec^4θ$ ?
a) $a^2(a^2-4) + 2$
b) $a(a-4) + 2$
c) $(a^3-3)^2+2$
d) $(a^4 – 2a) + 4$
Question 40: If $x + y = z$ and $tan z = 20$, $m = tan x + tan y$ and $n = tan x * tan y$, what is the value of $\frac{m}{1 – n}$? (Assume that $0 \leq x, y, z \leq 90$)
a) 40
b) 20
c) 10
d) 5
Answers & Solutions:
1) Answer (D)
Given : $\sec \theta +\tan \theta=2$ ———–(i)
Also, $sec^2\theta-tan^2\theta=1$
=> $(sec\theta-tan\theta)(sec\theta+tan\theta)=1$
=> $\sec \theta -\tan \theta=\frac{1}{2}$ ———-(ii)
Adding equations (i) and (ii), => $2sec\theta=2+\frac{1}{2}=\frac{5}{2}$
=> $sec\theta=\frac{5}{4}$
=> $cos\theta=\frac{4}{5}$
$\therefore$ $sin\theta=\sqrt{1-cos^2\theta}$
= $\sqrt{1-(\frac{4}{5})^2}=\sqrt{1-\frac{16}{25}}$
= $\sqrt{\frac{9}{25}}=\frac{3}{5}$
=> Ans – (D)
2) Answer (B)
3) Answer (C)
Given : $\cos\theta + \sec\theta = \sqrt{3}$ ———-(i)
Cubing both sides, we get :
=> $(\cos\theta + \sec\theta)^3 = (\sqrt{3})^3$
=> $cos^3\theta+sec^3\theta+3(cos\theta)(sec\theta)(cos\theta+sec\theta)=3\sqrt3$
=> $cos^3\theta+sec^3\theta+3(cos\theta\times sec\theta)(\sqrt3)=3\sqrt3$
$\because$ $cos\theta\times sec\theta=1$ and using equation (i),
=> $cos^3\theta+sec^3\theta=3\sqrt3-3\sqrt3=0$
=> Ans – (C)
4) Answer (A)
Expression : $\frac{\sin\theta\ \cos ec\theta\ \tan\theta\ \cot\theta}{\sin^{2}\theta+\cos^{2}\theta}$
= $\frac{(sin\theta\times\frac{1}{sin\theta})\times(tan\theta\times\frac{1}{tan\theta})}{1}$
= $1$
=> Ans – (A)
5) Answer (B)
Given is a right angled triangle at B.
Since it is a right angled triangle $AC^{2}=AB^{2}+BC^{2}$.
Given AB=BC.
$\therefore AC=\sqrt{2}\star BC $
sinA=$\frac{BC}{AC}$
cosC=$\frac{BC}{AC}$
sinA$\star$cosC=$\frac{BC^{2}}{AC^{2}}$
Substituting $AC$=$\sqrt{2}\star BC$
sinA$\star$cosC=$\frac{1}{2}$
6) Answer (A)
Given $\sin\theta +\frac{1}{\sin\theta}$=$\frac{5}{2}$
$\sin^{2}\theta+1=(\frac{5}{2})\sin\theta$
$2\sin^{2}\theta-5\sin\theta+1=0$
$(\sin\theta-2)(2\sin\theta-1)=0$
Therefore $\sin\theta=\frac{1}{2}$
$\therefore \theta=30$
$\tan30=\frac{1}{2}$.
7) Answer (C)
From $\triangle$ ACD,
tan $60^\circ = \frac{AD}{CD}$
$\Rightarrow AD = CD\sqrt{3}$
From $\triangle$ ABD,
tan $45^\circ = \frac{AD}{BD}$
$\Rightarrow$ AD = BD
$\Rightarrow$ AD = BC+CD
$\Rightarrow$ $AD = 60+\frac{AD}{\sqrt{3}}$
$\Rightarrow$ $AD-\frac{AD}{\sqrt{3}} = 60$
$\Rightarrow \frac{\sqrt{3}AD-AD}{\sqrt{3}} = 60$
$AD(\sqrt{3}-1) = 60\sqrt{3}$
$AD = \frac{60\sqrt{3}}{\sqrt{3}-1}$
Rationalising above equation
$AD = \frac{60\sqrt{3}}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$AD = \frac{60\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}}$
$\therefore AD = 30(\sqrt{3}+3)$
8) Answer (A)
Expression : $\sqrt{\frac{1+sin\ \theta}{1-sin\ \theta}}+\sqrt{\frac{1-sin\ \theta}{1+sin\ \theta}}$
= $\frac{(\sqrt{1+sin\theta})^2+(\sqrt{1-sin\theta})^2}{(\sqrt{1-sin\theta})(\sqrt{1+sin\theta})}$
= $\frac{(1+sin\theta)+(1-sin\theta)}{\sqrt{1-sin^2\theta}}$
= $\frac{2}{cos\theta}=2sec\theta$
=> Ans – (A)
9) Answer (A)
It is given that $sinA = \frac{12}{13}$
$\Rightarrow$ $\cos^2A = \sqrt{1 – \sin^2A }$
$\Rightarrow$ $\cos^2A = \sqrt{1 – (\frac{12}{13})^2}$
$\Rightarrow$ $\cos^2A = \frac{25}{169}$
$\Rightarrow$ $cosA$ = $\frac{5}{13}$ ($0 \leq A \leq 90$)
We have to find out $\frac{3sinA + 4cosA}{3sinA – 4cosA}$
$\Rightarrow$ $\frac{3\times \frac{12}{13} + 4\times \frac{5}{13}}{3\times \frac{12}{13} – 4\times \frac{5}{13}}$
$\Rightarrow$ $\frac{36+20}{36-20}$
$\Rightarrow$ $\frac{7}{2}$
Hence option A is the correct answer.
10) Answer (D)
We know that the minimum and maximum values of the expression $ASin\theta + Bcos\theta$ is given by:
Minimum value =$-\sqrt{A^2+B^2}$
Maximum value =$\sqrt{A^2+B^2}$
So in the given question: (A=12, B=-35)
Minimum value =$ -\sqrt{A^2+B^2} = -\sqrt{(12)^2+(-35)^2} = -37$
Maximum value =$ \sqrt{A^2+B^2} = \sqrt{(12)^2+(-35)^2} = 37$
So, the minimum and maximum value of the expression will be $-37$ and $37$.
Therefore, option D is the right answer.
11) Answer (D)
It is given that $cos\theta_{1} = \frac{35}{37}$
$\Rightarrow$ $\sin^2\theta_{1} = \sqrt{1 – \cos^2\theta_{1} }$
$\Rightarrow$ $\sin^2\theta_{1} = \sqrt{1 – (\frac{35}{37})^2}$
$\Rightarrow$ $\sin^2\theta_{1} = \frac{144}{1369}$
$\Rightarrow$ $\sin\theta_{1} = \frac{12}{37}$ ($0 \leq \theta \leq 90$)
Also $\tan\theta_{1} = \frac{\sin\theta_{1} }{cos\theta_{1} }$
$\Rightarrow$ $\tan\theta_{1} = \frac{\frac{12}{37}}{\frac{35}{37}}$
$\Rightarrow$ $\tan\theta_{1} = \frac{12}{35}$ … (1)
Also $\cos^2\theta_{2} = \sqrt{1 – \sin^2\theta_{2} }$
$\Rightarrow$ $\cos^2\theta_{2} = \sqrt{1 – (\frac{3}{5})^2}$
$\Rightarrow$ $\cos\theta_{2} = \frac{4}{5}$
Also $\tan\theta_{2} = \frac{\sin\theta_{1} }{cos\theta_{1}} = \frac{3}{4}$ …(2)
$\tan(\theta_{2} – \theta{1})$ = $\frac{\tan\theta_{2} – \tan\theta_{1}}{1 + \tan\theta_{2}*\tan\theta_{1}}$
$\Rightarrow$ $\frac{\frac{3}{4} – \frac{12}{35}}{1+\frac{3}{4}*\frac{12}{35}}$
$\Rightarrow$ $\frac{57}{176}$
Multiplying the numerator and denominator by 2
$\Rightarrow$ $\frac{114}{352}$
Hence option D is the correct answer.
12) Answer (A)
It is given that $tan\theta = \frac{11}{60}$ … (1)
$\Rightarrow$ $\frac{cos\theta – sin\theta}{cos\theta + sin\theta}$
Dividing both numerator and denominator by $cos\theta$
$\Rightarrow$ $\frac{1 – tan\theta}{1 + tan\theta}$
$\Rightarrow$ $\frac{1 – \frac{11}{60}}{1 + \frac{11}{60}}$
$\Rightarrow$ $\frac{\frac{60-11}{60}}{\frac{60+11}{60}}$
$\Rightarrow$ $\frac{49}{71}$
Hence option A is the correct answer.
13) Answer (B)
Given that $tan θ + cot θ = x$
Squaring on both sides
$\Rightarrow$ $(tan θ + cot θ)^2 = x^2$
$\Rightarrow$ $tan^2θ + cot^2θ+2tanθ\times cotθ = x^2$
$\Rightarrow$ $tan^2θ + cot^2θ = x^2 – 2$ (We know that $tanθ\times cotθ = 1$)
Squaring on both sides
$\Rightarrow$ $(tan^2θ + cot^2θ)^2 = (x^2 – 2)^2$
$\Rightarrow$ $tan^4θ + cot^4θ + 2tan^2θ\times cot^2θ = x^4-4x^2+4$
$\Rightarrow$ $tan^4θ + cot^4θ +2=x^4-4x^2+4$ (We know that $tanθ\times cotθ = 1$)
$\Rightarrow$ $tan^4θ + cot^4θ =x^2(x^2-4)+4-2$
$\Rightarrow$ $tan^4θ + cot^4θ =x^2(x^2-4)+2$.
Therefore, option B is the right answer.
14) Answer (C)
It is given that $sin^2\theta – cos^2\theta = \frac{16}{25}$ … (1)
We know that $sin^2\theta + cos^2\theta = 1$ … (2)
On multiplying both the equations, we get,
$( sin^2\theta – cos^2\theta)*(sin^2\theta + cos^2\theta) = \frac{16}{25} * 1$
$\Rightarrow$ $sin^4\theta – cos^4\theta = \frac{16}{25}$
$\Rightarrow$ $cos^4\theta – sin^4\theta = -\frac{16}{25}$
Therefore, option C is the right answer.
15) Answer (A)
It is given that $cosec^2\theta + cot^2\theta = \frac{11}{5}$ … (1)
We know that $cosec^2\theta – cot^2\theta = 1$ … (2)
On multiplying both the equations, we get,
$\Rightarrow$ $(cosec^2\theta + cot^2\theta)*( cosec^2\theta – cot^2\theta) = \frac{11}{5} * 1$
$\Rightarrow$ $cosec^4\theta – cot^4\theta = \frac{11}{5}$
Therefore, option A is the right answer.
16) Answer (B)
It is given that $tan^2\theta + sec^2\theta = \frac{9}{8}$ … (1)
We know that $sec^2\theta – tan^2\theta = 1$ … (2)
On multiplying both the equations , we get,
$( tan^2\theta + sec^2\theta)*( sec^2\theta – tan^2\theta) = \frac{9}{8} * 1$
$\Rightarrow$ $sec^4\theta – tan^4\theta = \frac{9}{8}$
$\Rightarrow$ $tan^4\theta – sec^4\theta = \frac{-9}{8}$
Therefore, option B is the right answer.
17) Answer (D)
Given that $cos θ + sec θ = a$
Squaring on both sides
$\Rightarrow$ $(cos θ + sec θ)^2 = a^2$
$\Rightarrow$ $cos^2θ + sec^2θ+2cosθ\times secθ = a^2$
$\Rightarrow$ $cos^2θ + sec^2θ = a^2 – 2$ … (1) (We know that $cosθ\times secθ = 1$)
Taking cube on both sides
$\Rightarrow$ $(cos^2θ + sec^2θ)^3 = (a^2 – 2)^3$
$\Rightarrow$ $cos^6θ + 3cos^4θ\times sec^2θ + 3cos^2θ\times sec^4θ + sec^6θ$ = $a^6 – 6a^4 + 12 a^2 – 8$
$\Rightarrow$ $cos^6θ + sec^6θ$ = $a^6 – 6a^4 + 12 a^2 – 8$ $- 3cos^2θ\times sec^2θ ( cos^2θ + sec^2θ )$
$\Rightarrow$ $cos^6θ + sec^6θ$ = $a^6 – 6a^4 + 12 a^2 – 8 – 3(a^2 – 2)$ (From equation 1)
$\Rightarrow$ $cos^6θ + sec^6θ = a^6 – 6a^4 + 6a^2 – 2$
Therefore option D is the correct answer.
18) Answer (B)
In the given diagram PR is the tower and RS is shadow.
$\Rightarrow$ tan60° = $\frac{PR}{QR}$
$\Rightarrow$ $\sqrt{3} = \frac{PR}{60}$
$\Rightarrow$ $PR = 60\sqrt{3}$ m
Therefore the height of the tower is $60\sqrt{3}$ m. Hence we can say that option B is the correct answer.
19) Answer (A)
Given that
$\Rightarrow$ $tanA-cotA =0$
Squaring on both sides
$\Rightarrow$ $(tanA-cotA)^2 = 0$
$\Rightarrow$ $tan^2A+cot^2A-2tanA*cotA = 0$ (We know that $\tan\theta\times \cot\theta = 1$)
$\Rightarrow$ $tan^2A+cot^2A = 2$
Now squaring again on both the sides
$\Rightarrow$ $(tan^2A+cot^2A)^2 = 2^2$
$\Rightarrow$ $tan^4A+cot^4A+2tan^2A\times cot^2A = 4$
$\Rightarrow$ $tan^4A+cot^4A+2(tanA\times cotA)^2 = 4$ (We know that $\tan\theta\times \cot\theta = 1$)
$\Rightarrow$ $tan^4A+cot^4A = 4 – 2$
$\Rightarrow$ $tan^4A+cot^4A = 2$
Therefore option A is the correct answer.
20) Answer (D)
Given that
$5tanA = 12$
$tanA = \frac{12}{5}$
We have to find out the value of $\frac{12sinA + 5cosA}{12sinA – 5cosA}$
$\Rightarrow$ $\frac{12sinA + 5cosA}{12sinA – 5cosA}$
Dividing numerator and denominator by cosA
$\Rightarrow$ $\frac{12(\frac{sinA}{cosA}) + 5(\frac{cosA}{cosA})}{12(\frac{sinA}{cosA}) – 5(\frac{cosA}{cosA})}$
$\Rightarrow$ $\frac{12tanA + 5}{12tanA – 5}$
$\Rightarrow$ $\frac{12(\frac{12}{5}) + 5}{12(\frac{12}{5}) – 5}$
$\Rightarrow$ $\frac{\frac{144 + 25}{5}}{\frac{144 – 25}{5}}$
$\Rightarrow$ $\frac{169}{119}$
Hence option D is the correct answer.
21) Answer (D)
Given that
sinA = $\frac{-\sqrt{3}}{2}$
Therefore A = (180°+ 60°) or (360° – 60°)
$\Rightarrow$ A = 240° or 300°
Similarly cosB = $\frac{\sqrt{3}}{2}$
Therefore B = 330°
Hence A+B = 240°+330° or 300°+330°
$\Rightarrow$ (A+B) = 570° or 630°
Hence option D is correct answer.
22) Answer (A)
Given that
cosA = -$\frac{1}{\sqrt{2}}$
Therefore A = 180°+ 45° = 225°
Similarly cosB = -$\frac{1}{2}$
Therefore B = 180° + 60° = 240°
Hence A+B = 225°+240 = 465°
Hence option A is correct answer.
23) Answer (A)
Given that
sinA = $\frac{\sqrt{3}}{2}$
Therefore A = 60° or 120°
Similarly cosB = $\frac{1}{2}$
Therefore B = 60°
Hence A+B = 60°+60° or 120°+60°
$\Rightarrow$ (A+B) = 120° or 180°
Hence option A is correct answer.
24) Answer (C)
Given that $\cos\theta_{1} = \frac{\sqrt{3}}{2}$
We know that $\sin\theta = \sqrt{1-\cos^2\theta}$
Therefore $\sin\theta_{1} = \sqrt{1-\cos^2\theta_{1}}$
$\Rightarrow$ $\sin\theta_{1} = \sqrt{1-(\frac{\sqrt{3}}{2})^2}$
$\Rightarrow$ $\sin\theta_{1} = \frac{1}{2}$
Therefore $\tan\theta_{1} = \frac{\sin\theta_{1}}{\cos\theta_{1}}$
$\Rightarrow$ $\tan\theta_{1} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$
$\Rightarrow$ $\tan\theta_{1} = \frac{1}{\sqrt{3}}$
Also $\sin\theta_{2} = \sqrt{1-\cos^2\theta_{2}}$
$\Rightarrow$ $\sin\theta_{2} = \sqrt{1-(\frac{24}{25})^2}$
$\Rightarrow$ $\sin\theta_{2} = \frac{7}{25}$
$\tan\theta_{2} = \frac{\sin\theta_{2}}{\cos\theta_{2}}$
$\Rightarrow$ $\tan\theta_{2} = \frac{\frac{7}{25}}{\frac{24}{25}}$
$\Rightarrow$ $\tan\theta_{2} = \frac{7}{24}$
We have to find out the value of $\tan(\theta_{1}-\theta_{2})$
$\Rightarrow$ $\frac{\tan\theta_{1}-\tan\theta_{2}}{1+\tan\theta_{1}\times \tan\theta_{2}}$
$\Rightarrow$ $\frac{\frac{1}{\sqrt{3}}-\frac{7}{24}}{1+\frac{1}{\sqrt{3}}\times \frac{7}{24}}$
$\Rightarrow$ $\frac{\frac{24 – 7\sqrt{3}}{24\sqrt{3}}}{\frac{24\sqrt{3} + 7}{24\sqrt{3}}}$
$\Rightarrow$ $\frac{24 – 7\sqrt{3}}{24\sqrt{3} + 7}$
Hence option C is the correct answer.
25) Answer (A)
Given that $\cos\theta_{1} = \frac{1}{\sqrt{2}}$
We know that $\sin\theta = \sqrt{1-\cos^2\theta}$
Therefore $\sin\theta_{1} = \sqrt{1-\cos^2\theta_{1}}$
$\Rightarrow$ $\sin\theta_{1} = \sqrt{1-(\frac{1}{\sqrt{2}})^2}$
$\Rightarrow$ $\sin\theta_{1} = \frac{1}{\sqrt{2}}$
Also $\sin\theta_{2} = \sqrt{1-\cos^2\theta_{2}}$
$\Rightarrow$ $\sin\theta_{2} = \sqrt{1-(\frac{40}{41})^2}$
$\Rightarrow$ $\sin\theta_{2} = \frac{9}{41}$
We have to find out the value of $\cos(\theta_{1}-\theta_{2})$
$\Rightarrow$ $\cos\theta_{1}\cos\theta_{2} + \sin\theta_{1}\sin\theta_{2}$
$\Rightarrow$ $\frac{1}{\sqrt{2}}\times \frac{40}{41} + \frac{1}{\sqrt{2}}\times \frac{9}{41}$
$\Rightarrow$ $\frac{40 + 9}{41\sqrt{2}}$
$\Rightarrow$ $\frac{49\sqrt{2}}{82}$
Hence option A is the correct answer.
26) Answer (B)
Given that $\sin\theta_{1} = \frac{3}{5}$
We know that $\cos\theta = \sqrt{1-\sin^2\theta}$
Therefore $\cos\theta_{1} = \sqrt{1-\sin^2\theta_{1}}$
$\Rightarrow$ $\cos\theta_{1} = \sqrt{1-(\frac{3}{5})^2}$
$\Rightarrow$ $\cos\theta_{1} = \frac{4}{5}$
Also $\sin\theta = \sqrt{1-\cos^2\theta}$
Therefore $\sin\theta_{2} = \sqrt{1-\cos^2\theta_{2}}$
$\Rightarrow$ $\sin\theta_{2} = \sqrt{1-(\frac{7}{25})^2}$
$\Rightarrow$ $\sin\theta_{2} = \frac{24}{25}$
We have to find out the value of $\sin(\theta_{1}+\theta_{2})$
$\Rightarrow$ $\sin\theta_{1}\cos\theta_{2} + \cos\theta_{1}\sin\theta_{2}$
$\Rightarrow$ $\frac{3}{5}\times \frac{7}{25} + \frac{4}{5}\times \frac{24}{25}$
$\Rightarrow$ $\frac{21 + 96}{125}$
$\Rightarrow$ $\frac{117}{125}$
Hence option B is the correct answer.
27) Answer (B)
$\frac{\sin x}{1 + \cos x} + \frac{1 +\ cos x}{\sin x} = 4$
=>$\frac{\sin^2x + (1 + \cos x)^2}{\sin x(1 + \cos x)} = 4$
=>$\frac{\sin^2x + 1 + \cos^2x + 2\cos x}{\sin x(1 + \cos x)} = 4$
=>$\frac{2 + 2\cos x}{\sin x(1 + \cos x)} = 4$
=>$\frac{2 + 2\cos x}{\sin x(1 + \cos x)} = 4$
=>$\frac{2(1 + \cos x)}{\sin x(1 + \cos x)} = 4$
=>$2\cosec x = 4$
=>$\cosec x = 2$
=>$\sin x = \frac{1}{2}$
=> $x = 30$°
so, $\cot x = \cot 30$° = $\sqrt{3}$
Hence, option B is the correct answer.
28) Answer (A)
$\cos 4x = \sin x$
=>$\cos 4x = \cos(90$°$ – x)$
=>$4x = 90$°$ – x$
=>$x = 18$°$$
So, $\sin 5x = \sin 90$°$ = 1$
Hence, option A is the correct answer.
29) Answer (B)
$2a + a^2\tan 2x = \tan 2x$
=>$2a = \tan 2x – a^2\tan 2x$
=>$\tan 2x = \frac{2a}{1 – a^2}$
we know that $\tan 2x$ = $\frac{2\tan x}{1 – \tan^2x}$
On comparing we get $\tan x = a$
$\sec^2x = 1 + \tan^2x = 1 + a^2$
Hence, option B is the correct answer.
30) Answer (C)
We are given that $\cos\theta = \frac{3}{5}$
$\Rightarrow$ $\sin^2\theta = \sqrt{1 – \cos^2\theta}$
$\Rightarrow$ $\sin^2\theta = \sqrt{1 – (\frac{3}{5})^2}$
$\Rightarrow$ $\sin^2\theta = \frac{16}{25}$
$\Rightarrow$ $\sin\theta = \frac{4}{5}$ ($0 \leq \theta \leq 90$)
Therefore $\tan\theta = \frac{\sin\theta}{cos\theta}$
$\Rightarrow$ $\tan\theta = \frac{\frac{4}{5}}{\frac{3}{5}}$
$\Rightarrow$ $\tan\theta = \frac{4}{3}$ … (1)
We can easily start $\cot\theta=\frac{3}{4}$ … (2)
We have to find out, $\frac{\frac{4}{3} – \frac{3}{4}}{\frac{4}{3} + \frac{3}{4} }$
$\Rightarrow$ $\frac{16 – 9}{16 + 9}$
$\Rightarrow$ $\frac{7}{25}$
Hence option C is the correct answer.
31) Answer (D)
Given that $\frac{\tan^2\theta+3}{\tan^2\theta-3} = 2$.
Assuming $\tan^2\theta = t $, we get,
$\frac{t + 3}{t – 3} = 2$
$\Rightarrow$ $t = 9$
$\Rightarrow$ $\tan^2\theta=9$
$\Rightarrow$ $\tan\theta = 3$ ($\theta$ is in 1st quadrant)
We know that $\sec^2\theta – \tan^2\theta = 1$
$\Rightarrow$ $\sec^2\theta = 1 + \tan^2\theta $
Also $\cot^2\theta = \frac{1}{\tan^2\theta}$
We have to find out $\frac{\sec^2\theta+\cot^2\theta}{\sec^2\theta-\cot^2\theta}$
$\Rightarrow$ $\frac{1 + \tan^2\theta+\frac{1}{\tan^2\theta}}{1 + \tan^2\theta – \frac{1}{\tan^2\theta}}$
$\Rightarrow$ $\frac{1 + 9 +\frac{1}{9}}{1 + 9 – \frac{1}{9}}$
$\Rightarrow$ $\frac{91}{89}$
Hence option D is the correct answer.
32) Answer (A)
Given that $cos\theta = \frac{\sqrt{2+\sqrt{3}}}{2}$
Then,
$\Rightarrow$ $cos\theta +sec\theta$
$\Rightarrow$ $cos\theta +\frac{1}{cos\theta}$
$\Rightarrow$ $\frac{cos^2\theta +1}{cos\theta}$
$\Rightarrow$ $\frac{(\frac{\sqrt{2+\sqrt{3}}}{2})^2 + 1}{\frac{\sqrt{2+\sqrt{3}}}{2}}$
$\Rightarrow$ $\frac{\frac{2+\sqrt{3}}{4}+ 1}{\frac{\sqrt{2+\sqrt{3}}}{2}}$
$\Rightarrow$ $\frac{6 +\sqrt{3}}{2\sqrt{2+\sqrt{3}}}$
33) Answer (B)
Given that $\frac{\sec^2\theta+1}{\sec^2\theta-1} = \frac{37}{35}$.
Assuming $\sec^2\theta = t $, we get,
$\frac{t + 1}{t – 1} = \frac{37}{35}$
$\Rightarrow$ $t = 36$
$\Rightarrow$ $\sec^2\theta=36$
$\Rightarrow$ $\sec\theta = 6$ ($\theta$ is in 1st quadrant)
We have to find out value of $\frac{\tan\theta+\sin\theta}{\tan\theta-\sin\theta}$.
$\frac{\tan\theta+\sin\theta}{\tan\theta-\sin\theta}$ = $\frac{\sec\theta+1}{\sec\theta-1}$ (Dividing numerator and denominator by $\sin\theta$)
$\Rightarrow$ $\frac{6 + 1}{6 – 1}$
$\Rightarrow$ $\frac{7}{5}$.
Therefore, option B is the right answer.
34) Answer (C)
Let us assume that height of flag is h meters.
In triangle AEF,
tan(45) = $\frac{AE}{EF}$ = $1$
AE =EF =600 m
We can see that BC = EF
Hence BC = 600m … (1)
In triangle ABC,
tan(30) = $\frac{AB}{BC}$ = $\frac{1}{\sqrt{3}}$
tan(30) = $\frac{600-h}{BC}$ =$\frac{1}{\sqrt{3}}$
$\Rightarrow$ $\frac{600-h}{600}$ =$\frac{1}{\sqrt{3}}$
$\Rightarrow$ $h$ =$600 – 200\sqrt{3}$
$\Rightarrow$ $h$ =$200\sqrt{3}(\sqrt{3} – 1)$
Hence, option C is the right choice.
35) Answer (A)
Given that $\cos\theta – \sec\theta = 7$
Squaring on both sides
$\Rightarrow$ $(\cos\theta – \sec\theta)^2 = 7^2$
$\Rightarrow$ $\cos^2\theta – 2\times\cos\theta\sec\theta + \sec^2\theta = 49$ (We know that $\cos\theta\sec\theta = 1$)
$\Rightarrow$ $\cos^2\theta – 2\times1 + \sec^2\theta = 49$
$\Rightarrow$ $\cos^2\theta + \sec^2\theta = 49 + 2$
$\Rightarrow$ $\cos^2\theta + \sec^2\theta = 51$
Therefore, option A is the right answer.
36) Answer (D)
Given that $\sin\theta + \cosec\theta = 2$ (We know that $\cosec\theta = \frac{1}{\sin\theta}$)
$\Rightarrow$ $\sin\theta + \frac{1}{\sin\theta}= 2$
$\Rightarrow$ $\sin^2\theta – 2\times\sin\theta + 1 = 0$
$\Rightarrow$ $(\sin\theta – 1)^2 = 0$
$\therefore$ $\sin\theta = +1 or -1$
Since $\sin\theta = +1 or -1$, $\cosec\theta$ is also equal to $-1 or +1$.
Hence, the value of $\cosec^{32}\theta + \sin^{32}\theta$ = $1 + 1$ = $2$.
Therefore, option D is the right answer.
37) Answer (C)
Given that $\sec\theta$ = $\frac{15}{9}$
$\cos\theta$ = $\frac{9}{15}$
$\Rightarrow$ $\sin\theta = \sqrt{1- \cos^2\theta}$
$\Rightarrow$ $\sin\theta = \sqrt{1- (\frac{9}{15})^2}$
$\Rightarrow$ $\sin\theta = \frac{12}{15}$
$\Rightarrow$ $\cosec\theta = \frac{15}{12}$ (Inverse of $\sin\theta$)
$\Rightarrow$ $\cot\theta = \frac{\cos\theta}{\sin\theta}$
$\Rightarrow$ $\cot\theta = \frac{\frac{9}{15}}{\frac{12}{15}}$
$\Rightarrow$ $\cot\theta = \frac{9}{12}$
We have to find
$\Rightarrow$ $\frac{\cot\theta + \cosec\theta}{\cot\theta – \cosec\theta }$
$\Rightarrow$ $\frac{\frac{9}{12} + \frac{15}{12}}{\frac{9}{12} – \frac{15}{12}}$
$\Rightarrow$ $\frac{9 + 15 }{9 – 15}$
$\Rightarrow$ $-4$
Therefore, option C is the right answer.
38) Answer (C)
Given that: $sec\theta=\frac{25}{24}$
$cos\theta=\frac{24}{25}$
Using, $sin^2\theta+cos^2\theta =1$
$\Rightarrow sin^2\theta=1-(\frac{24}{25})^2= \frac{625-576}{625}$
$\Rightarrow sin^2\theta=\frac{49}{625}$
$\Rightarrow sin\theta=\frac{7}{25}$
We know that $cot\theta = \frac{cos\theta}{sin\theta}$
$\Rightarrow cot \theta =\frac{\frac{24}{25}}{\frac{7}{25}}$
$\Rightarrow cot \theta =\frac{24}{7} $.
Therefore, option C is the right answer.
39) Answer (A)
Given that cos θ + sec θ = a
Squaring both sides,
$\Rightarrow$ $(cos θ + sec θ)^2=(a)^2$
$\Rightarrow$ $cos^2\theta+sec^2\theta+2(cos\theta)(sec\theta)=a^2$
$\Rightarrow$ $cos^2\theta+sec^2\theta+2=a^2$ [$\because cos\theta sec\theta=1$]
$\Rightarrow$ $cos^2\theta+sec^2\theta=a^2-2$
Again squaring both sides, we get :
$\Rightarrow$ $(cos^2\theta+sec^2\theta)^2=(a^2-2)^2$
$\Rightarrow$ $cos^4\theta+sec^4\theta+2(cos^2\theta)(sec^2\theta)=a^4-4a^2+4$
$\Rightarrow$ $cos^4\theta+sec^4\theta+2=a^4-4a^2+4$
$\Rightarrow$ $cos^4\theta+sec^4\theta=a^2(a^2-4)+4-2$
$\Rightarrow$ $cos^4\theta+sec^4\theta=a^2(a^2-4)+2$.
Therefore, option A is the right answer.
40) Answer (B)
$x + y = z$
$tan z = tan (x + y)$ = $\frac{tan x + tan y}{1 – tan x * tan y}$
$\Rightarrow$ $tan z = \frac{m}{1 -n}$
$\Rightarrow$ $\frac{m}{1 -n}$ = 20
Hence, option B is correct.
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