Trigonometry Questions for SSC CHSL set-3 PDF
Download SSC CHSL Trigonometry Questions with answers set-3 PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Questions for SSC Exams.
Download Trigonometry Questions for SSC CHSL set-3 PDF
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Question 1: If $sin θ + sin^{2} θ = 1$ then $cos^2 θ + cos^4 θ$ is equal to
a)Â None
b)Â 1
c) $Sin θ /cos^{2} θ$
d) $cos^{2}θ / Sin θ$
Question 2:Â The numerical value of $\frac{cos^2 45\circ}{sin^2 60\circ}+\frac{cos^2 60\circ}{sin^2 45\circ}-\frac{tan^230\circ}{cot^245\circ}-\frac{sin^230\circ}{cot^230\circ}$ is
a)Â $1 \frac{1}{4}$
b)Â $\frac{3}{4}$
c)Â $\frac{1}{4}$
d)Â $\frac{1}{2}$
Question 3: The value of tan1°tan2°tan3° ……………tan89° is
a)Â 1
b)Â -1
c)Â 0
d)Â None of the options
Question 4:Â If $\frac{\cos\alpha}{\sin\beta}=n$ and $\frac{\cos\alpha}{\cos\beta}=m$ then the value of $\cos^{2} \beta$ is
a)Â $\frac{m^2}{m^2+n^2}$
b)Â $\frac{1}{m^2+n^2}$
c)Â $\frac{n^2}{m^2+n^2}$
d)Â 0
Question 5: If 0° ≤ A ≤ 90°, 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
Question 6: If θ is an acute angle and $\tan^2\theta+\frac{1}{\tan^2\theta}=2$ then the value of θ is :
a) 60°
b) 45°
c) 15°
d) 30°
Question 7: If tan θ + cot θ = 5, then $tan^2 θ + cot^2 θ$ is
a)Â 23
b)Â 25
c)Â 26
d)Â 24
Question 8: A person of height 6ft. wants to pluck a fruit which is on a 26/3 ft. high tree. If the person is standing 8/√3 ft. away from the base of the tree, then at what angle should he throw a stone so that it hits the fruit ?
a) 75°
b) 30°
c) 45°
d) 60°
Question 9: The value of $sin^{2}$ 22° + $sin^{2}$ 68° + $cot^{2}$ 30° is
a)Â $4$
b)Â $3$
c)Â $\frac{3}{4}$
d)Â $\frac{5}{4}$
Question 10: The minimum value of $2sin^{2}$ θ + $3cos^{2}$ θ is
a)Â 3
b)Â 4
c)Â 2
d)Â 1
Question 11: If θ be acute angle and tan (4θ – 50°) = cot(50° – θ), then the value of 9 in degrees is:
a)Â 20
b)Â 50
c)Â 40
d)Â 30
Question 12: If secθ + tanθ = p, (p ≠0) then secθ is equal to
a)Â $[p-\frac{1}{p}], p\neq 0$
b)Â $2[p-\frac{1}{p}], p\neq 0$
c)Â $[p+\frac{1}{p}], p\neq 0$
d)Â $\frac{1}{2}[p+\frac{1}{p}], p\neq 0$
Question 13:Â The value of $sin^{2} 30^{\circ} cos^{2} 45^{\circ}$ + $5tan^{2} 30^{\circ}$ + $\frac{3}{2} sin^{2} 90^{\circ}$ – $3 cos^{2} 90^{\circ}$ is
a)Â $3\frac{7}{24}$
b)Â $3\frac{3}{24}$
c)Â $3\frac{1}{24}$
d)Â $3\frac{5}{24}$
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Question 14: If $cos^{2} θ – sin^{2} θ = \frac{1}{3}$ , where 0 ≤ θ ≤ Ï€/2 then the value of $cos^{4} θ – sin^{4} θ$ is
a)Â $\frac{1}{3}$
b)Â $\frac{2}{3}$
c)Â $\frac{1}{9}$
d)Â $\frac{2}{9}$
Question 15: If tanθ = 1/√11 0 < θ < π/2, then the value of $\frac{cosec^{2}\theta-\sec^2\theta}{cosec^2\theta+\sec^2\theta}$
a)Â $\frac{3}{4}$
b)Â $\frac{4}{5}$
c)Â $\frac{5}{6}$
d)Â $\frac{6}{7}$
Answers & Solutions:
1) Answer (B)
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
2) Answer (B)
Expression : $\frac{cos^2 45}{sin^2 60}+\frac{cos^2 60}{sin^2 45}-\frac{tan^230}{cot^245}-\frac{sin^230}{cot^230}$
= $\frac{(\frac{1}{\sqrt{2}})^2}{(\frac{\sqrt{3}}{2})^2} + \frac{(\frac{1}{2})^2}{(\frac{1}{\sqrt{2}})^2} – \frac{(\frac{1}{\sqrt{3}})^2}{(1)^2} – \frac{(\frac{1}{2})^2}{(\sqrt{3})^2}$
= $(\frac{1}{2} \times \frac{4}{3}) + (\frac{1}{4} \times 2) – (\frac{1}{3} \times 1) – (\frac{1}{4 \times 3})$
= $\frac{2}{3} + \frac{1}{2} – \frac{1}{3} – \frac{1}{12}$
= $\frac{9}{12} = \frac{3}{4}$
3) Answer (A)
Expression : tan1°tan2°tan3° ……………tan88°tan89°
$\because$ $tan(90^{\circ}-\theta) = cot\theta$
=> tan 89° = tan(90°-1) = cot 1°
Similarly, tan 88° = cot 2° and so on till tan 46° = cot 44°
=> (tan1°tan2°tan3°…….tan45°……cot3°cot2°cot1°)
Using, $tan\theta cot\theta$ = 1 and $tan45^{\circ}$ = 1
=> 1*1 = 1
4) Answer (C)
$\frac{\cos\alpha}{\sin\beta}=n$
=> $cos\alpha = nsin\beta$
and $\frac{\cos\alpha}{\cos\beta}=m$
=> $cos\alpha = mcos\beta$
Comparing above equations, we get :
=> $nsin\beta = mcos\beta$
Squaring both sides :
=> $n^2sin^2\beta = m^2cos^2\beta$
=> $n^2(1-cos^2\beta) = m^2cos^2\beta$
=> $n^2 = (n^2+m^2)cos^2\beta$
=> $cos^2\beta = \frac{n^2}{m^2+n^2}$
5) Answer (C)
Expression : $sin A cos A (tan A – cot A)$
= $sin A cos A (\frac{sin A}{cos A} – \frac{cos A}{sin A})$
= $sin A cos A (\frac{sin^2 A – cos^2 A}{sin A cos A})$
= $sin^2 A – cos^2 A$
= $sin^2 A – (1 – sin^2 A)$
= $2sin^2 A – 1$
6) Answer (B)
Expression : $\tan^2\theta+\frac{1}{\tan^2\theta}=2$
=> $(tan\theta + \frac{1}{tan\theta})^2 – 2 = 2$
=> $(tan\theta + \frac{1}{tan\theta})^2 = 4$
=> $tan\theta + \frac{1}{tan\theta} = 2$
[It can’t be -2 as $\theta$ is in 1st quadrant, and $tan\theta$ is positive in 1st quadrant.]
=> $tan^2\theta + 1 = 2tan\theta$
=> $(tan\theta – 1)^2 = 0$
=> $tan\theta = 1$
=> $\theta = 45^{\circ}$
7) Answer (A)
Expression : $tan\theta + cot\theta = 5$
Squaring both sides, we get :
=> $tan^2\theta + cot^2\theta + 2tan\theta cot\theta = 25$
We know that, $tan\theta cot\theta = 1$
=> $tan^2\theta + cot^2\theta = 25-2 = 23$
8) Answer (B)
Height of person = CD = 6 ft
Height of tree = AB = $\frac{26}{3}$ ft
Distance between them = BD = $\frac{8}{\sqrt{3}}$ ft
To find : $\angle$ACE = $\theta$ = ?
Solution : AE = AB – BE = $\frac{26}{3}$ – 6
=> AE = $\frac{8}{3} ft$
and BD = CE = $\frac{8}{\sqrt{3}}$ ft
Now, in $\triangle$AEC
=> $tan\theta$ = $\frac{AE}{CE}$
=> $tan\theta$ = $\frac{\frac{8}{3}}{\frac{8}{\sqrt{3}}}$
=> $tan\theta$ = $\frac{1}{\sqrt{3}}$
=> $\theta$ = 30°
9) Answer (A)
Expression : $sin^{2}$ 22° + $sin^{2}$ 68° + $cot^{2}$ 30°
We know that $sin(90^{\circ}-\theta) = cos\theta$
=> $sin 22^{\circ} = sin (90^{\circ}-68^{\circ}) = cos 68^{\circ}$
=> $cos^2 68^{\circ} + sin^2 68^{\circ} + (\sqrt{3})^2$
=> 1 + 3 = 4
10) Answer (C)
Expression : $2sin^{2}$ θ + $3cos^{2}$ θ
We know that, $sin^2 \theta = 1 – cos^2 \theta$
=> $2 (1- cos^2 \theta) + 3cos^2 \theta$
= $cos^2 \theta + 2$
Using the formula, $cos^2 \theta = \frac{cos 2\theta + 1}{2}$
=> $\frac{cos 2\theta + 1}{2} + 2$
= $\frac{cos 2\theta}{2} + \frac{5}{2}$
$\because$ minimum value of $cos 2\theta = -1$
=> min value = $\frac{5}{2} – \frac{1}{2} = 2$
11) Answer (D)
tan(90-θ)=cotθ
tan(4θ-50) = cot(90 – (4θ-50))
=cot(140-4θ)
Given tan (4θ – 50°) = cot(50° – θ)
tan (4θ – 50°) = cot(140° – 4θ)
cot(50° – θ) = cot(140° – 4θ)
50° – θ = 140° – 4θ
3θ = 90°
θ = 30°
Option D is the correct answer.
12) Answer (D)
=> $sec\theta + tan\theta = p$ ————–Eqn(1)
$\because$ $sec^2\theta – tan^2\theta = 1$
=> $(sec\theta + tan\theta)(sec\theta – tan\theta) = 1$
=> $(sec\theta – tan\theta) = \frac{1}{p}$ ————–Eqn(2)
Adding eqns(1)&(2)
=> $2sec\theta = p + \frac{1}{p} = \frac{p^2 + 1}{p}$
=> $sec\theta = \frac{1}{2}[p + \frac{1}{p}]$
13) Answer (A)
Expression : $sin^{2} 30^{\circ} cos^{2} 45^{\circ}$ + $5tan^{2} 30^{\circ}$ + $\frac{3}{2} sin^{2} 90^{\circ}$ – $3 cos^{2} 90^{\circ}$
= $[(\frac{1}{2})^2 * (\frac{1}{\sqrt{2}})^2]$ + $5(\frac{1}{\sqrt{3}})^2 + \frac{3}{2}(1^2) – 3(0)^2$
= $ (\frac{1}{4}*\frac{1}{2}) + \frac{5}{3} + \frac{3}{2}$
= $ \frac{3+40+36}{24}$
= $\frac{79}{24} = 3\frac{7}{24}$
14) Answer (A)
Expression : $cos^{2} θ – sin^{2} θ = \frac{1}{3}$
We know that $cos^2 \theta + sin^2 \theta = 1$
Adding the above two equations, we get :
=> $2cos^2 \theta = \frac{4}{3}$
=> $cos^2 \theta = \frac{2}{3}$
Squaring both sides,
=> $cos^4 \theta = \frac{4}{9}$
Similarly, subtracting those two equations, we get :
=> $sin^2 \theta = \frac{1}{3}$
=> $sin^4 \theta = \frac{1}{9}$
Now, to find : $cos^{4} θ – sin^{4} θ$
= $\frac{4}{9} – \frac{1}{9}$
= $\frac{3}{9} = \frac{1}{3}$
15) Answer (C)
Expression : $tan\theta = \frac{1}{\sqrt{11}}$
We know that, $sec\theta = \sqrt{1 + tan^2 \theta}$
=> $sec\theta = \sqrt{1 + \frac{1}{11}} = \sqrt{\frac{12}{11}}$
Now, $cosec\theta = \frac{sec\theta}{tan\theta}$
=> $cosec\theta = \sqrt{12}$
To find : $\frac{cosec^{2}\theta-\sec^2\theta}{cosec^2\theta+\sec^2\theta}$
= $\frac{12 – \frac{12}{11}}{12 + \frac{12}{11}}$
= $\frac{1 – \frac{1}{11}}{1 + \frac{1}{11}}$
= $\frac{10}{12} = \frac{5}{6}$
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