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# 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.

Question 1: If $sin θ + sin^{2} θ = 1$ then $cos^2 θ + cos^4 θ$ is equal to

a) None

b) 1

c) $Sin θ /cos^{2} θ$

d) $cos^{2}θ / Sin θ$

Question 2: 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) $1 \frac{1}{4}$

b) $\frac{3}{4}$

c) $\frac{1}{4}$

d) $\frac{1}{2}$

Question 3: The value of tan1°tan2°tan3° ……………tan89° is

a) 1

b) -1

c) 0

d) None of the options

Question 4: If $\frac{\cos\alpha}{\sin\beta}=n$ and $\frac{\cos\alpha}{\cos\beta}=m$ then the value of $\cos^{2} \beta$ is

a) $\frac{m^2}{m^2+n^2}$

b) $\frac{1}{m^2+n^2}$

c) $\frac{n^2}{m^2+n^2}$

d) 0

Question 5: If 0° ≤ A ≤ 90°, the simplified form of the given expression sin A cos A (tan A – cot A) is

a) 1

b) 1 – 2 $sin^2$ A

c) 2 $sin^2$ A – 1

d) 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) 23

b) 25

c) 26

d) 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) $4$

b) $3$

c) $\frac{3}{4}$

d) $\frac{5}{4}$

Question 10: The minimum value of $2sin^{2}$ θ + $3cos^{2}$ θ is

a) 3

b) 4

c) 2

d) 1

Question 11: If θ be acute angle and tan (4θ – 50°) = cot(50° – θ), then the value of 9 in degrees is:

a) 20

b) 50

c) 40

d) 30

Question 12: If secθ + tanθ = p, (p ≠ 0) then secθ is equal to

a) $[p-\frac{1}{p}], p\neq 0$

b) $2[p-\frac{1}{p}], p\neq 0$

c) $[p+\frac{1}{p}], p\neq 0$

d) $\frac{1}{2}[p+\frac{1}{p}], p\neq 0$

Question 13: 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) $3\frac{7}{24}$

b) $3\frac{3}{24}$

c) $3\frac{1}{24}$

d) $3\frac{5}{24}$

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Question 14: If $cos^{2} θ – sin^{2} θ = \frac{1}{3}$ , where 0 ≤ θ ≤ π/2 then the value of $cos^{4} θ – sin^{4} θ$ is

a) $\frac{1}{3}$

b) $\frac{2}{3}$

c) $\frac{1}{9}$

d) $\frac{2}{9}$

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) $\frac{3}{4}$

b) $\frac{4}{5}$

c) $\frac{5}{6}$

d) $\frac{6}{7}$

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 : $\frac{cos^2 45}{sin^2 60}+\frac{cos^2 60}{sin^2 45}-\frac{tan^230}{cot^245}-\frac{sin^230}{cot^230}$

= $\frac{(\frac{1}{\sqrt{2}})^2}{(\frac{\sqrt{3}}{2})^2} + \frac{(\frac{1}{2})^2}{(\frac{1}{\sqrt{2}})^2} – \frac{(\frac{1}{\sqrt{3}})^2}{(1)^2} – \frac{(\frac{1}{2})^2}{(\sqrt{3})^2}$

= $(\frac{1}{2} \times \frac{4}{3}) + (\frac{1}{4} \times 2) – (\frac{1}{3} \times 1) – (\frac{1}{4 \times 3})$

= $\frac{2}{3} + \frac{1}{2} – \frac{1}{3} – \frac{1}{12}$

= $\frac{9}{12} = \frac{3}{4}$

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

$\frac{\cos\alpha}{\sin\beta}=n$

=> $cos\alpha = nsin\beta$

and $\frac{\cos\alpha}{\cos\beta}=m$

=> $cos\alpha = mcos\beta$

Comparing above equations, we get :

=> $nsin\beta = mcos\beta$

Squaring both sides :

=> $n^2sin^2\beta = m^2cos^2\beta$

=> $n^2(1-cos^2\beta) = m^2cos^2\beta$

=> $n^2 = (n^2+m^2)cos^2\beta$

=> $cos^2\beta = \frac{n^2}{m^2+n^2}$

Expression : $sin A cos A (tan A – cot A)$

= $sin A cos A (\frac{sin A}{cos A} – \frac{cos A}{sin A})$

= $sin A cos A (\frac{sin^2 A – cos^2 A}{sin A cos A})$

= $sin^2 A – cos^2 A$

= $sin^2 A – (1 – sin^2 A)$

= $2sin^2 A – 1$

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.]

=> $tan^2\theta + 1 = 2tan\theta$

=> $(tan\theta – 1)^2 = 0$

=> $tan\theta = 1$

=> $\theta = 45^{\circ}$

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, $tan\theta cot\theta = 1$

=> $tan^2\theta + cot^2\theta = 25-2 = 23$

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°

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

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 = $\frac{5}{2} – \frac{1}{2} = 2$

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
= 90°
θ = 30°
Option D is the correct answer.

=> $sec\theta + tan\theta = p$ ————–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)

=> $2sec\theta = p + \frac{1}{p} = \frac{p^2 + 1}{p}$

=> $sec\theta = \frac{1}{2}[p + \frac{1}{p}]$

Expression : $sin^{2} 30^{\circ} cos^{2} 45^{\circ}$ + $5tan^{2} 30^{\circ}$ + $\frac{3}{2} sin^{2} 90^{\circ}$ – $3 cos^{2} 90^{\circ}$

= $[(\frac{1}{2})^2 * (\frac{1}{\sqrt{2}})^2]$ + $5(\frac{1}{\sqrt{3}})^2 + \frac{3}{2}(1^2) – 3(0)^2$

= $(\frac{1}{4}*\frac{1}{2}) + \frac{5}{3} + \frac{3}{2}$

= $\frac{3+40+36}{24}$

= $\frac{79}{24} = 3\frac{7}{24}$

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 : $cos^{4} θ – sin^{4} θ$

= $\frac{4}{9} – \frac{1}{9}$

= $\frac{3}{9} = \frac{1}{3}$

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 : $\frac{cosec^{2}\theta-\sec^2\theta}{cosec^2\theta+\sec^2\theta}$

= $\frac{12 – \frac{12}{11}}{12 + \frac{12}{11}}$

= $\frac{1 – \frac{1}{11}}{1 + \frac{1}{11}}$

= $\frac{10}{12} = \frac{5}{6}$

We hope this Trigonometry Questions of set-3 pdf for SSC CHSL Exam preparation is so helpful to you.