0
400

# SSC CGL Trigonometry Questions:

Download Top-20 Trigonometry questions for SSC CGL exam 2020. Most important  Trigonometry questions based on asked questions in previous exam papers for SSC CGL.

Question 1: $5tan\theta = 4$, then the value of $(\frac{5sin\theta – 3cos\theta}{5sin\theta + 3cos\theta})$ is

a) $\frac{1}{7}$

b) $\frac{2}{7}$

c) $\frac{5}{7}$

d) $\frac{2}{5}$

Question 2: The least value of $(4sec^2\theta + 9cosec^2\theta)$ is

a) 1

b) 19

c) 25

d) 7

Question 3: If $x=cosec\theta-sin\theta$ and $y=sec\theta-cos\theta$, then the value of $x2y2(x2 + y2 + 3)$

a) 0

b) 1

c) 2

d) 3

Question 4: If $0 \leq \theta \leq \frac{\pi}{2}$, $2ycos\theta=sin\theta$ and $\frac{x}{2cosec\theta}=y$, then the value of $x^2-4y^2$ is

a) 1

b) 2

c) 3

d) 4

Question 5: The value of $\frac{1}{cosec\theta – cot\theta} – \frac{1}{sin\theta}$

a) $cot\theta$

b) $cosec\theta$

c) $tan\theta$

d) $1$

Question 6: If $\cos\theta + \sin\theta = \sqrt{2}\cos\theta$, then $\cos\theta – \sin\theta$ is

a) -$\sqrt{2}\cos\theta$

b) -$\sqrt{2}\sin\theta$

c) $\sqrt{2}\sin\theta$

d) $\sqrt{2}\tan\theta$

Question 7: If $cos^4\theta-sin^4\theta=\frac{2}{3}$, then the value of $1-2sin^2\theta$ is,

a) 0

b) $\frac{2}{3}$

c) $\frac{1}{3}$

d) $\frac{4}{3}$

Question 8: The value of $\frac{1}{1 + tan^2\theta}$ + $\frac{1}{1 + cot^2\theta}$ is

a) 1

b) 2

c) $\frac{1}{2}$

d) $\frac{1}{4}$

Question 9: Maximum value of $(2sin\theta+3 cos\theta)$ is

a) 2

b) $\sqrt{13}$

c) $\sqrt{15}$

d) 1

Question 10: The value of $cos^2 30^{\circ} + sin^2 60^{\circ} + tan^2 45^{\circ} + sec^2 60^{\circ} + cos0^{\circ}$ is

a) $4\frac{1}{2}$

b) $5\frac{1}{2}$

c) $6\frac{1}{2}$

d) $7\frac{1}{2}$

Question 11: If $cos x + cos^{2} x = 1,$ then $sin^{8} x + 2 sin^{6} x + sin^{4}$ x is equal to

a) 0

b) 3

c) 2

d) 1

Question 12: From an aeroplane just over a straight road, the angles of depression of two consecutive kilometre stones situated at opposite sides of the aeroplane were found to be 60° and 30° respectively. The height (in km) of the aeroplane from the road at that instant was (Given √3 = 1.732)

a) 0.433

b) 8.66

c) 4.33

d) 0.866

Question 13: In ΔABC, ∠C = 90° and AB = c, BC = a, CA = b; then the value of (cosec B – cos A) is

a) $\frac{c^2}{ab}$

b) $\frac{b^2}{ca}$

c) $\frac{a^2}{bc}$

d) $\frac{bc}{a^{2}}$

Question 14: The maximum value of sin θ + cos θ is

a) $1$

b) $\sqrt{2}$

c) $2$

d) $3$

Question 15: Find the value of tan 4° tan 43° tan 47 tan 86°

a) $\frac{2}{3}$

b) $1$

c) $\frac{1}{2}$

d) $2$

Question 16: The value of tan1°tan2°tan3° ……………tan89° is

a) 1

b) -1

c) 0

d) None of the options

Question 17: 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 18: If tan θ + cot θ = 5, then $tan^2 θ + cot^2 θ$ is

a) 23

b) 25

c) 26

d) 24

Question 19: 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 20: The angle of elevation of a tower from a distance of 100 metre from its foot is 30°. Then the height of the tower is

a) $50\sqrt{3}$ metre

b) $100\sqrt{3}$ metre

c) $\frac{50}{\sqrt{3}}$ metre

d) $\frac{100}{\sqrt{3}}$ metre

Taking $cos\theta$ outside in numerator and in denominator and making $tan\theta$
hence eq will be  $(\frac{5tan\theta – 3}{5tan\theta + 3})$
As it is given that $5tan\theta$ = 4
after putting values and solving we will get the equation reduced to 1/7.

$4sec^2\theta+9cosec^2\theta$
or $4+4tan^2\theta+9+9cot^2\theta$
or $13+4tan^2\theta+9cot^2\theta$
or $13+4tan^2\theta+\frac{9}{tan^2\theta}$
or $13-12+(2tan\theta+\frac{3}{tan\theta})^2$    (eq. (1) )
or now above expression to be minimum, equation $(2tan\theta+\frac{3}{tan\theta})^2$ should be minimum.
So applying $A.M.\geq G.M.$
$\frac{(2tan\theta +\frac{3}{tan\theta})}{2} \geq \sqrt{6}$
or ${(2tan\theta+\frac{3}{tan\theta})}=2\sqrt{6}$ ( for value to be minimum)
After putting above value in eq.(1) , we will get least value of expression as 25.

$x=cosec\theta – sin\theta=\frac{cos^2\theta}{sin\theta}=cot\theta cos\theta$
Similarly $y=tan\theta sin\theta$
$xy=sin\theta cos\theta$
$x^2+y^2+3=(sec^2\theta +cosec^2\theta )$
Now putting above values in given equation, and after solving it will be reduced to 1

$2y=tan\theta$
$x=2ycosec\theta$
Hence value of $x^2 – 4y^2$ = $4y^2(cosec^2\theta – 1)$
or $tan^2\theta cot^2\theta$ = 1

$\frac{sin\theta}{1-cos\theta} – \frac{1}{sin\theta}$

or $\frac{cos\theta – cos^2\theta}{(1-cos\theta)sin\theta}$ = $cot\theta$

$\sin^2 \theta + \cos^2 \theta = 1$
So, $\sin^2 \theta + \cos^2 \theta + 2\sin\theta * \cos \theta = 2 \cos^2\theta$
Hence, $\cos^2 \theta – \sin^2 \theta = 2 \sin\theta*\cos\theta$
So, $\cos\theta – \sin\theta = \sqrt{2}\sin\theta$

$cos^4\theta-sin^4\theta=(cos^2\theta-sin^2\theta)(cos^2\theta+sin^2\theta)=cos^2\theta-sin^2\theta=\frac{2}{3}$
$cos^2\theta-sin^2\theta =1-2sin^2\theta=\frac{2}{3}$

$1 + \tan ^2 \theta = \sec ^2 \theta$
$1 + \cot ^2 \theta = \csc ^2 \theta$
So, the given fraction becomes,

$\frac{1}{\sec ^2 \theta} + \frac{1}{\csc ^2 \theta} = \sin^2\theta + \cos^2 \theta = 1$

$\because$ Maximum Value of $a\sin{\theta}+b\cos{\theta}=\sqrt{a^{2}+b^{2}}$
$\therefore$ Maximum Value of $2\sin{\theta}+3\cos{\theta}=\sqrt{2^{2}+3^{2}}$
$=\sqrt{13}$
Hence, Correct option is B.

Substituting values of angles, we get,
3/4 + 3/4+ 1 + 4 + 1 = 7.5. Option D is the right answer.

$cos x + cos^2 x = 1$

=> $cos x = 1 – cos^2 x$

=> $cos x = sin^2 x$

$\therefore$ $sin^{8} x + 2 sin^{6} x + sin^{4} x$

= $(sin^4 x + sin^2 x)^2$

= $((cos x)^2 + sin^2 x)^2$

= $(cos^2 x + sin^2 x)^2 = 1$

OC = height of plane = $h$

$\angle$OAC = $\angle$DOA = 60°

$\angle$OBC = $\angle$BOE = 30°

AB = 2 and let AC = $x$

=> BC = $(2-x)$

From, $\triangle$OAC

$tan60^{\circ} = \frac{OC}{AC}$

=> $\sqrt{3} = \frac{h}{x}$

=> $x = \frac{h}{\sqrt{3}}$ ————Eqn(1)

From, $\triangle$OBC

$tan30^{\circ} = \frac{OC}{BC}$

=> $\frac{1}{\sqrt{3}} = \frac{h}{2-x}$

=> $\sqrt{3}h = 2 – \frac{h}{\sqrt{3}}$ [From eqn(1)]

=> $\frac{3h+h}{\sqrt{3}} = 2$

=> $h = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$

= $\frac{1.732}{2}$ = 0.866

In $\triangle$$ABC, AB^2 = AC^2 + BC^2$

=> $c^2 = a^2 + b^2 => c^2 – b^2 = a^2$

$cosecB = \frac{AB}{BC} = \frac{c}{b}$

$cosA = \frac{AC}{AB} = \frac{b}{c}$

$\therefore cosecB – cosA = \frac{c}{b} – \frac{b}{c}$

= $\frac{c^2-b^2}{bc} = \frac{a^2}{bc}$

for  asin θ + bcos θ + c,

maximum value = $c+\sqrt{a^{2}+b^{2}}$

minimum value =  $c-\sqrt{a^{2}+b^{2}}$

for  sin θ + cos θ , 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.

Expression : tan 4° tan 43° tan 47 tan 86°

$\because$ $tan(90-\theta) = cot\theta$

=> $tan 4^{\circ} = tan(90^{\circ}-86^{\circ}) = cot 86^{\circ}$

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

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

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°

In $\triangle$ABC
=> $tan\theta = \frac{AB}{BC}$
=> $tan30^{\circ} = \frac{AB}{100}$
=> $\frac{1}{\sqrt{3}} = \frac{AB}{100}$
=> $AB = \frac{100}{\sqrt{3}}$