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# SSC CGL Trigonometry Questions:

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

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

a)Â $\frac{1}{7}$

b)Â $\frac{2}{7}$

c)Â $\frac{5}{7}$

d)Â $\frac{2}{5}$

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

a)Â 1

b)Â 19

c)Â 25

d)Â 7

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

a)Â 0

b)Â 1

c)Â 2

d)Â 3

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

a)Â 1

b)Â 2

c)Â 3

d)Â 4

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

a)Â $cot\theta$

b)Â $cosec\theta$

c)Â $tan\theta$

d)Â $1$

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

a)Â -$\sqrt{2}\cos\theta$

b)Â -$\sqrt{2}\sin\theta$

c)Â $\sqrt{2}\sin\theta$

d)Â $\sqrt{2}\tan\theta$

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

a)Â 0

b)Â $\frac{2}{3}$

c)Â $\frac{1}{3}$

d)Â $\frac{4}{3}$

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

a)Â 1

b)Â 2

c)Â $\frac{1}{2}$

d)Â $\frac{1}{4}$

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

a)Â 2

b)Â $\sqrt{13}$

c)Â $\sqrt{15}$

d)Â 1

Question 10: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 11: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 12:Â 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 60Â° and 30Â° respectively. The height (in km) of the aeroplane from the road at that instant was (Given âˆš3 = 1.732)

a)Â 0.433

b)Â 8.66

c)Â 4.33

d)Â 0.866

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

a)Â $\frac{c^2}{ab}$

b)Â $\frac{b^2}{ca}$

c)Â $\frac{a^2}{bc}$

d)Â $\frac{bc}{a^{2}}$

Question 14:Â The maximum value of sin Î¸ + cos Î¸ is

a)Â $1$

b)Â $\sqrt{2}$

c)Â $2$

d)Â $3$

Question 15: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 16: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 17:Â If Î¸ is an acute angle and $\tan^2\theta+\frac{1}{\tan^2\theta}=2$ then the value of Î¸ is :

a)Â 60Â°

b)Â 45Â°

c)Â 15Â°

d)Â 30Â°

Question 18:Â If tan Î¸ + cot Î¸ = 5, then $tan^2 Î¸ + cot^2 Î¸$ is

a)Â 23

b)Â 25

c)Â 26

d)Â 24

Question 19:Â 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/âˆš3 ft. away from the base of the tree, then at what angle should he throw a stone so that it hits the fruit ?

a)Â 75Â°

b)Â 30Â°

c)Â 45Â°

d)Â 60Â°

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

a)Â $50\sqrt{3}$ metre

b)Â $100\sqrt{3}$ metre

c)Â $\frac{50}{\sqrt{3}}$ metre

d)Â $\frac{100}{\sqrt{3}}$ metre

Taking $cos\theta$ outside in numerator and in denominator and making $tan\theta$
hence eq will beÂ Â $(\frac{5tan\theta – 3}{5tan\theta + 3})$
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 $Â 13-12+(2tan\theta+\frac{3}{tan\theta})^2$ Â  Â (eq. (1) )
or now above expression to be minimum, equation $(2tan\theta+\frac{3}{tan\theta})^2$ should be minimum.
So applying $A.M.\geq G.M.$
$\frac{(2tan\theta +\frac{3}{tan\theta})}{2} \geq \sqrt{6}$
orÂ ${(2tan\theta+\frac{3}{tan\theta})}=2\sqrt{6}$ ( 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 = 60Â°

$\angle$OBC = $\angle$BOE = 30Â°

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$

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

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

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

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

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

forÂ Â asin Î¸ + bcos Î¸ + c,

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

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

forÂ Â 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.

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

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

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

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 : 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 : $\tan^2\theta+\frac{1}{\tan^2\theta}=2$

=> $(tan\theta + \frac{1}{tan\theta})^2 – 2 = 2$

=> $(tan\theta + \frac{1}{tan\theta})^2 = 4$

=> $tan\theta + \frac{1}{tan\theta} = 2$

[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

=> $tan\theta$ = $\frac{AE}{CE}$

=> $tan\theta$ = $\frac{\frac{8}{3}}{\frac{8}{\sqrt{3}}}$

=> $tan\theta$ = $\frac{1}{\sqrt{3}}$

=> $\theta$ = 30Â°

Height of tower = AB

In $\triangle$ABC

=> $tan\theta = \frac{AB}{BC}$

=> $tan30^{\circ} = \frac{AB}{100}$

=> $\frac{1}{\sqrt{3}} = \frac{AB}{100}$

=> $AB = \frac{100}{\sqrt{3}}$