# Trigonometry Questions for SSC CGL Tier 2 PDF

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## Trigonometry Questions for SSC CGL Tier 2 PDF

Download SSC CGL Tier 2 Trigonometry Questions PDF. Top 15 SSC CGL Tier 2 Trigonometry questions based on asked questions in previous exam papers very important for the SSC exam.

Question 1: 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 2: If a cos θ + b sin θ = p and a sin θ – b cos θ = q, then the relation between a, b, p and q is

a) $a^{2}- b^{2} = p^{2} – q^{2}$

b) $a^{2} + b^{2} = p^{2}+ q^{2}$

c) $a+b=p+q$

d) $a- b=p-q$

Question 3: If $2\sin\theta+\cos\theta-\frac{7}{3}$ then the value of $(\tan^{2}\theta-\sec^{2}\theta)$ is

a) 0

b) -1

c) $\frac{3}{7}$

d) $\frac{7}{3}$

Question 4: If tan θ + cot θ = 2, then the value of $tan^n \theta + cot^{n}\theta$(0° < θ < 90°, n is an integer) is

a) 2

b) 2n+1

c) 2n

d) 0

Question 5: If sin θ + cosec θ = 2 then the value of $sin^{5}\theta+cosec^{5}\theta$ is

a) 1/2

b) 1

c) 0

d) 2

Question 6: Find the value of $1 – 2 sin^{2} θ + sin^{4} θ.$

a) $sin^{4} θ$

b) $cos^{4} θ$

c) $cosec^{4} θ$

d) $sec^{4} θ$

Question 7: If sin (x + y) = cos [3(x + y)], then the value of tan[2(x + y)] is :

a) $\sqrt{3}$

b) 1

c) 0

d) $\frac{1}{\sqrt{3}}$

Question 8: The height of a tower is h and the angle of elevation of the top of the tower is a. On moving a distance h/2 towards, the tower, the angle of elevation becomes 0. The value of cotα – cot β is

a) 1

b) 2

c) 1/2

d) 2/3

Question 9: If A and B are positive acute angles such that sin (A — B) =1/2 and cos (A+ B) = 1/2 , then A and B are given by

a) A = 45°, B = 15°

b) A = 15°, B = 45°

c) A = 30°, B = 30°

d) None of these

Question 10: If 2secA – (1 + sinA)/cosA = x, then the value of x is

a) cosecA/(1+sinA)

b) cosA/(1+sinA)

c) cosA(1+sinA)

d) cosecA(1+sinA)

Question 11: If 1/(cosecA ­+ cotA) = x, then the value of x is

a) cosecA + cotA

b) cosecA ­- cotA

c) $cosec^{2}$ A + cot2A

d) √[$cosec^{2}$ A + cot2A]

Question 12: cos3A is equal to

a) $cos^{3}A – 3sin^{2}cosA$

b) $cos^{3}A + 4sin^{2}cosA$

c) $cos^{3}A + 3sin^{2}cosA$

d) $cos^{3}A – 4sin^{2}cosA$

Question 13: What is the value of tan $(-\frac{5\pi}{6})$

a) $-\frac{1}{\sqrt{3}}$

b) $\frac{1}{\sqrt{3}}$

c) ${\sqrt{3}}$

d) $-{\sqrt{3}}$

Question 14: If cos 135° = x, then the value of x is

a) -1/√2

b) -√3/2

c) -1/2

d) 2

Question 15: 2cos[(C+D)/2].cos[(C-D)/2] is equal to

a) cosC – cosD

b) sinC + sinD

c) cosC + cosD

d) sinC – sinD

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 1 : a cos θ + b sin θ = p

Squaring both sides, we get :

=> $a^2 cos^2 \theta + b^2 sin^2 \theta + 2ab sin\theta cos\theta = p^2$ ——–Eqn(1)

Expression 2 : a sin θ – b cos θ = q

Squaring both sides, we get :

=> $a^2 sin^2 \theta + b^2 cos^ \theta – 2ab sin\theta cos\theta = q^2$ ———-Eqn(2)

=> $a^2 (sin^2 \theta+cos^2 \theta) + b^2 (sin^2 \theta + cos^2 \theta) = p^2 + q^2$

=> $a^{2} + b^{2} = p^{2}+ q^{2}$

tan2 θ – sec2θ

⇒ – (sec2θ – tan2 θ)

⇒ -1 because sec2θ – tan2 θ = 1

Given tan θ + cot θ = 2
Then $(tan θ + cot θ)^2 = 4$
$(tan ^2θ + cot ^2θ + 2tan θ cot θ) = 4$
$(tan ^2θ + cot ^2θ ) = 2$
Option A is the correct answer.

Since $cosecθ = \frac{1}{sinθ}$
$sin θ + cosec θ = 2$ becomes
$sin θ + \frac{1}{sinθ} =2$
$sin^2θ – 2sinθ +1 =0$
which is $(sinθ – 1)^2 = 0$
$sin θ =1$
$sin^{5}\theta+cosec^{5}\theta = 1 + 1 = 2$
Hence Option D is th correct answer.

Here,

$1 – 2 sin^{2} θ + sin^{4} θ.$ = $1^2 + (sin^2 \theta)^2 – 2 \times 1 \times sin^2 \theta$

it is similar to $(a-b)^2$ = $a^2 + b^2 – 2ab$

So,

$1^2 + (sin^2 \theta)^2 – 2 \times 1 \times sin^2 \theta$ = $(sin^2 \theta – 1)^2$……(1)

Now $sin^2 \theta + cos^2 \theta$ = 1…..(2)

From equation 1 and 2

$(sin^2 \theta – 1)^2$ = $(sin^2 \theta – sin^2 \theta – cos^2 \theta)^2$

= $(cos^2 \theta)^2$

= $cos^4 \theta$

Given,

sin (x + y) = cos [3(x + y)]
Using: cos θ = sin (90° – θ)
sin (x + y) = sin [90° – 3(x + y)]
sin [90° – 3(x + y)] – sin (x + y) = 0
sinCsinD =2sin[(CD)/2] cos[(C+D)/2]
=2sin$\frac{(90-3(x+y)-(x+y))}{2} cos \frac{90-3(x+y)+(x+y)}{2}$=0

=2sin(45-2(x+y)) cos (45-(x+y))=0

∴ sin 45° – 2(x + y)} = 0
45° – 2(x + y) = 0
2(x + y) = 45°
OR
Cos{45°- (x + y)} = 0
45°- (x + y)} = 90°
x + y = – 45°
2(x + y) = – 90°
Putting 2(x + y) = 45°
tan 2(x + y) = tan 45° = 1
Again, Putting 2(x + y) = – 90°, we will not get any answer among given options
Option B is the correct answer.

Here, $\angle$ACB = $\alpha$ and $\angle$ADB = $\beta$

AB = tower = $h$ metre

and CD = $\frac{h}{2}$ metre

From $\triangle$ABC

=> $tan \alpha = \frac{AB}{BC} = \frac{h}{BC}$

=> $BC = h cot \alpha$ ———-Eqn(1)

From $\triangle$ABD

=> $tan \beta = \frac{AB}{BD} = \frac{h}{BC – CD}$

=> $tan \beta = \frac{h}{h cot \alpha – \frac{h}{2}}$

=> $h cot \alpha – \frac{h}{2} = h cot \beta$

=> $h (cot \alpha – cot \beta) = \frac{h}{2}$

=> $cot \alpha – cot \beta = \frac{1}{2}$

$sin (A – B) = \frac{1}{2} = sin 30^{\circ}$

=> $A – B = 30^{\circ}$ ———-Eqn(1)

Again, $cos (A + B) = \frac{1}{2} = cos 60^{\circ}$

=> $A + B = 60^{\circ}$ ———Eqn(2)

$2A = 90^{\circ}$

=> $A = 45^{\circ}$ and $B = 15^{\circ}$

Expression : 2secA – (1 + sinA)/cosA = x

= $\frac{2}{sec A} – \frac{(1 + sin A)}{cos A}$

= $\frac{2 – 1 – sin A}{cos A} = \frac{1 – sin A}{cos A}$

Multiplying both numerator and denominator by $(1 + sin A)$

= $\frac{1 – sin A}{cos A} \times \frac{(1 + sin A)}{(1 + sin A)}$

= $\frac{1 – sin^2 A}{cos A(1 + sin A)} = \frac{cos^2 A}{cos A(1 + sin A)}$

= $\frac{cos A}{1 + sin A}$

=> Ans – (B)

Expression : $\frac{1}{cosec A + cot A}$

= $\frac{1}{\frac{1}{sin A} + \frac{cos A}{sin A}}$

= $\frac{1}{\frac{1 + cos A}{sin A}} = \frac{sin A}{1 + cos A}$

Multiplying both numerator and denominator by $(1 – cos A)$

= $\frac{sin A}{1 + cos A} \times \frac{(1 – cos A)}{(1 – cos A)}$

= $\frac{sin A(1 – cos A)}{1 – cos^2 A} = \frac{sin A(1 – cos A)}{sin^2 A}$

= $\frac{1 – cos A}{sin A} = \frac{1}{sin A} – \frac{cos A}{sin A}$

= $cosec A – cot A$

=> Ans – (B)

Using triple angle formula, we know that : $cos(3A) = 4cos^3A – 3cosA$

= $cos^3A + (3cos^3A – 3cosA)$

= $cos^3A + 3cosA(cos^2A – 1)$

= $cos^3A – 3cosA(1 – cos^2A)$

= $cos^3A – 3sin^2AcosA$

=> Ans – (A)

Expression : tan $(-\frac{5\pi}{6})$

= $- tan(\frac{5 \pi}{6})$

= $-tan (\pi – \frac{\pi}{6}) = -(-tan \frac{\pi}{6})$

= $tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$

=> Ans – (B)

Expression : cos 135° = x

= $cos (180 – 45) = -cos (45$°$)$

= $\frac{-1}{\sqrt{2}}$

=> Ans – (A)

Expression : 2cos[(C+D)/2]cos[(C-D)/2]

Using the formula : $cos x . cos y = \frac{1}{2} [cos (x + y) + cos (x – y)]$ ————–(i)

Substituting $(x + y) = C$ and $(x – y) = D$

=> $x = \frac{C + D}{2}$ and $y = \frac{C – D}{2}$ in equation (i),

=> $cos (\frac{C + D}{2}) . cos (\frac{C – D}{2}) = \frac{1}{2} [cos C + cos D]$

=> $2 . cos (\frac{C + D}{2}) . cos (\frac{C – D}{2}) = cos C + cos D$

=> Ans – (C)

We hope this Trogonometry Questions pdf for SSC CGL Tier 2 exam will be highly useful for your Preparation.