Top 30 Trigonometry Questions and Answers

Top 30 Trigonometry Questions and Answers: Trigonometry is a branch of mathematics and by using some mathematical techniques, we can determine distances or heights. The word “Trigonometry” is derived from the Greek words “tri” (that means three), ‘gon’ (that means sides) and ‘metron’ (means measure). In short, Trigonometry is the study of relationships between the sides and angles of a triangle. Trigonometry is one of the most ancient subjects studied by scholars across the world. Trigonometry was used by astronomers to calculate distance from the Earth to the planets and stars. In geography, trigonometry is also used to create maps and establish an island's location in relation to longitudes and latitudes, among other things.

Top 30 Trigonometry Questions and their Detail Solutions

In this article we are going to post an excellent collection of Top 30 Trigonometry questions along with their detail and tricky solutions asked in various Government examination like SSC CGL, CHSL, CPO, MTS, JE, GD Constable, Stenographer, IBPS, RRB etc. These type of questions are repeated every time so you can practice by solving it. Go through the below Top 30 Trigonometry questions and answers-

Trigonometry Formula Basic to Advance Level

Q. (1) What is the value of cos20^ocos40^ocos60^ocos80^o?

(a) 1/4

(b) 1/16

(c) 3/4

(d) 3/16

^{Detail Solution:}

Trick to solve these type of questions
Remember Below Trick

a) cos θ x cos 2θ x cos 4θ = ¼cos3θ 
b) Sin θ x Sin 2θ x Sin 4θ = ¼ Sin3θ
c) tanθ x tan2θ x tan4θ = tan3θ

We have value
=cos20^ocos40^ocos60^ocos80^o
=(cos20^ocos40^ocos80)cos60^o
Here, θ = 20^o
So, ¼ (Cos 3x20)xcos60^o =¼ Cos²60^o
= ¼ x (½)² = 1/16
Hence, Option (b) is correct

Q. (2) What is the value of Sin20^oSin40^oSin60^oSin80^o?

(a) 1/4

(b) 1/16

(c) 3/16

(d) 3/4

^{Detail Solution:}

Trick to solve these type of questions

Sin θ x Sin 2θ x Sin 4θ = ¼ Sin 3θ

We have value
=Sin20^oSin40^oSin60^oSin80^o
=(Sin20^oSin40^oSin80)Sin60^o
Here, θ = 20^o
So, ¼ (Sin 3x20)Sin60^o =¼ Sin²60^o
= ¼ x (√3/2)² = ¼ x 3/4= 3/16
Hence, Option (c) is correct

Q. (3) The value of cot18⁰[cot72⁰cos²22⁰+1/(tan72⁰sec²68⁰)] is,

(a) 0

(b) 1

(c) 3

(d) 2

^{Detail Solution:}

cot72⁰= cot(90⁰-18⁰)=tan18⁰

tan72⁰= tan(90⁰-18⁰)=cot18⁰ 

sec68⁰= sec(90⁰-22⁰)=cosec22⁰.

cos68⁰= cos(90⁰-22⁰)=sin22⁰

We have,

cot18⁰[cot72⁰cos²22⁰+1/(tan72⁰sec²68⁰)]

=cot18⁰[tan18⁰cos²22⁰+cot72⁰cos²68⁰)]

=cot18⁰[tan18⁰cos²22⁰+tan18⁰cos²68⁰)]

=tan18⁰cot18⁰[cos²22⁰+sin²22⁰)]

=1.[1]=1

Hence, Option (b) is correct

Q. (4) The value of tan1⁰tan2⁰tan3^{0.......................} tan89⁰ ?

(a) 1

(b) 0

(c) 3

(d) 2

^{Detail Solution:}

As  tan89⁰= tan(90⁰ - 1⁰)=cot1⁰,

tan88⁰= tan(90⁰ - 2⁰)=cot2⁰,

tan87⁰= tan(90⁰ - 3⁰)=cot3⁰

tan46⁰= tan(90⁰ - 44⁰)=cot44⁰

Now substituting these values in original equation we get

=tan1⁰tan2⁰tan3⁰.... tan44⁰ tan45⁰ tan46⁰..... tan87⁰ tan88⁰tan89⁰

=tan1⁰tan2⁰tan3⁰......tan44⁰ tan45⁰ cot44⁰...........cot3⁰ cot2⁰cot1⁰

=tan1⁰cot1⁰tan2⁰ cot2⁰tan3⁰cot3⁰.........tan44⁰ cot44⁰ tan45⁰1.1.1.....................1.1                [As cot45⁰=1]

Trick to solve these type of questions:

Remember Below Tricks:

a) tan1⁰tan2⁰tan3^{0.......................} tan89=1

b) cot1⁰cot2⁰cot3^{0.......................} cot89=1

c) cos1⁰cos2⁰cos3^{0.......................} cos90=0

d) cos1⁰cos2⁰cos3^{0.......................} (Greater than cos90)=0

e) sin1⁰sin2⁰sin3^{0.......................}sin180=0

e) sin1⁰sin2⁰sin3^{0.......................} (Greater than sin180)=0

Q. (5) The greatest value of sin⁴θ+cos⁴θ is ?

(a) 2

(b) 0

(c) 3

(d) 1

^{Detail Solution:}

sin⁴θ+cos⁴θ=(sin²θ+cos²θ)²−2sin²θcos²θ

=1−2sin²θcos²θ

The maximum of this expression can only be 1 when, the second term is zero, or when either sinθ or cosθ is 0.

Hence, Option (b) is correct.

Q. (6) The minimum value of tan²θ+cot²θ is ?

(a) 2

(b) 0

(c) 3

(d) 1

Detail Solution:

We know a² + 1/a² has minimum value as 2,

So, tan²θ+cot²θ = tan²θ+1/tan²θ has min value as 2

Hence, Option (a) is correct.

Q. (7) Which one is greater  sin1° or sin1 ?

(a) sin1°> sin1

(b) sin1°< sin1

(c) sin1°= sin1°

(d) None of these

Detail Solution:

We have 1=57.32⁰ and we know

sinx>siny if x>y for acute angles 

So 1>1⁰

=sin1 > sin1⁰

Hence, Option (b) is correct.

Q. (8) If tan(x+y)tan(x-y) = 1, then find tan(2x/3) ?

(a) 1/4

(b) 1/√3

(c) 3

(d) 1

Detail Solution:

tanA.tanB = 1
tanA = CotB
tanA=tan(90०-B)
A = 90०-B
A + B= 90०
So, (x+y) + (x-y)= 90०
2x =90०
x = 45०
tan(2x/3) = tan30०= 1/√3

Hence, Option (b) is correct.

Q. (9) The minimum value of cos²θ+sec²θ is ?

(a) 2

(b) 0

(c) 3

(d) 1

Detail Solution:

We know that a² + 1/a² has minimum value as 2,

So, cos²θ+sec²θ = cos²θ+1/cos²θ has min value as 2

Hence, Option (a) is correct.

Q. (10) The minimum value of sin²θ+cosec²θ is ?

(a) 2

(b) 0

(c) 3

(d) 1

Detail Solution:

We know that a² + 1/a² has minimum value as 2,                   

So, sin²θ+cosec²θ = sin²θ+1/sin²θ has min value as 2

Hence, Option (a) is correct.

Q. (11) If tanθ+cotθ=2 (0⁰≤θ≤90⁰) then the value of tan¹⁰θ+cot¹¹θ is ?

(a) 2

(b) 0

(c) -1

(d) 1

Detail Solution:

We have 

tanθ+cotθ=2

Put θ= 45, The equation will be satisfy

⇨1 + 1 = 2

So, tan¹⁰θ+cot¹¹θ = 1 + 1 = 2 

Hence, Option (a) is correct.

Q. (12) If tanθ+cotθ=2 (0⁰≤θ≤90⁰) then the value of tanθ+cotθ is ?

(a) 2

(b) 0

(c) -1

(d) 1

Detail Solution:

We have 

tanθ+cotθ=2

Put θ= 45, The equation will be satisfy

⇨1 + 1 = 2

tanθ+cotθ=1 + 1 =2

Hence, Option (a) is correct.

Q. (13) If tanθ+cotθ=2 (0⁰≤θ≤90⁰) then the value of sinθ+cosθ is ?

(a) 2

(b) 0

(c) √2

(d) None of these

Detail Solution:

We have 

tanθ+cotθ=2

By putting θ= 45, The equation will be satisfy

⇨1 + 1 = 2

So, sinθ+cosθ = 1/√2 + 1/√2 =2/√2 = √2 

Hence, Option (c) is correct.

Q. (14) If 0⁰<θ<90⁰ and 2sin²θ+3cosθ=3 then the value of θ is ?

(a) 0⁰

(b) 30⁰

(c) 60⁰

(d) A or C

^{Detail Solution:}

As, 2sin²θ+3cosθ=3

⇨2(1−cos²θ)+3cosθ=3                                       

⇨2cos²θ−3cosθ+1=0

⇨(2cosθ−1)(cosθ−1)=0.

⇨cosθ = ½ Or cosθ =1  

 Thus, θ = 60⁰ Or θ = 0⁰  

Hence, Option (d) is correct.

Q. (15) If sinθ=a/√(a²+b²) then the value of tanθ will be

(a) a/b

(b) b/a

(c) a+1/b+1

(d) ab/a+b

Detail Solution:

We have, cos²θ=1−sin²θ=1 – a²/(a²+b²)

cos²θ=( a² + b² a²)/(a²+b²)=b²/(a²+b²)                 

cosθ=b/√(a²+b²)

tanθ=sinθ/cosθ= *a/√(a²+b²)]/[ b/√(a²+b²)]=a/b.

Hence, Option (a) is correct.

Q. (16) If sin21⁰=x/y then sec21⁰−sin69⁰ is ?

(a) x²/*y√(y²−x²)]

(b) x/*y√(y²−x²)]

(c) xy/*√(y²−x²)]

(d) y/*x√(y²−x²)]

Detail Solution:

sin21⁰=cos69⁰,          

⇨⇨sin21⁰=cos(90⁰ – 21⁰) = cos69⁰              

⇨sin21⁰=cos69⁰=x/y,         

⇨1−cos²69⁰=sin²69⁰=1−x²/y²=(y²−x²)/y²             

sin69⁰=√(y²−x²)/y.

We have

sec21⁰−sin69⁰=cosec69⁰−sin69⁰

=1−sin²69⁰/sin69⁰

=cos²69⁰/sin69⁰

=x²/y²×y/√(y²−x²)

=x²/*y√(y²−x²)]

Hence, Option (a) is correct.

Q. (17) If sin θ + cos θ = 2 , then find the value of cosec θ – sec θ?

(a) 1/3

(b) 2/3

(c) 4/3

(d) 3

Detail Solution:

If sin θ + cos θ = p                      

cosec θ – sec θ = q

Then, p - 1/q = 2/q

⇨2-(1/2) = 3/2 = 2/q

q= 4/3 

or cosec θ – sec θ = 4/3

Hence, Option (c) is correct.

Q. (18) The minimum value of 2sin²θ+3cos²θ is ?

(a) 0

(b) 1

(c) 2

(d) 3

Detail Solution:

2sin²θ+3cos²θ

=2sin²θ+2cos²θ + cos²θ

=2(sin²θ+cos²θ) + cos²θ

=2x1 + cos²θ

=2 + cos²θ

This expression will be minimum for Minimum value of cos²θ that is 0

=2

Hence, Option (c) is correct.

Q. (19) If tanθ=3/4 and θ is acute then, sinθ is equal to,

(a) 1/3

(b) 3/5

(c) 5/7

(d) 1

Detail Solution:

We have:

tanθ=3/4

cotθ=4/3

cot²θ=cosec²θ−1=16/9              

cosec²θ=25/9

cosecθ=5/3.

sinθ = 3/5

Hence, Option (b) is correct.

Q. (20) If y = 36cos²x + 16cosec²x - 4 then Ymin ?

(a) 0

(b) 11

(c) 22

(d) 44

Detail Solution:

Trick:

If below equations are given

y = a²sin²x + b² cosec2x + c

y = a² cos²x + b²sec²x + c

y = a²tan²x + b²cot²x + c 

Then,   y_min = 2ab + c

y_max = infinite.

For the given sum,

y_min = 2x √36 x √16 - 4

= 2x6x4 + 4 = 48 - 4 = 44

Hence, Option (d) is correct.

Q. (21) If (1+sinA)(1+sinB)(1+sinC)=(1−sinA)(1−sinB)(1−sinC), then the expression on each side of the equation equals,

(a) 1

(b) sinA.sinB.sinC

(c) cosA.cosB.cosC

(d) tanA.tanB.tanC

Detail Solution:

We have given 

(1+sinA)(1+sinB)(1+sinC)=(1−sinA)(1−sinB)(1−sinC)=k 

So,(1+sinA)(1+sinB)(1+sinC)=k---------------------- -- (i) &,

(1−sinA)(1−sinB)(1−sinC)=k.------------------------------ (ii)

Multiplying equation (i) and (ii) we get,

k²=(1−sin²A)(1−sin²B)(1−sin²C) 

k² =cos²A.cos²B.cos²C

k = cosA.cosB.cosC

Hence, Option (c) is correct.

Q. (22) If tanθ=1, then the value of (8sinθ+5cosθ)/(sin³θ−2cos³θ+7cosθ) is,

(a) 1

(b) 2

(c) 4

(d) 3

Detail Solution:

tanθ=1,

then sinθ=cosθ.

So, (8sinθ+5cosθ)/(sin³θ−2cos³θ+7cosθ)

=13/(7−sin²θ) 

[putting sinθ=cosθ]                

Or, cot²θ=1,

Or, cosec²θ−1=1,

Or, sin²θ=1/2.

Substituting, in 13/(7−sin²θ) 

We get,

=13/(7−1/2)

=13/(13/2)

=2

Hence, Option (b) is correct.

Q. (23) Find maximum value of 4tanx + 3cot x + 10 is:

(a) 10

(b) 15

(c) 17

(d) 31

Detail Solution:

Q. (24) (secθ−cosθ)²+(cosecθ−sinθ)²−(cotθ−tanθ)² is,

(a) 0

(b) 1

(c) -1

(d) 2

Detail Solution:

(secθ−cosθ)²=sec²θ(1−cos²θ)²= (sin²θ)²/ cos²θ =sin²θ. sin²θ / cos²θ =sin²θtan²θ. 

Similarly,

(cosecθ−sinθ)²=cos²θcot²θ. 

Adding all we get:

2−tan²θ(1−sin²θ)−cot²θ(1−cos²θ)

=2−(sin²θ+cos²θ)

= 2-1=1

Hence, Option (b) is correct.

Q. (25) If If tan2θ.tan4θ=1, then the value of tan3θ is,

(a) 0

(b) 1

(c) √3

(d) 2

Detail Solution:

tan2θ.tan4θ=1,

tan2θ=1/tan4θ=cot4θ.

tan2θ= tan(90⁰−4θ), where θ is acute. 

2θ= 90⁰−4θ

6θ=90⁰

3θ=45⁰

tan3θ=tan45⁰

⁼¹

Hence, Option (b) is correct.

Q. (26) If sinθ+cosecθ=2 (0⁰≤θ≤90⁰) then the value of sin¹⁰⁰θ - cosec¹¹¹θ is,

(a) 0

(b) 1

(c) -1

(d) 2

Detail Solution:

Trick:

We have
sinθ+cosecθ=2
Put θ=90^{0 
Then equation will be satisfy
⇒ 1 + 1 = 2}

^Hence, sin¹⁰⁰θ - cosec¹¹¹θ = 1 - 1 = 0 

By putting value of θ=90⁰ 

Hence, Option (b) is correct.

Q. (27) If (rcosθ−√3)²+(rsinθ−1)²=0, then the value of (rtanθ+secθ)/(rsecθ+tanθ) is,

(a) 4/5

(b) √34

(c) √54

(d) 54

Detail Solution:

(rcosθ−√3)²+(rsinθ−1)²=0

(rcosθ−√3)=0 ⇒ rcosθ=√3.......(i)

(rsinθ−1)=0⇒ rsinθ=1........(ii)

(rcosθ)² + (rsinθ)² = √3² + 1² =3+1=4  

r²(cos²θ + sin²θ)=4 and r²(1)=4

r=2

By putting these values in (i) (rtanθ+secθ)/(rsecθ+tanθ) we get 4/5

Hence, Option (a) is correct.

Q. (28) If tanθ=3/4 and 0<θ<π/2 and 25xsin²θcosθ=tan²θ, then the value of x is

(a) 1/4

(b) 3/16

(c) 5/64

(d) 7/256

Detail Solution:

25xsin²θcosθ=tan²θ 

25xsin²θcosθ=sin²θ/cos²θ 

25x=sec³θ

As given, tanθ=3/4

sec²θ=1+tan²θ=1 + 9/16= 25/16

secθ = 5/4

By substituting the value of secθ= 5/4, We get

25x=sec³θ ⇒ x = 5/64

Hence, Option (c) is correct.

Q. (29) If tanθ+cotθ=2 (0⁰≤θ≤90⁰) then the value of tan¹⁰⁰θ - cot¹¹¹θ is ?

(a) 0

(b) 1

(c) -1

(d) -2

Detail Solution:

tanθ+cotθ=2

Put θ= 45, then equation will be satisfy

⇒1+1=2

Now, by putting value of θ=45 , we get

So, tan¹⁰⁰θ - cot¹¹¹θ =1 - 1 = 0

Hence, Option (a) is correct.

Q. (30) If 4 sin θ + 3 cos θ = 2 , then find the value of 4 cos θ – 3 sin θ:

(a) √21

(b) 3

(c) √3

(d) 1

Detail Solution:

Trick:

If a sin θ + b cos θ = m     

& a cos θ – b sin θ = n  Then a² + b² = m² + n²

Let 4 cos θ – 3 sin θ = x

By using above Trick

We get 

⇒4² + 3² = 2² + x²

⇒16 + 9 = 4 + x² 

⇒X = √21

Hence, Option (a) is correct.