Free Introduction to Trigonometry 02 Practice Test - 10th Grade

If cos 3θ=32, 0° < 3θ < 90°, then find the value of θ.

A.

15°

B.

10°

C.

D.

12°

SOLUTION

Solution : B

Given: cos3θ=32
We know that cos30=32
Comparing the two we get,
3θ=30.... (given 0 < 3θ < 90)
θ=10

sin18cos72=

SOLUTION

Solution : sin18cos72
=sin(9072)cos72
=cos72cos72=1   as  sin(90A)=cosA

(1+tanθ+secθ)(1+cotθcosecθ)=

A. 0
B. 1
C. 2
D. -1

SOLUTION

Solution : C

(1+tanθ+secθ)(1+cotθcosecθ)
=(1+sinθcosθ+1cosθ)(1+cosθsinθ1sinθ)
=(cosθ+sinθ+1)cosθ×(sinθ+cosθ1)sinθ
=(cosθ+sinθ)212cosθsinθ
=(cos2θ+sin2θ+2cosθsinθ1)cosθsinθ
=(1+2cosθsinθ1)cosθsinθ
=2cosθsinθcosθsinθ=2

(1+tanA tanB)2+(tanA  tanB)2sec2A sec2B= ___

A.

1

B.

-tan A

C.

2

D.

cot A

SOLUTION

Solution : A

(1+tan A tan B)2+(tan A-tan B)2sec2 A sec2 B

=1+2 tan A tan B+tan2 A tan2 B+tan2 A2 tan A tan B+tan2 Bsec2 A sec2 B

=1+tan2 A tan2 B+tan2 A+tan2 Bsec2 A sec2 B

=1+tan2 A+tan2 B+tan2 A tan2 Bsec2 A sec2 B

=1(1+tan2 A)+tan2 B(1+tan2 A)sec2 A sec2 B

=(1+tan2 A)(1+tan2 B)sec2 A sec2 B

=(sec2 A)(sec2 B)sec2 A sec2 B.(1+tan2 θ=sec2 θ)

=1

Which of the following trigonometric expressions is equal to sec6 θ ?

A.

tan6 θ+3 tan2 θ sec2 θ+1

B.

tan6 θ3 tan2 θ sec2 θ+1

C.

tan6 θ1

D.

tan6 θ+1

SOLUTION

Solution : A

sec6 θ=(sec2 θ)3

=(tan2 θ+1)3(1+tan2 θ=sec2 θ)

=tan6 θ+3 tan4 θ+3 tan2 θ+1

=tan6 θ+3 tan2 θ(tan2 θ)+3 tan2 θ+1

=tan6 θ+3 tan2 θ(sec2 θ1)+3 tan2 θ+1

=tan6 θ+3 tan2 θ sec2 θ3 tan2 θ+3 tan2 θ+1

=tan6 θ+3 tan2 θ sec2 θ+1

Which of the following options are equal to tan Asec A1+tan Asec A+1 ?

A.

2 cosec A

B.

1sinA

C.

2sinA

D.

2

SOLUTION

Solution : A and C

tanAsecA1+tanAsecA+1

=tan A(1sec A1+1sec A+1)

=tan A(sec A+1+sec A1sec2A1)

=tan A(2secAtan2A)...(1+tan2A=sec2A)

=2secAtanA

=2cosA(sinAcosA)....(tanA=sinAcosA)

=2sinA=2cosecA     ...(1sinA=cosecA)

Which of the following options are equal to 1+cos A1cos A ?

A.

1cos Asin A

B.

1+cos Asin A

C.

cosec A+cot A

D.

sec A+tan A

SOLUTION

Solution : B and C

1+cos A1cos A=(1+cos A1cos A)(1+cos A1+cos A)

=(1+cos A)21cos2A

=(1+cos A)2sin2 A....(sin2A+cos2A=1)

=1+cos AsinA

=1sin A+cos Asin A

=cosec A+cot A

If x=a sec A cos B, y=b sec A sin B and z=c tan A, then x2a2+y2b2z2c2=

A.

0

B.

3

C.

2

D.

1

SOLUTION

Solution : D

x=a sec A cos Bxa=sec A cos B

y=b sec A sin Byb=sec A sin B

z=c tan Azc=tan A

Squaring each of the equations we get,

x2a2=sec2 A cos2 B, y2b2=sec2 A sin2 B,z2c2=tan2 A

x2a2+y2b2z2c2=sec2 A cos2 B+sec2 A sin2 Btan2 A

=sec2 A(cos2 B+sin2 B)tan2 A

=sec2 Atan2 A(cos2 B+sin2 B=1)

=1(sec2 Atan2 A=1)

If x=a cosec θ and y=b cot θ, then which of the following equations is true ?

A.

x2y2=a2b2

B.

x2a2y2b2=1

C.

x2+y2=a2+b2

D.

x2a2+y2b2=1

SOLUTION

Solution : B

x=a cosec θxa=cosec θ.....(1)

y=b cot θyb=cot θ.....(2)

Squaring the equations and subtracting (2) from (1) we get,

x2a2y2b2=cosec2θcot2θ

x2a2y2b2=1....(1+cot2θ=cosec2θ)

Which of the following options is equal to cot θ+cosec θ1cot θcosec θ+1

A.

1

B.

sin θ1cos θ

C.

1+cos θsin θ

D.

cotθ+cosecθ

SOLUTION

Solution : B, C, and D

=cot θ+cosec θ1cot θcosec θ+1

=cot θ+cosec θ(cosec2θcot2θ)cot θcosec θ+1

=(cot θ+cosec θ)(cosec θcot θ)(cosec θ+cot θ)cot θcosec θ+1

=(cot θ+cosec θ)(1cosec θ+cot θ)(cot θcosec θ+1)

=cot θ+cosec θ

=cos θsin θ+1sin θ

=1+cos θsin θ

=(1+cos θ)(1cos θ)sinθ(1cos θ)   ....[Multiplying numerator and denominator by 1cosθ ]

=1cos2θsinθ(1cos θ)

=sin2θsinθ(1cos θ)

=sin θ1cos θ