# Free Introduction to Trigonometry 01 Practice Test - 10th Grade

Find the value of  1tan2 301+tan2 30.

A.

0.5

B.

1

C.

2

D.

4

#### SOLUTION

Solution : A

We know that tan 30=13.

1tan2 301+tan2 30=1(13)21+(13)2

=1(13)1+(13)

=313+1

=24 = 12=0.5

(1+tanθ1+cotθ)2 = ____ .

A.

1

B.

tanθ

C. tan2θ
D.

4

#### SOLUTION

Solution : C

The given expression is

(1+tanθ1+cotθ)2  =(1+tanθ1+1tanθ)2

=(1+tanθtanθ+1tanθ)2

=((tanθ)(1+tanθ)tanθ+1)2

=tan2θ

3tanθ + cotθ = 5cosecθ. Solve for θ, 0θ90.

__

#### SOLUTION

Solution :

3sinθcosθ + cosθsinθ5sinθ

3sin2θ + cos2θsinθcosθ=5sinθ
3sin2θ+cos2θ=5cosθ

3(1 - cos2θ) + cos2θ = 5 cosθ

2cos2θ + 5cosθ - 3 = 0

2cosθ  [cosθ + 3] - 1(cosθ + 3) = 0

(cosθ + 3) (2cosθ - 1) = 0

cosθ = -3     or     cosθ = 12

Note that cosθ = -3  is not possible as 1cosθ1
Thus, θ = 60

Which of the following options is/are equal to tan2A+cot2A+2 ?

A.

1cos2A sin2A

B.

sec2A cosec2A

C.

tan2Acot2A

D.

sec2A+cosec2A

#### SOLUTION

Solution : A, B, and D

tan2A+cot2A+2 can be written as (1+tan2A)+(1+cot2A)

We know that, 1+tan2A=sec2A and 1+cot2A=cosec2A

tan2A+cot2A+2=sec2A+cosec2A

=1cos2A+1sin2A

=sin2A+cos2Acos2A sin2A...(sin2A+cos2A=1)

=1cos2A sin2A

=sec2A cosec2A

Which of the following options are equal to cos2Asin2A ?

A.

12 sin2A

B.

2 sin2A

C.

2 cos2A

D.

2 cos2A1

#### SOLUTION

Solution : A and D

We know that, sin2A+cos2A=1.
1) cos2Asin2A=(1sin2A)sin2A
=12 sin2A

2) cos2Asin2A=cos2A(1cos2A)
=cos2A1+cos2A
=2 cos2A1

If A, B, C are the angles of a triangle, then
sin(B+C2)= ___.

A.

cotA2

B.

cosA2

C.

sinA2

D.

tanA2

#### SOLUTION

Solution : B

A+B+C=180
(sum of angles of a triangle)

B+C=180A

B+C2=180A2

B+C2=90A2

Taking sine on both sides we get,

sin(B+C2)=sin(90A2)

sin(B+C2)=cosA2

sin A2 sin3A2 cos3Acos A=

A.

secA

B.

cotA

C.

tanA

D.

1

#### SOLUTION

Solution : C

sin A2 sin3A2 cos3Acos A=sin A(12 sin2A)cos A(2 cos2A1)

=sin A(sin2A+cos2A2 sin2A)cos A(2 cos2A(sin2A+cos2A))

=sin A(cos2Asin2A)cosA(cos2Asin2A)

=tan A

Which of the following options is equal to sin θ+1cos θcos θ1+sin θ?

A.

cosθ1sin θ

B.

1+sinθcosθ

C.

1+sinθ1sinθ

D.

tanθ+secθ

#### SOLUTION

Solution : A, B, and D

sinθ+1cosθcosθ1+sinθ

Dividing the numerator and denominator by cos θ we get,

sinθcosθ+1cos θcosθcosθcosθcosθ1cosθ+sinθcosθ

=tanθ+secθ11secθ+tanθ

(tanθ+secθ)(sec2θtan2θ)1secθ+tanθ....(1+tan2θ=sec2θ)

=(tanθ+secθ)(secθtanθ)(secθ+tanθ)1secθ+tanθ

=(tanθ+secθ)(1secθ+tanθ)(1secθ+tanθ)

=tanθ+secθ

=sinθcosθ+1cosθ

=sinθ+1cosθ

[Multiplying the numerator and denominator by (1sinθ)]
=1+sinθcosθ×1sinθ1sinθ

=1sin2θcosθ(1sinθ)

=cos2θcosθ(1sinθ)

=cosθ1sinθ

1+sin(90θ)cos2(90θ)cos(90θ) [1+sin(90θ)]=

A.

cot θ

B.

tan θ

C.

1

D.

0

#### SOLUTION

Solution : A

1+sin(90θ)cos2(90θ)cos(90θ)[1+sin(90θ)]

=1+cos θsin2 θsin θ(1+cos θ)

=(1sin2 θ)+cos θsin θ(1+cos θ)

=cos2 θ+cos θsin θ(1+cos θ)

=cos θ(cos θ+1)sin θ(1+cos θ)=cot θ

Which of the following options is equal to sec2θ+cosec2θ ?

A.

tanθ+cotθ

B.

1

C.

cosec θ+sec θ

D.

cosec θ sec θ

#### SOLUTION

Solution : A and D

sec2 θ+cosec2 θ=1+tan2 θ+1+cot2 θ

=tan2 θ+2+cot2 θ

=tan2 θ+2(tan θ)(cot θ)+cot2 θ(tan θ cot θ=1)

=(tan θ+cot θ)2

=tan θ+cot θ

=sin θcos θ+cos θsin θ

=sin2 θ+cos2 θsin θ cos θ

=1sinθcosθ

=cosec θ sec θ