# Free Introduction to Trigonometry 03 Practice Test - 10th Grade

cos 1 × cos 2 × cos 3 ×……..× cos 180 is equal to:

A.

1

B.

0

C.

12

D.

-1

#### SOLUTION

Solution : B

We know that, cos 90 = 0
The given expression
cos 1 × cos 2 × cos 3  ×....× cos 90  ×……..× cos 180 reduces to zero as it contains  cos 90 which is equal to 0.

sinAsinBcosA+cosB+cosAcosBsinA+sinB=

A.

- 1

B.

0

C.

1

D.

2

#### SOLUTION

Solution : B

sinAsinBcosA+cosB+cosAcosBsinA+sinB

= (sinAsinB)(sinA+sinB) + (cosAcosB)(cosA+cosB)(cosA+cosB)(sinA+sinB)
=sin2Asin2B+cos2Acos2B(cosA + cosB)(sinA + sinB)
= sin2A  + cos2A   (sin2B +  cos2B)(cosA + cosB)(sinA + sinB)
= 11(cosA + cosB)(sinA + sinB)
= 0

Which of the following options is equal to tan A1cot A+cot A1tan A ?

A.

1+cosec A sec A

B.

1+tan A+cot A

C.

cosec A  sec A

D.

tan A + cot A

#### SOLUTION

Solution : A and B

tan A1cot A+cot A1tan A

=tan A1(1tan A)+1tan A1tan A

=tan Atan A1tan A+1tan A(1tan A)

=tan2 Atan A11tan A(tan A1)(ab=(ba))

=tan3 A1tan A(tan A1)

=(/tan A1)(tan2 A+tan A+1)tan A(/tan A1)[a3b3=(ab)(a2+ab+b2)]

=tan2 Atan A+tan Atan A+1tan A

=tan A + 1 + cot A

=1+sin AcosA+cos AsinA

=1+(sin2 A+cos2 Asin A cos A)

=1+1sin A cos A

=1+cosec A sec A

(sec A+tan A1)(sec Atan A+1)tan A=

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

#### SOLUTION

Solution : D

(sec A+tan A1)(sec Atan A+1)tan A

=[sec A+(tan A1)]  [sec A(tan A1)]tan A

=sec2 A(tan A1)2tan A

=sec2 A(tan2 A2 tan A+1)tan A

=sec2 Atan2 A+2 tan A1tan A

=1+2 tan A1tan A(1+tan2 A=sec2 A)

=2 tan Atan A

=2

If  x=a sec θ+b tan θ and y=a tan θ+b sec θ, then x2y2a2b2= ___

A.

2

B.

1

C.

0

D.

- 1

#### SOLUTION

Solution : B

x=a sec θ+b tan θ, y=a tan θ+b sec θ

Squaring both sides we get,

x2=(a sec θ+b tan θ)2, y2=(a tan θ+b sec θ)2

x2=a2 sec2 θ+2ab sec θ tan θ+b2 tan2 θy2=a2 tan2 θ+2ab sec θ tan θ+b2 sec2 θ                                                           –––––––––––––––––––––––––––––––––––––––––––Subtracting, x2y2=a2(sec2 θtan2 θ)+b2(tan2 θsec2 θ)

x2y2=a2(sec2 θtan2 θ)b2(sec2 θtan2 θ)

x2y2=a2b2(sec2 θtan2 θ=1)

x2y2a2b2=1

tan2 θ(sec θ1)2=

A.

1+sin θ1sin θ

B.

1

C.

tan θ

D.

1+cos θ1cos θ

#### SOLUTION

Solution : D

tan2 θ(sec θ1)2=sec2 θ1(sec θ1)2

=(sec θ1)(sec θ+1)(sec θ1)(sec θ1)

=1cos θ+11cos θ1

=1+cos θcos θ1cos θcos θ

=1+cos θ1cos θ

sec 17cosec 73+tan 68cot 22[cos2 44+cos2 46]=___.

A.

0

B.

-1

C.

2

D.

1

#### SOLUTION

Solution : D

sec 17cosec 73+tan 68cot 22[cos2 44+cos2 46]

=sec 17cosec (9017)+tan 68cot (9068)[cos2 44+cos2 (9044)]

=sec 17sec 17+tan 68tan 68[cos2 44+sin2 44]....(complementary angles)

=1+11

=1

(1+tan A)2+(1tan A)2=

A.

2

B.

1

C.

2 sec2 A

D.

2 cos2 A

#### SOLUTION

Solution : C

(1+tanA)2+(1tanA)2
=(1+tan2A+2tanA)+(1+tan2A2tanA)
=(2+2tan2A+2tanA2tanA)
=(2+2tan2A)
=2(1+tan2A)
=2sec2A

11+tan2θ+11+cot2θ=

A.

-1

B.

0

C.

1

D.

2

#### SOLUTION

Solution : C

We know that,1+tan2θ=sec2θ and 1+cot2θ=cosec2θ

11+tan2θ+11+cot2θ=1sec2θ+1cosec2θ

=cos2θ + sin2θ

=1

(sinA+cosecA)2(sinAcosecA)2=

A.

0

B.

1

C.

2

D.

4

#### SOLUTION

Solution : D

(sinA+cosec A)2(sinAcosec A)2

=4sinA cosec A

as [(a+b)2(ab)2=4ab]

now, (sin A×cosec A=sinA×1sinA=1)

4.sinA cosec A=4×1=4