# Free Objective Test 01 Practice Test - 11th and 12th

1+cos 56+cos 58cos 66=

A.

2cos28 cos29 cos33

B.

4cos28 cos29 cos33

C.

4cos28 cos29 sin33

D.

2cos28 cos29 sin33

#### SOLUTION

Solution : C

1+cos 56+cos 58cos 66=
=2cos228+2sin 62.sin 4
=2cos228+2cos 28.sin 4
=2cos28(cos 28+cos 86)
=2cos28.2cos 57.cos 29
=4cos 28cos 29sin 33
Aliter: Apply the condition identify

cos A+cos Bcos C=1+4cosA2cosB2sinc2
[56+58+66=180]
We get the value of required expression equal to 4cos28 cos29 sin33.

3cosec 20sec 20=

A.

2

B.

2sin20sin40

C.

4

D.

4sin20sin40

#### SOLUTION

Solution : C

3cosec 20sec 20=3sin 201cos 20
=3 cos20sin20sin20 cos20=2[32cos2012sin20]22sin20 cos20
=4cos(20+30)sin40=4cos50sin40=4sin40sin40=4.

sin 20sin 40sin 60sin 80=

A.

-316

B.

516

C.

316

D.

-516

#### SOLUTION

Solution : C

sin 20osin 40sin 60sin 80
=12sin 20sin 60(2sin 40sin 80)
=12sin 20sin 60(cos 40cos 120)
=12.32sin 20(12sin220+12)
=34sin 20(322sin220)
=38(3sin 204sin320)
=38sin60=38.32=316

If sinx+cosecx=2, then sinnx+cosecnx is equal to

A.

2

B.

1

C.

2n1

D.

2n2

#### SOLUTION

Solution : A

sinx+cosecx=2(sinx1)2=0sinx=1
sinnx+cosecnx=1+1=2

If sin A = n sin B, then  n1n+1 tan  A+B2 =

A.

sin (AB2)

B.

tan (AB2)

C.

cot (AB2)

D.

cos (AB2)

#### SOLUTION

Solution : B

We have sin A = n sin B   n1sinAsinB

n1n+1  sinAsinBsinA+sinB2cosA+B2sinAB22sinA+B2cosAB2

= tan  AB2 cot  A+B2

n1n+1 tan(A+B2) = tan  AB2.

sin2A1+cos2AcosA1+cosA =

A.

tan A2

B.

cot  A2

C.

sec A2

D.

cosec A2

#### SOLUTION

Solution : A

(sin2A1+cos2A)(cosA1+cosA)

= 2sinAcosA2cos2A  cosA1+cosAsinA1+cosA = tan  A2

If a tan θ = b, then a cos 2θ + b sin 2θ

A.

a

B.

b

C.

-a

D.

-b

#### SOLUTION

Solution : A

Given that tan θba.

Now, a cos 2θ + b sin 2 θ = a(1tan2θ1+tan2θ) + b(2tanθ1+tan2θ)

Putting tanθba, we get

= a1b2a21+b2a2 + b2ba1+b2a2 = a(a2b2a2+b2) + b(2baa2+b2)

1(a2+b2) a3ab2+2ab2a(a2+b2)a2+b2 = a.

If cos 2 B =  cos(A+C)cos(AC), then tan A, tan B, tan C are in

A.

A.P.

B.

G.P.

C.

H.P.

D.

None of these

#### SOLUTION

Solution : B

cos 2B =  cos(A+C)cos(AC)cosAcosCsinAsinCcosAcosC+sinAsinC

1tan2B1+tan2B1tanAtanC1+tanAtanC

1+tan2BtanAtanCtanAtanCtan2B

= 1tan2B+tanAtanCtanAtanCtan2B

2tan2B=2tanAtanC  tan2B = tan A tan C

Hence, tan A, tan B and tan C will be in G.P.

If cos A =  32, then tan 3A =

A.

0

B.

12

C.

1

D.

#### SOLUTION

Solution : D

We have cos A =  32 A = 30

tan 3A = tan90 = .

If tan  A232, then  1+cosA1cosA =

A.

-5

B.

5

C.

94

D.

49

#### SOLUTION

Solution : D

Given that tan  A232.

1+cosA1cosA2cos2A22sin2A2 = cot2A2 = (23)249

If cos θ35 and cos ϕ =   45, where θ and ϕ are

positive acute angles, then cos  θϕ2

A.

72

B.

752

C.

75

D.

725

#### SOLUTION

Solution : B

We have cos θ35 and cos ϕ45.

Therefore cos(θϕ)=cosθcosϕ+sinθsinϕ

=  354545352425

But 2cos2(θϕ2)=1+cos(θϕ) = 1 +  24254950

cos2(θϕ2)4950. Hence, cos(θϕ2)752.

The maximum value of cos2(π3x)cos2(π3+x) is

A.

- 32

B.

12

C.

32

D.

32

#### SOLUTION

Solution : C

cos2(π3x)cos2(π3+x)

= {cos(π3x)+cos(π3+x)}  {cos(π3x)cos(π3+x)}

= {2cosπ3cosx} {2sinπ3sinx}

= sin 2π3 sin 2x =  32 sin 2x

Its maximum value is  32, {- 1 sin 2x 1}.

14[3cos23sin23]

A.

cos43

B.

cos7

C.

cos53

D.

None of these

#### SOLUTION

Solution : D

143cos23sin23

= 12cos30cos23sin30sin23

=  12 cos(30+23) =  12 cos53.

The circular wire of diameter 10cm is cut and placed along the circumference of a circle of diameter 1 metre. The angle subtended by the wire at the centre of the circle is equal to

A.

B.

C.

D.

#### SOLUTION

Solution : C

Given, diameter of circular wire = 10cm, therefore length of wire = 10π.

The value of sin25+sin210+sin215+..............

+sin285+sin290 is equal to

A.

7

B.

8

C.

9

D.

192

#### SOLUTION

Solution : D

Given expression is

sin25+sin210+sin215+..............+sin285+sin290.

We know that sin90=1 or sin290=1

Similarly, sin45= 12 or sin245=12 and the angles are in A.P. of 18 terms. We also know that

sin285=[sin(905)]2=cos25.

Therefore from the complementary rule, we find sin25+sin285=sin25+cos25=1

Therefore,

sin25+sin210+sin215+.............+sin285+sin290

=(1+1+1+1+1+1+1+1)+1+12=912.