Free Objective Test 01 Practice Test - 11th and 12th
Question 1
1+cos 56∘+cos 58∘−cos 66∘=
2cos28∘ cos29∘ cos33∘
4cos28∘ cos29∘ cos33∘
4cos28∘ cos29∘ sin33∘
2cos28∘ cos29∘ sin33∘
SOLUTION
Solution : C
1+cos 56∘+cos 58∘−cos 66∘=
=2cos228∘+2sin 62∘.sin 4∘
=2cos228∘+2cos 28∘.sin 4∘
=2cos28∘(cos 28∘+cos 86∘)
=2cos28∘.2cos 57∘.cos 29∘
=4cos 28∘cos 29∘sin 33∘
Aliter: Apply the condition identify
cos A+cos B−cos C=−1+4cosA2cosB2sinc2
[∴56∘+58∘+66∘=180∘]
We get the value of required expression equal to 4cos28∘ cos29∘ sin33∘.
Question 2
√3cosec 20∘−sec 20∘=
2
2sin20∘sin40∘
4
4sin20∘sin40∘
SOLUTION
Solution : C
√3cosec 20∘−sec 20∘=√3sin 20∘−1cos 20∘
=√3 cos20∘−sin20∘sin20∘ cos20∘=2[√32cos20∘−12sin20∘]22sin20∘ cos20∘
=4cos(20∘+30∘)sin40∘=4cos50∘sin40∘=4sin40∘sin40∘=4.
Question 3
sin 20∘sin 40∘sin 60∘sin 80∘=
-316
516
316
-516
SOLUTION
Solution : C
sin 20osin 40∘sin 60∘sin 80∘
=12sin 20∘sin 60∘(2sin 40∘sin 80∘)
=12sin 20∘sin 60∘(cos 40∘−cos 120∘)
=12.√32sin 20∘(1−2sin220∘+12)
=√34sin 20∘(32−2sin220∘)
=√38(3sin 20∘−4sin320∘)
=√38sin60∘=√38.√32=316
Question 4
If sinx+cosecx=2, then sinnx+cosecnx is equal to
2
1
2n−1
2n−2
SOLUTION
Solution : A
sinx+cosecx=2⇒(sinx−1)2=0⇒sinx=1
∴sinnx+cosecnx=1+1=2
Question 5
If sin A = n sin B, then n−1n+1 tan A+B2 =
sin (A−B2)
tan (A−B2)
cot (A−B2)
cos (A−B2)
SOLUTION
Solution : B
We have sin A = n sin B ⇒ n1 = sinAsinB
⇒ n−1n+1 sinA−sinBsinA+sinB = 2cosA+B2sinA−B22sinA+B2cosA−B2
= tan A−B2 cot A+B2
⇒ n−1n+1 tan(A+B2) = tan A−B2.
Question 6
sin2A1+cos2A . cosA1+cosA =
tan A2
cot A2
sec A2
cosec A2
SOLUTION
Solution : A
(sin2A1+cos2A)(cosA1+cosA)
= 2sinAcosA2cos2A cosA1+cosA = sinA1+cosA = tan A2
Question 7
If a tan θ = b, then a cos 2θ + b sin 2θ =
a
b
-a
-b
SOLUTION
Solution : A
Given that tan θ = ba.
Now, a cos 2θ + b sin 2 θ = a(1−tan2θ1+tan2θ) + b(2tanθ1+tan2θ)
Putting tanθ = ba, we get
= a⎛⎝1−b2a21+b2a2⎞⎠ + b⎛⎝2ba1+b2a2⎞⎠ = a(a2−b2a2+b2) + b(2baa2+b2)
= 1(a2+b2) a3−ab2+2ab2 = a(a2+b2)a2+b2 = a.
Question 8
If cos 2 B = cos(A+C)cos(A−C), then tan A, tan B, tan C are in
A.P.
G.P.
H.P.
None of these
SOLUTION
Solution : B
cos 2B = cos(A+C)cos(A−C) = cosAcosC−sinAsinCcosAcosC+sinAsinC
⇒1−tan2B1+tan2B = 1−tanAtanC1+tanAtanC
⇒1+tan2B−tanAtanC−tanAtanCtan2B
= 1−tan2B+tanAtanC−tanAtanCtan2B
⇒ 2tan2B=2tanAtanC ⇒ tan2B = tan A tan C
Hence, tan A, tan B and tan C will be in G.P.
Question 9
If cos A = √32, then tan 3A =
0
12
1
∞
SOLUTION
Solution : D
We have cos A = √32 ⇒ A = 30∘
⇒ tan 3A = tan90∘ = ∞.
Question 10
If tan A2 = 32, then 1+cosA1−cosA =
-5
5
94
49
SOLUTION
Solution : D
Given that tan A2 = 32.
1+cosA1−cosA = 2cos2A22sin2A2 = cot2A2 = (23)2 = 49
Question 11
If cos θ = 35 and cos ϕ = 45, where θ and ϕ are
positive acute angles, then cos θ−ϕ2 =
7√2
75√2
7√5
72√5
SOLUTION
Solution : B
We have cos θ = 35 and cos ϕ = 45.
Therefore cos(θ−ϕ)=cosθcosϕ+sinθsinϕ
= 35 . 45 + 45 . 35 = 2425
But 2cos2(θ−ϕ2)=1+cos(θ−ϕ) = 1 + 2425 = 4950
∴ cos2(θ−ϕ2) = 4950. Hence, cos(θ−ϕ2) = 75√2.
Question 12
The maximum value of cos2(π3−x)−cos2(π3+x) is
- √32
12
√32
32
SOLUTION
Solution : C
cos2(π3−x) - cos2(π3+x)
= {cos(π3−x)+cos(π3+x)} {cos(π3−x)−cos(π3+x)}
= {2cosπ3cosx} {2sinπ3sinx}
= sin 2π3 sin 2x = √32 sin 2x
Its maximum value is √32, {- 1 ≤ sin 2x ≤ 1}.
Question 13
14[√3cos23∘−sin23∘] =
cos43∘
cos7∘
cos53∘
None of these
SOLUTION
Solution : D
14√3cos23∘−sin23∘
= 12cos30∘cos23∘−sin30∘sin23∘
= 12 cos(30∘+23∘) = 12 cos53∘.
Question 14
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
π4 radian
π3 radian
π5 radian
π10 radian
SOLUTION
Solution : C
Given, diameter of circular wire = 10cm, therefore length of wire = 10π.
Hence required angle = arcradius=10π50=π5 radian.
Question 15
The value of sin25∘+sin210∘+sin215∘+..............
+sin285∘+sin290∘ is equal to
7
8
9
192
SOLUTION
Solution : D
Given expression is
sin25∘+sin210∘+sin215∘+..............+sin285∘+sin290∘.
We know that sin90∘=1 or sin290∘=1
Similarly, sin45∘= 1√2 or sin245∘=12 and the angles are in A.P. of 18 terms. We also know that
sin285∘=[sin(90∘−5∘)]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.