If (1-tan 20 cot220)/(tan 1520- cot 880)= k√3, then value of k is
1
-1
1/2
-1/2
The value of cos θ+3√2 sin( θ+π/4)+6 lies between
2 and 12
-2 and 12
1 and 11
-1 and 11
cos2A+cos2 (1200+A)+ cos2 (1200-A)=
3/2
3/18
1/4
If (sin x+ cos x)/(cos3 x)= a tan3 x+ b tan2 x+c tan x+d then a+b+c+d=
0
2
4
- 2
If sin x+sin y=1/4, sin x- siny=1/5, then 4 cot(x-y/2)=
5cot(x-y/2)
5cot(x+y/2)
5tan(x-y/2)
5tan(x+y/2)
If sin θ+ cos θ=a then sin4 θ+ cos 4 θ=
1-(1/2) (a2+1)2
1-(1/2) (a2-1)2
1+ (1/2) (a2+1)2
1+ (1/2) (a2-1)2
4(cos3 200 +cos34 00) =
3(cos 200+cos 400)
3(cos 100+sin 200)
3(cos 200+sin 400)
3(cos 100+cos 200)
cos 10+ cos 20+ cos 30+...+ cos 1790=
89
If tan2 A= 2 tan2 B+1, then cos 2A+ sin2 B=
3
If A+B+C= 1800 then sin2 A/2+ sin 2 B/2 - sin2 C/2=
1- 2 sin A/2 sin B/2 sin C/2
1- 2 cos A/2 cos B/2 cos C/2
1- 2 cos A/2cos B/2 sin C/2
1- 2 sin A/2cos B/2 sin C/2
If A+B+C=1800 then sin A+sin B+sin C=
4sin A/2 sin B/2 cos C/2
4 cos A/2 cos B/2 cos C/2
4sin A/2 cos B/2 cos C/2
4 sin A/2 cos B/2 sin C/2
If A+B+C= 1800 then sin 2A+ sin 2B- sin 2C=
4sin A cos B sin C
4 cos A sin B sin C
4 cos Acos B sinC
4 cos Asin B cos C
If A+B+C+D= 2π, then -4 cos (A+B/2) sin (A+C/2) cos (A-D/2)=
sin A+ sin B+ sin C- sin D
sin A- sin B+ sin C- sin D
sin A+ sin B+sin C+ sin D
sin A- sin B+ sin c+ sin D
The maximum value of 5 cos x+ 3 cos (x- 600)+7 is
11
13
14
17
8 sin θ cos θ cos 2θ cos 4θ=
sin 8θ
cos 8θ
sin 4θ
cos 4θ
2 cos 540. Sin 660=
√3/2+ sin120
√3/2-sin120
√3/2+ cos 120
√3/2-cos 120
If 4 sin(600+ θ) sin(600-θ)-1= kcos 2θ, the value of k is
none
If tan A=18/17, tan B= 1/35 then tan(A-B)=
(sin(n+1)α- sin(n-1)α)/ (cos(n+1)α+ 2 cos nα+ cos(n-1)α)=
tan α
cot α
tan α/2
cot α/2
cos2 360+ cos2 720=
2/3
-2/3
3/4
cos (θ + α).cos (θ - α)+ sin (θ + α). sin(θ - α)=
cos 2α
sin 2α
cot 2α
tan 2α
√3 cosec 200- sec 200 =
(sin 4θ)/(sin θ)=
8cos3θ-4cosθ
8sin3θ-4sinθ
4cos3θ-8cosθ
4sin3θ-8sinθ
1- cos A+ cos B- cos (A+B)/1+cos A- cos B- cos(A+B)=
sin A/2. Cos B/2
tan A/2.cot B/2
sec A/2.cosec B/2
2 sec2 θ- sec4 θ- 2 cosec2 θ+ cosec4θ=
cot4θ- tan4 θ
sec4θ- cosec4 θ
sin4θ- cos4θ
sec4θ- cot4 θ
If A=cos 150- cos 750,B= tan 150+ tan 750, C= cos2450 – sin2 150 then ascending order is
A,B,C
C,A,B
B,C,A
C,B,A
In ΔABC, 1+4 sin(π-A/4)sin(π-B/4)sin(π-C/4)=
sin A/2+ sin B/2+ sin C/2
cos A/2+ cos B/2+ cos C/2
sin A/2+ sin B/2- sin C/2
cos A/2+ cos B/2- cos C/2
sin5θ/sinθ is equal to
16cos4θ – 12cos2θ + 1
16cos4θ + 12cos2θ + 1
16cos4θ – 12cos2θ – 1
16cos4θ + 12cos2θ – 1
If A+B+C =900 then (cot A+cot B+cot C)/(cot A cot B cot C) =
-2
A student has to answer 10 out of 13 questions in an examination choosing at least 5 questions from the first 6 questions. The number of choice available to the student is
63
91
161
196