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The period of sin^{2}θ is [2002]
; Period
The number of solution of tan x + sec x = 2cos x in [0, 2π) is [2002]
The given equation is tanx + secx = 2 cos x;
⇒ sin x + 1 = 2cos^{2 }x ⇒ sin x + 1 = 2(1 – sin^{2} x);
⇒ 2sin^{2}x + sin x – 1= 0;
⇒ (2sin x – 1)(sin x + 1) = 0 ⇒ sin x = , –1.;
⇒ x = 30°, 150°, 270°
Which one is not periodic [2002]
∵ cos is non periodic
∴ cos + cos ^{2}x can not be periodic.
Let a, b be such that p < a  b < 3p. If and then the value of [2004]
.....(1)
.....(2)
Square and add (1) and (2)
If then the difference between the maximum and minimum values of u^{2} is given by [2004]
u^{2} = a^{2} + b^{2 }
Now (a^{ 4} + b^{4} ) cos^{2}θ sin ^{2}θ + a^{ 2}b ^{2} (cos ^{4}θ+ sin^{4}θ)
= (a^{4} + b^{4} ) cos ^{2}θ sin^{2}θ + a^{ 2} b ^{2} (1  2 cos ^{2}θ sin^{2}θ)
= (a^{4} + b^{4}  2a ^{2 }b^{2}) cos^{2}θ sin ^{2}θ+ a ^{2} b^{2}
…(2)
∴ from (1) , (2) and (3)
Minimum value of u^{2} = a^{ 2} + b^{2}
Maximum value of u^{2}
= a^{ 2} + b^{2}
= a^{ 2} + b^{2} = 2(a ^{2}+b^{2})
∴ Max value  Min value
= 2(a^{2} + b^{2} )  (a +b^{2} ) = (ab)^{2}
A line makes the same angle θ, with each of the x and z axis.
If the angle β, which it makes with yaxis, is such that sin ^{2}β = 3 sin ^{2}θ, then cos^{2}θ equals [2004]
The direction cosines of the line are cosθ, cosβ, cosθ
∴ cos ^{2}θ + cos ^{2}β + cos^{ 2}θ = 1
⇒ 2 cos^{ 2}θ = sin ^{2}β = 3 sin ^{2}θ (given)
⇒ 2 cos ^{2}θ = 3  3 cos^{ 2}θ
The number of values of x in the interval [0, 3π] satisfying the equation 2 sin^{ 2} x + 5 sinx  3 =0 is [2006]
2 sin ^{2} x + 5 sinx  3=0
⇒ (sin x + 3)(2 sinx  1) = 0
⇒ sin x = and sin x ≠3
∴ In [0, 3π] , x has 4 values.
If 0 < x < π and cosx + sin x = , then tan x is [2006]
⇒ 3 tan 2 x + 8 tanx + 3= 0
⇒ 3 tan 2 x + 8 tanx + 3= 0
as tan x <0
Let A and B denote the statements A : cosα + cosβ + cosγ = 0 B : sinα + sinβ + sinγ = 0
If cos (β – γ) + cos (γ – α) + cos (α – β) = then : [2009]
We have
cos (β – γ) + cos (γ – α) + cos (α – β) =
⇒ 2 [cos (β – γ) + cos (γ – α) + cos (α – β)] + 3 = 0
⇒ 2 [cos (β – γ) + cos (γ – α) + cos (α – β)]
+ sin^{2}α + cos^{2}α + sin^{2}β + cos^{2}β + sin^{2}γ + cos^{2}γ = 0
⇒ [sin^{2}α + sin^{2}β + sin^{2}γ + 2 sinα sinβ + 2 sinβ sinγ
+ 2 sinγ sinα ] + [cos^{2}α + cos^{2}β + cos^{2}γ + 2cosα cosβ
+ 2 cosβ cosγ + 2cosγ cosα] = 0
⇒ [sinα + sinβ + sinγ]^{2} + (cos2α + cosβ + cos2γ )^{2}= 0
⇒ sinα + sinβ + sinγ = 0 and cosα + cosβ + cosγ =0
∴ A and B both are true.
Let and sin where Then tan 2α = [2010]
If A = sin^{2} x + cos^{4}x, then for all real x : [2011]
A = sin^{ 2} x+ cos^{4}x = sin^{ 2} x + cos ^{2} x(1 sin^{2}x)
Now
In a ΔPQR, If 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R is equal to : [2012]
Given 3 sin P + 4cos Q = 6 ...(i)
4 sin Q + 3cos P = 1 ...(ii)
Squaring and adding (i) & (ii) we get
9 sin^{2} P + 16cos^{2}Q + 24 sin P cos Q
+ 16 sin^{2}Q + 9cos^{2 }P + 24 sin Q cos P = 36 + 1 = 37
⇒ 9 (sin^{2}P + cos^{2}P) + 16 (sin^{2} Q + cos^{2} Q) + 24 (sinP cosQ + cosP sinQ) = 37
⇒ 9 + 16 + 24 sin ( P + Q) = 37
[∵ sin^{2}θ + cos^{2}θ = 1 and sinα cos B + cos2α sinβ = sin (A + B)]
⇒ (∵P + Q + R = p)
If R =then 0 < P,Q <
⇒ cos Q <1 and sin P <
⇒ 3 sin P + 4 cos Q < which is not true.
So R =
A B C D is a trapezium such that A B and CD are parallel and BC ⊥ CD. If ΔADB = θ, BC = p and CD = q, then AB is equal to : [JEE M 2013]
From Sine Rule
The expression can be written as :[JEE M 2013]
Given expression can be written as
= 1 + sec A cosec AA
Let where x ∈ R and k ≥ 1.Then f_{4} (x)  f_{6} (x) equals [JEE M 2014]
Let Consider
If 0 ≤ x < 2p, then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0 is: [JEE M 2016]
cos x + cos 2x + cos 3x + cos 4x = 0
⇒ 2 cos 2x cos x + 2 cos 3x cos x = 0
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