Test: Trigonometric Functions - 2 - JEE MCQ

Let the angles of the triangle are (a - d), a, (a + d)
Now, in a triangle, the sum of all angles equal to 180 degree
⇒ (a - d) + a + (a + d) = 180
⇒ a - d + a + a + d = 180
⇒ 3a = 180
⇒ a = 180/3
⇒ a = 60
Now, the angle are (60 - d), 60, (60 + d)
Now, 60 - d is the least and 60 + d is the greatest angle.
Now, (60 + d)° = {(60 + d) * (π/180)}c
Given that, number of radians in the greatest angle/number of degrees in the least one = π/60
⇒ {(60 + d) * (π/180)}c /(60 - d) = π/60
⇒ (60 - d)/{(60 + d) * (π/180)}c = 60/π
⇒ 180(60 - d)/{(60 + d) * π} = 60/π
⇒ 180(60 - d)/(60 + d) = 60
⇒ (60 - d)/(60 + d) = 60/180
⇒ (60 - d)/(60 + d) = 1/3
⇒ 3(60 - d) = (60 + d)
⇒ 180 - 3d = 60 + d
⇒ 180 - 60 = 3d + d
⇒ 4d = 120
⇒ d = 120/4
⇒ d = 30

1/root 2 ; root 2 is largest value of sin theta and cos thetha

sin(120 – α) = sin(120 – β), 0 < α and β < π
⇒ either 120 – α = 120 – β ⇒ α = β]
or 120 – α = 180 – (120 – β)
∴ 120 – α = 60 + β 
∴ α + β = 60° = (π/3)
∴ α = β or α + β = (π/3).

Discriminant of the given equation = (cos p) 2 − 4(cosp − 1) sin p 
= cos2p + 4 (1 − cosp) sin p ≥ 0, if p ∈ (0, π) 
[∵ cos2p ≥ 0, 0 ≤ 1 − cosp ≤ 2 and sin p > 0 for all p ∈ (0, π)]

sinθ+cscθ=2
⇒ sinθ+1/sinθ=2
⇒ sin²θ−2sinθ+1=0
⇒ (sinθ−1)² = 0
⇒ sinθ=1
⇒ sin²θ + csc²θ
= sin²θ + 1/sin²θ
= 1+1
= 2

tan75° - cot75°
= sin75°/cos75° - cos75°/sin75°
= 2(sin² 75° - cos² 75°)/2sin75°cos75°
⇒ -2cos150°/sin150°
= -2cot150° 
= -2cot(180-30)°  = 2cot30°
= 2(3)^1/2

Let the angles of a triangle are x° ,2x°. 3x°.acordingly:-

x°+2x° +3x°=180°

6x°=180°

or. x=30°

Angles are 30° , 60°. and. 90°.

Let a. , b and c units be the length of the sides of the triangle. Applying

sine rule

a/sin30° = b /sin60° = c/sin90°

2.a/1= 2b/√3. = c/1

a= c/2 , b= (c.√3)/2.

a : b : c = c/2. : c√3/2. : c. = 1/2. : √3/2 : 1. or 1 : √3. : 2. Answer.

As the angles A, B, C of ∆ ABC are in AP
∴ Let A = x – d, B = x, C = x + d
But A + B + C = 180° (∠ Sum prop. of ∆)
∴ x – d + x + x + d = 180°
⇒ 3x = 180° ⇒ x = 60° ∴ ∠B = 60°
Now by sine law in ∆ ABC, we have
b/sin B = c/sin C ⇒ sin B/sin C
⇒ √3/√2 = sin 60°/sin C [using b : c = √3 : √2 and ∠B = 60°]
⇒ √3/√2 = √3/2 sin C ⇒ sin C = 1/√2 = sin 45°
∴ ∠C = 45° ⇒ ∠A = 180° - (∠B + ∠C)
= 180° - (60° + 45°) = 75°

sinθ = sin(π−θ)
sin(9/14π) = sin(π−9/14π) = sin5/14π
(sin11/14)π = (sin3/14)π
(sin 13/14)π = (sinπ/14)
= (sin² π/14) (sin² 3/14)π (sin² 5/14)π
= [(sinπ/14) (sin3/14)π (sin5/14)π]²
= 1/[(4sin²π)/7][sin2/7 π soc 2/7π ⋅ cos(π−4/7π)]²
= 1/16sin² π/7[1/4(sin²8/7)π]
= 1/64{sin²(π+π/7)sin² π/7}
= 1/64{sin²(π/7)sin²(π/7)}
= 1/64

cos(x+π/6) + sin(x+π/6)
√2[1/√2 cos(x+π/6) + 1/√2 sin(x+π/6)]
= √2 cos[x + π/6 - π/4]
= √2 cos[x - π/12]
For maximum value x = π/12

sinx+cosx=2—(1)sinx+cosx=2—(1)

Divide by 2–√2on both LHSandRHS=>sinx∗12√+cosx∗12√=22√=2–√—(2)LHSandRHS=>sinx∗12+cosx∗12=22=2—(2)

Since sin45=cos45=12√sin45=cos45=12, the equation(2)(2) can be written assinxcos45+cosxsin45=2–√—(3)sinxcos45+cosxsin45=2—(3)

SincesinAcosB+cosAsinB=sin(A+B)sinAcosB+cosAsinB=sin(A+B), equation (3)(3)can be written as sin(x+45)=2–√—(4)sin(x+45)=2—(4)

The known facts are:

• The value for sinAsinAis always between−1−1and11
• 2–√=1.4142=1.414, which is greater than 1

From the above facts, LHS of the equation (4) is always between -1 and 1 whereas RHS is > 1. This is not possible and hence six+cosxsix+cosx can never be equal to 2.

CosA value can vary between -1 and +1..here any value of "t" except "0" will fall out of this range

Thus, the value of xy + yz + zx is 0.