Eamcet - Maths - Trigonometric Ratios And Identities Test

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In a ΔABC, cos[(B+2C+3A)/2]+cos[(A-B)/2] is equal to

If a, b, c form a geometric progression with common ratio r,then the sum of the ordinates of the Points of intersection of the line ax + by + c = 0 and the curve x + 2y2=0

P(-1, -3) is a centre of similitude for the two circles x2+y2=1 and x2+y2-2x-6y+6=0. The length of the common tangent through P to the circle is

sin 100 +sin 200 +sin  400 +sin 500 -sin 700 -sin 800=

cosec2 θ. cot2 θ- sec2 θ. tan2 θ-( cot2 θ- tan2 θ)( sec2 θ. cosec2θ-1)=

cos (θ + α).cos (θ - α)+ sin (θ + α). sin(θ - α)=

The period of cot(5x+3)+sin(3x+4)/ sec(3-4x)-cos(4-6x) is

sin 120 sin 240 sin 480 sin 840 =

The range of sinn-1 √x is

If sin-1x+sin-1(1-x)=cos-1x, then x ε to

sin (θ/2)sin (7θ/2)+sin(3θ/2). Sin(11θ/2)- sin 2θsin 5θ=

(sin θ+ cosec θ)2+(cos θ+ sec θ)2 =

The function f(x) = xe-x (x ∈R) attains a maximum value at x = …….

If Sinx Sinhy= cos θ and Cosx Coshy= Sinθ, then Cosh2y + Cos2x

If the area of the triangle formed by the pair of lines 8x2-6xy+y2=0 and the line 2x + 3y = a is 7 then

If tan400= λ then (tan 1400- tan 1300)/(1+ tan 1400 tan 1300) =

cos θ- cos (600+ θ)-cos (600- θ)=

tan 82(10/2)=

sec 2550

If A+B+C= 1800 then 4 cos(π-A/4)cos (π-B/4) cos(π-C/4)=

I: If tan A + tan B = P and Cot A + Cot B = q then cot (A + B) = (q-p) / pq II : If 2 tan A + cot A = tan B then cot A +2 tan (A –B) = 0

If cos θ= 3/5 and θ is not in the first quadrant,then (5tan(π+ θ)+4 cos(π- θ))/(5sec(2π-θ)- 4 cot(2π+θ) )

The period of sin x sin(1200+x) sin(1200-x)is

If A+B+C = 3600 then tan A/2+ tan B/2+ tan C/2=

2(sin6 x+ cos6 x)-3(sin4 x+ sin2 x)+1=

The minimum of cos2(1200+x)+ cos2(1200-x)is

The maximum value of sin2 θ+ cos4 θ is

If A=cos 150- cos 750,B= tan 150+ tan 750, C= cos2450 – sin2 150 then ascending order is

1- cos A+ cos B- cos (A+B)/1+cos A- cos B- cos(A+B)=

tan (5π/32)+2 tan (5π/16)+ 4 tan (5π/8)- cot (5π/32)=

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