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Well I don't exactly get "the solution", but simplifying the equation is quite simple. Maybe that's what you're looking for. Here are the steps for simplifying it.
costancsc = 1
1. Change tan to sin/cos
2. Change csc to 1/sin
cos(sin/cos)(1/sin) = 1
And as you can now see, the first cos cancels with the second one under the sin/cos fraction, and the first sin cancels with the second one under the 1/sin fraction consequentially leaving you with 1 = 1.
For a better look, notice this fraction when all three parts are combined
cos * sin * 1
----------------
cos * sin
See how the cos and sin cancel each other leaving you with 1 * 1 * 1 which is just 1. Therefore the final simplification is just 1 = 1.
I hope this helps!
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Ross
Beau
LaoTan = Sin/Cos
Csc = 1/Sin
Hence
Cos Tan Csc =
Cos * Sin/Cos * 1/ Sin
Cancel down by 'cos'.
1 * Sin/1 * 1 / Sin
Cancel down by 'son'
Hence
1* 1/1 * 1/1/ = 1/1/ = 1
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How do you simplify csc theta cot theta?
There are 6 basic trig functions.sin(x) = 1/csc(x)cos(x) = 1/sec(x)tan(x) = sin(x)/cos(x) or 1/cot(x)csc(x) = 1/sin(x)sec(x) = 1/cos(x)cot(x) = cos(x)/sin(x) or 1/tan(x)---- In your problem csc(x)*cot(x) we can simplify csc(x).csc(x) = 1/sin(x)Similarly, cot(x) = cos(x)/sin(x).csc(x)*cot(x) = (1/sin[x])*(cos[x]/sin[x])= cos(x)/sin2(x) = cos(x) * 1/sin2(x)Either of the above answers should work.In general, try converting your trig functions into sine and cosine to make things simpler.
What is the Sine reciprocal?
Sine Its reciprocal is Cosecant Algebraically Sin ; Reciprocal is '1/ Sin' known as 'Cosecant(Csc)'. Similarly Cos(Cosine) ; 1/ Cos (Secant(Sec)) Tan(Tangent) ; 1/ Tan ( Cotangent(Cot)).
What is sin cos tan csc sec cot of 30 45 60 degrees?
Sin(30) = 1/2 Sin(45) = root(2)/2 Sin(60) = root(3)/2 Cos(30) = root(3)/2 Cos(45) = root(2)/2 Cos(60) = 1/2 Tan(30) = root(3)/3 Tan(45) = 1 Tan(60) = root(3) Csc(30) = 2 Csc(45) = root(2) Csc(60) = 2root(3)/3 Sec(30) = 2root(3)/3 Sec(45) = root(2) Sec(60) = 2 Cot(30) = root(3) Cot(45) = 1 Cot(60) = root(3)/3
What is the reciprocal function of sin a?
Sin(A) = Opposite/Hypotenuse Its reciprotcal is 1/Sin(A) = Cosecant(A) = Csc(A) = Hypotenuse / Opposite. Similarly Cos(A) = Adjacent/Hypotenuse Its reciprotcal is 1/Cos(A) = Secant(A) = Sec(A) = Hypotenuse / Adjacent Tan(A) = Opposite/Adjacent Its reciprotcal is 1/Tan(A) = Cotangent(A) = Cot(A) = Adjacent / Opposite.
How tan9-tan27-tan63 tan81 equals 4?
tan(9) + tan(81) = sin(9)/cos(9) + sin(81)/cos(81)= {sin(9)*cos(81) + sin(81)*cos(9)} / {cos(9)*cos(81)} = 1/2*{sin(-72) + sin(90)} + 1/2*{sin(72) + sin(90)} / 1/2*{cos(-72) + cos(90)} = 1/2*{sin(-72) + 1 + sin(72) + 1} / 1/2*{cos(-72) + 0} = 2/cos(72) since sin(-72) = -sin(72), and cos(-72) = cos(72) . . . . . (A) Also tan(27) + tan(63) = sin(27)/cos(27) + sin(63)/cos(63) = {sin(27)*cos(63) + sin(63)*cos(27)} / {cos(27)*cos(63)} = 1/2*{sin(-36) + sin(90)} + 1/2*{sin(72) + sin(36)} / 1/2*{cos(-36) + cos(90)} = 1/2*{sin(-36) + 1 + sin(36) + 1} / 1/2*{cos(-36) + 0} = 2/cos(36) since sin(-36) = -sin(36), and cos(-36) = cos(36) . . . . . (B) Therefore, by (A) and (B), tan(9) - tan(27) - tan(63) + tan(81) = tan(9) + tan(81) - tan(27) - tan(63) = 2/cos(72) – 2/cos(36) = 2*{cos(36) – cos(72)} / {cos(72)*cos(36)} = 2*2*sin(54)*sin(18)/{cos(72)*cos(36)} . . . . . . . (C) But cos(72) = sin(90-72) = sin(18) so that sin(18)/cos(72) = 1 and cos(36) = sin(90-36) = sin(54) so that sin(54)/cos(36) = 1 and therefore from C, tan(9) – tan(27) – tan(63) + tan(81) = 2*2*1*1 = 4
