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What is the answer to cos square x divide by 1 minus sin x?
cos2 x /(1 - sin x)= (1 - sin2 x )/(1 - sin x)= (1 + sin x)(1 - sin x)/(1 - sin x)= 1 + sin x
What is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
Does cotangent plus one equal cosecant?
Cotangent = 1/Tangent : Cosecant = 1/Sine Then, cot + 1 = (1/tan) + 1 = (cos/sin) + (sin/sin) = (cos + sin)/ sin. Now, cos² + sin² = 1 so for the statement to be valid the final expression would have to be : (cos² + sin² ) / sin = 1/sin. As this is not the case then, cot + 1 ≠ cosec. In fact, the relationship link is cot² + 1 = cosec²
What is the value of sin 1?
-- The sin of 1 degree is 0.01745. (rounded) -- The sin of 1 radian is 0.84147. (rounded) -- The sin of 1 grad is 0.01571. (rounded)
Prove each Indentity tanx mins sinx divided by tanxsinx equals tanxsinx divided by tanx plus sinx?
(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
What is the answer to cos square x divide by 1 minus sin x?
cos2 x /(1 - sin x)= (1 - sin2 x )/(1 - sin x)= (1 + sin x)(1 - sin x)/(1 - sin x)= 1 + sin x
What does cosx divided by 1-sinx equal?
cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x
What is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
Does cotangent plus one equal cosecant?
Cotangent = 1/Tangent : Cosecant = 1/Sine Then, cot + 1 = (1/tan) + 1 = (cos/sin) + (sin/sin) = (cos + sin)/ sin. Now, cos² + sin² = 1 so for the statement to be valid the final expression would have to be : (cos² + sin² ) / sin = 1/sin. As this is not the case then, cot + 1 ≠ cosec. In fact, the relationship link is cot² + 1 = cosec²
What is the value of sin 1?
-- The sin of 1 degree is 0.01745. (rounded) -- The sin of 1 radian is 0.84147. (rounded) -- The sin of 1 grad is 0.01571. (rounded)
Prove each Indentity tanx mins sinx divided by tanxsinx equals tanxsinx divided by tanx plus sinx?
(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True
How do you simplify this expression... cot2x over csc2x-cscx... please note that cot2x and csc2x are really just raised to the second power?
I assume the expression is cot^2 x / ( csc^2 x - csc x) express it in terms of sin x and cos x: =(cos^2 x / sin^2 x) / (1/sin^2 x - 1/sin x) =(cos^2 x / sin^2 x) / [(1 - sin x)/sin^2 x] =cos^2 x / (1 - sin x) = (1 - sin^2 x) / (1 - sin x) = (1 + sin x)(1 - sin x) / (1 - sin x) = 1 + sin x
What is the solution for cos tan csc equals 1?
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!
What is the integral of 1 divided by sin x plus cos x?
The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
What is sin 90?
sin 90 is 1
