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Cos x = 1 / Sec x so 1 / Cos x = Sec x


Tan x = Sin x / Cos x = Sin x * (1 / Cos x) = Sin x * Sec x

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โˆ™ 2010-03-02 13:02:55
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: How do you solve Sin x sec x equals tan x?
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How do you identify sec x sin x equals tan x?

Rewrite sec x as 1/cos x. Then, sec x sin x = (1/cos x)(sin x) = sin x/cos x. By definition, this is equal to tan x.

How do you solve the following identity sec x - cos x equals sin x tan x?

sec x - cos x = (sin x)(tan x) 1/cos x - cos x = Cofunction Identity, sec x = 1/cos x. (1-cos^2 x)/cos x = Subtract the fractions. (sin^2 x)/cos x = Pythagorean Identity, 1-cos^2 x = sin^2 x. sin x (sin x)/(cos x) = Factor out sin x. (sin x)(tan x) = (sin x)(tan x) Cofunction Identity, (sin x)/(cos x) = tan x.

Sec x times sin x divided by tan x?

1 (sec x)(sin x /tan x = (1/cos x)(sin x)/tan x = (sin x/cos x)/tan x) = tan x/tan x = 1

What is sec x - tan x?

In algebra and trigonometry we can have various functions such as sin, cosine , tan and sec and to solve trigonometric equations we should know relation between them . sec x = 1 / cos x. tan x = sin x/ cos x. (1- sinx )/ cos x.

What is the derivative of y equals sec x?


Sec - cos equals tansin?

Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D

How do you verify the identity of cos ฮธ tan ฮธ equals sin ฮธ?

To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED

How to solve trigonometry problems easily?

The most easiest method to solve trigonometric problems is to be place the values of the sin/cos/tan/cot/sec/cosec . The values will help to solve the trigonometric problems with less difficulty.

Sin x Tan x equals Sin x?

No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)

Tan equals 0.3421 sin equals 3237 Which quadrant does it terminate?

Assuming sin equals 0.3237, the angle is in quadrant I.

How do you prove the following identity sec x - cos x equals sin x tan x?

you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx

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.

Tan equals 0.3421 sin equals 0.3237 Which quadrant does it terminate?

The value of tan and sin is positive so you must search quadrant that tan and sin value is positive. The only quadrant fill that qualification is Quadrant 1.

How do you prove tan x plus tan x sec 2x equals tan 2x?

tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x

What are solutions to the equation tan squared x minus three equals zero?

Solve for x, where tan² x - 3 = 0. tan² x = 3; then, sec² x = tan² x + 1 = 4, sec x = ±2, and cos x = 1 /sec x = ±½. Now, we know that sin 30° = ½; whence cos 60° = ½. Therefore, if 0 ≤ x < 2π, then x = 60°, 120°, 240°, or 300°; or, in radians, x = ⅓π, ⅔π, 1⅓π, or 1⅔π.

The expression sec x - sin x tan x is equivalent to?

cos x

Sec squared theta plus tan squared theta equals to 13 12 and sec raised to 4 theta minus tan raised to 4 theta equals to how much?

It also equals 13 12.

Solution for tan x plus cot x divided by sec x csc x?

(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.

What is the solution to cos tan-sin over cot equals 0?

either cos OR tan-sin equals zero socos=0 at pi/2 and 3pi/2ortan=sin which is impossibleim not sure though

Csc divided by cot squared equals tan multiplied by sec?


How do you simplify csc theta tan theta?

With all due respect, you don't really want to know howto solve it.You just want theΘ) = 1/sin(Θ)tan(Θ) = sin(Θ)/cos(Θ)csc(Θ) x tan(Θ) = 1/sin(Θ) x sin(Θ)/cos(Θ) = 1/cos(Θ) = sec(Θ)

How do you prove sin x tan x equals cos x?

You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.

How do you solve secant-tangent and tangent-tangent angles?

you solve secant angles when you have the hypotenuse and adjacent sides. sec=1/cos or, cos^-1 (reciprocal identity property) Tangent is solved when you have adjacent and opposite sides, or you can look at it as its what you use when you dont have the hypotenuse. tan=sin/cos or tan=opp/adj or tan=y/x

What is sec theta - 1 over sec theta?

Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta

How do you prove that the derivative of sec x is equal to sec x tan x?

Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.