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sinx*secx ( secx= 1/cos )

sinx*(1/cosx)

sinx/cosx=tanx

tanx=tanx

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โˆ™ 2011-04-27 21:39:26
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Q: How do you Prove sin x times sec x equals tan x?
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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


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 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 Sin x sec x equals tan x?

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


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

This would be a real bear to prove, mainly because it's not true.


What is the derivative of y equals sin times x to the power of 2?

D(y)= sin 2x


How do you solve sec x sin x?

There is nothing to solve in this equation because there is no =. If you accidentally omitted what the expression equals then resubmit it and I'll be happy to look at it


Csc squared divided by cot equals csc x sec. can someone make them equal?

cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)


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.


What is the solution to cos equals sec-sintan?

I'm not really sure what you mean by "the solution", but that equation cos = sec - sintan does simplify down to sin^2 + cos^2 = 1 which so happens to be an identity. I'm not sure if that's what you're looking for, but if it is, here are the steps in simplifying it. 1. Convert sec to 1/cos 2. Convert tan into sin/cos and multiply it by sin sintan = sin(sin/cos) = (sin^2)/cos You then have cos = 1/cos - (sin^2/cos) 3. Multiply everything by cos cos^2 = 1 - sin^2 4. And finally, send the sin^2 over to the left side by adding it (since it is being subracted on the right) You should see this sin^2 + cos^2 = 1 which is an identity.


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.


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.


How do I prove that sec A tan A sin A csc2 A?

It helps to convert everything to sines and cosines. Then you can often do lots of simplifications.Reminder: sec A = 1 / cos A tan A = sin A / cos A cosec A = 1 / sin A If you are unsure whether the two actually ARE equal, try evaluating both expressions (left and right) for some arbitrary value (for example, 10 degrees). If the two are NOT equal, then the expression are of course NOT equal. But if they ARE equal, you still need to prove that - since the claim is basically that the expressions are equal for ALL values of the variable.


How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?

sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)


Prove that sin 90 equals cos 50sin 40- cos 40 sin 50?

Sorry, but cos(50)sin(40) - cos(40)sin(50) is -0.1736, which is not even close to sin(90) which is 1.This does not work in radians, either. Please restate your question.


What is the range of y equals 3 sin x?

sin x can have any value between -1 and 1; therefore, 3 sin x has three times this range (from -3 to 3).


How do you simplify sec x cot x?

sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(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 cot squared theta plus cos squared theta plus sin squared theta scs squared theta?

Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.


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.


H ow do you verify that csc theta tan theta sec theta?

csc[]tan[] = sec[]. L: Change csc[] into one over sin[]. Change tan[] into sin[] over cos[]. R: Change sec[] into one over cos[]. 1/sin[] times sin[]/cos[] = 1/cos[]. L: To multiply 2 fractions, multiply the numerators, and multiply the denominators, and put the numerators' product over the denominators' product. R: Nothing more to do. sin[]/sin[]cos[] = 1/cos[]. L: You have a sin[] on both top and bottom. Cross them off to get a one on the top. 1/cos[] = 1/cos[]. Done. [] is theta. L is the left side of the equation. R is the right side.


How do you calculate Sin theta equals 13 times sin 32 degrees divided by 8?

sin (theta) = [13* sin (32o)]/8 = 13*0.529919264/8 = 0.861118804 [theta] = sin-1 (0.861118804) [theta] = 59.44o


Y equals 4 sin x for x equals pi?

sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0


When does cos x equal -sin x?

The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).