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Start on the left-hand side.
cos(x) + tan(x)sin(x)
Put tan(x) in terms of sin(x) and cos(x).
cos(x) + [sin(x)/cos(x)]sin(x)
Multiply.
cos(x) + sin2(x)/cos(x)
Make the denominators equal.
cos2(x)/cos(x) + sin2(x)/cos(x)
Add.
[cos2(x) + sin2(x)]/cos(x)
Use the Pythagorean Theorem to simplify.
1/cos(x)
Since 1/cos(x) is the same as sec(x)- the right-hand side- the proof is complete.
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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
1 over cos x equals what?
sec(x)=1/cos(x), by definition of secant.
Should we find cos theta if sec theta equals -10?
No.
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
Is cos 2 x sec x equals 2 cos x - sec x an identity?
Yes, it is. the basic identity is for a double angle relation: cos 2x = 2 cosx cos x -1 since sec x =1/cos x if we multiply both sides by sec x we get cos2xsec x = 2cosxcos x/cos x -1/cos x = 2cos x - sec x
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.
What is sec x cos x?
sec x = 1/cos x sec x cos x = [1/cos x] [cos x] = 1
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 you find sec x cos x?
sec x = 1/cos x so sec x * cos x = 1
What is the reciprocal function of sec A?
The answer is cos A . cos A = 1/ (sec A)
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 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
