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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

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

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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.

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.

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

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.

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

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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.

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.

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

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.

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

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.

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

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.

sec(x)tan(x)

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

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 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