To prove the identity ( \tan^2 x - \sin^2(\tan^2 x) \sin^2 x = 0 ), we start with the definitions of tangent and sine. Recall that ( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} ). By substituting this into the equation and simplifying using the Pythagorean identity, we can show that both sides of the equation are equal. Thus, the identity holds true for all ( x ) where the functions are defined.
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
In a right triangle, if we know (\sin A) and (\tan A), we can find (\cos A) using the identity (\tan A = \frac{\sin A}{\cos A}). Rearranging this gives us (\cos A = \frac{\sin A}{\tan A}). Therefore, if you have specific values for (\sin A) and (\tan A), you can substitute them into this equation to find (\cos A).
The identity for tan(theta) is sin(theta)/cos(theta).
You use the identity sin2x + cos2x = 1 (to simplify the expression in parentheses), and convert all functions to sines and cosines. sec x tan x (1 - sin2x) = (1/cos x) (sin x / cos x) (cos2x) = (sin x / cos2x) cos2x = sin x
(sin(x)cot(x) - cos(x))/tan(x)(Multiply by tan(x)/tan(x))sin(x) - cos(x)tan(x)(tan(x) = sin(x)/cos(x))sinx - cos(x)(sin(x)/cos(x))(cos(x) cancels out)sin(x) - sin(x)0
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tan^2(x) Proof: cos^2(x)+sin^2(x)=1 (Modified Pythagorean theorem) sin^2(x)=1-cos^2(x) (Property of subtraction) cos^2(x)-1/cos^2(x)=? sin^2(x)/cos^2(x)=? (Property of substitution) sin(x)/cos(x) * sin(x)/cos(x) = tan(x) * tan(x) (Definition of tanget) = tan^2(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
tan θ = sin θ / cos θ sec θ = 1 / cos θ sin ² θ + cos² θ = 1 → sin² θ - 1 = - cos² θ → tan² θ - sec² θ = (sin θ / cos θ)² - (1 / cos θ)² = sin² θ / cos² θ - 1 / cos² θ = (sin² θ - 1) / cos² θ = - cos² θ / cos² θ = -1
Tan^2
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
Yes. Both expressions are the same.
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
Tan Siew Sin died on 1988-03-17.
Tan Siew Sin was born on 1916-05-21.
The value of tan A is not clear from the question.However, sin A = sqrt[tan^2 A /(tan^2 A + 1)]
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.