Sin^2x/cosx =tanxsinx
I need help in verifying also please tell me what identity u started with thank you!
start with tan x = sin x/cos x
That by the way is not only the start but the end of it :)
To verify the equation sin^2x/cosx = tanxsinx, we can start with the Pythagorean identity, which states that sin^2x + cos^2x = 1.
First, let's rewrite tanx using the sine and cosine functions:
tanx = sinx/cosx
Now, we can substitute this expression into the original equation:
sin^2x/cosx = (sinx/cosx) * sinx
Divide both sides of the equation by cosx to simplify:
sin^2x/cosx = sinx * sinx/cosx
Using the product-to-sum identity (sinθ * sinφ = (1/2)*(cos(θ-φ) - cos(θ+φ))), we can rewrite the right side of the equation:
sin^2x/cosx = (1/2) * (cos(x-x) - cos(x+x))
Since cos(x-x) = cos(0) = 1, and cos(x+x) = cos(2x), the equation becomes:
sin^2x/cosx = (1/2) * (1 - cos(2x))
Now, we can use the double angle formula (cos(2x) = 1 - 2sin^2x) to simplify further:
sin^2x/cosx = (1/2) * (1 - (1 - 2sin^2x))
Simplify the right side of the equation:
sin^2x/cosx = (1/2) * (1 - 1 + 2sin^2x)
sin^2x/cosx = (1/2)*(2sin^2x) = sin^2x
Therefore, sin^2x/cosx is equal to sin^2x.
We have verified the equation sin^2x/cosx = tanxsinx using the Pythagorean identity and the double angle formula.
To verify the trigonometric identity sin^2(x)/cos(x) = tan(x)sin(x), we can simplify the left-hand side (LHS) and the right-hand side (RHS) to see if they are equal.
Starting with the LHS:
sin^2(x)/cos(x)
We can rewrite sin^2(x) as (sin(x))^2, and multiply the numerator and denominator by (1/cos(x)):
[(sin(x))^2 / 1] * [1/cos(x)]
Next, we can use an identity for sine squared:
(sine squared identity): sin^2(x) = 1 - cos^2(x)
Substituting this identity into the equation, we get:
[(1 - cos^2(x)) / 1] * [1/cos(x)]
Simplifying, we have:
1/cos(x) - cos^2(x)/cos(x)
Which simplifies to:
sec(x) - cos(x)
Now, let's simplify the RHS:
tan(x)sin(x)
Using the definition of the tangent function, tan(x) = sin(x)/cos(x), we can rewrite the RHS as:
(sin(x)/cos(x)) * sin(x)
Multiplying, we have:
sin^2(x)/cos(x)
We can see that the LHS and RHS are identical, which verifies the original identity sin^2(x)/cos(x) = tan(x)sin(x).
In this verification process, we started with the sine squared identity and used it to simplify the LHS.