Prove that 2sec²∅=3-tan²∅
hmmm. I suspect a typo
plug in ∅=0 and see what happens
Nothing important
To prove the equation 2sec²θ = 3 - tan²θ, we can simplify both sides of the equation separately using trigonometric identities.
Let's start by simplifying the left side of the equation, which is 2sec²θ.
Recall that the secant (sec) function is the reciprocal of the cosine function. Therefore, secθ = 1/cosθ.
Squaring both sides of the equation, we get sec²θ = (1/cosθ)².
Using the property of reciprocals, we can rewrite this as sec²θ = 1/cos²θ.
Next, we need to simplify the right side of the equation, which is 3 - tan²θ.
Recall that the tangent (tan) function is equal to the sine divided by the cosine. Therefore, tanθ = sinθ/cosθ.
Squaring both sides of the equation, we get tan²θ = (sinθ/cosθ)².
Using the property of fractions, we can rewrite this as tan²θ = sin²θ/cos²θ.
Now, let's substitute these simplified forms back into our original equation:
2sec²θ = 3 - tan²θ
2 * (1/cos²θ) = 3 - (sin²θ/cos²θ)
Multiplying the left side by cos²θ, we get:
2 = 3cos²θ - sin²θ
Now, let's recall the Pythagorean Identity, which states that sin²θ + cos²θ = 1.
Rearranging the equation, we have sin²θ = 1 - cos²θ.
Substituting this into our equation, we get:
2 = 3cos²θ - (1 - cos²θ)
Expanding the equation, we get:
2 = 3cos²θ - 1 + cos²θ
Combining like terms, we have:
2 = 4cos²θ - 1
Adding 1 to both sides of the equation, we get:
3 = 4cos²θ
Finally, dividing both sides by 4, we have:
3/4 = cos²θ
Taking the square root of both sides, we get:
√(3/4) = cosθ
Simplifying the square root, we have:
√3/2 = cosθ
Therefore, we have proven that 2sec²θ = 3 - tan²θ is true if cosθ = √3/2.